ALGEBRAIC EXPRESSIONS & FACTORI ZATION
3 CHAPTER
Monomial : An algebraic expression containing only one term is called a monomial.
CONTENTS Constant & Variable Algebraic Expressions Monomial & Binomial Identities Factorization Polynomials
Ex.5
– 5,3y,7xy,
2 2 5 2 3 x yz, a bc etc. are all monomials. 3 3
Binomial : An algebraic expression containing two terms is called a binomial. Ex.6
The expression 2x – 3, 3x + 2y, xyz –5 etc. are all binomials.
Trinomial : An algebraic expression containing three terms is called a trinomial.
Constant : A symbol having a fixed numerical value is called a constant. Variable : A symbol which takes various numerical values is called a variabale. Ex.1
Ex.2
We know that the perimeter P of a square of side s is given by P = 4 × s. Here, 4 is a constant and P and s are variables. The perimeter P of a rectangle of sides l and b is given by P = 2 (l + b). Here, 2 is a constant and l and b are variables.
Factors : Each terms in an algebraic expression is a product of one or more numbers (s) and /or literal (s). These number(s) and liteal(s) are known as the factors of that terms. A constant factor is called a numerical factor, while a variable factor is known as a literal factor.
Coefficient : In a term of an algebraic expression any of the factors with the sign of the term is called the coefficient of the other factors.
Algebraic Expressions : A combination of constants and variables connected by the signs of fundamental operation of addition, subtraction, multiplication and division is called an algebraic expression.
Terms : Various parts of an algebraic expression which are separated by the signs of + or – are called the ‘terms’ of the expression. Ex.3
Ex.4
2x2 – 3xy + 5y2 is an algebraic expression consisting of three terms, namely, 2x2,–3xy and 5y2. The expression 2x3 – 3x2 + 4x – 7 is an algebraic expression consisting of four terms, namely, 2x3, –3x2 , 4x and – 7.
The expressions a – b + 2x2 + y2 – xy, x3 – 2y3 – 3x2y2z etc. are trinomial.
Ex.7
Ex.8
In – 5xy, the coefficient of x is – 5y; the coefficient of y is –5x and the coefficient of xy is –5. In –x, the coefficient of x is –1.
Ex.9
Constant Term : A term of the expression having no literal factor is called a constant term. Ex.11
In the algebraic expression x2 – xy + yz – 4, the constant term is –4.
Like and Unlike Terms : The terms having the same literal factors are called like or similar terms, otherwise they are called unlike terms. Ex.12
In the algebraic exspression 2a2b + 3ab2 – 7ab –4ba2, we have 2 a2b and –4ba2 as like terms, whereas 3ab2 and –7ab are unlike terms.
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EXAMPLES Ex.13 Sol.
Ex.14
Sol.
Ex.15
Add : 7x 2 – 4x + 5, – 3x 2 + 2x – 1 and 5x2 –x + 9. We have, Required sum = (7x2 – 4x +5) + (–3x2 + 2x–1) + (5x2 – x+9) 2 2 2 = 7x – 3x + 5x – 4x + 2x – x + 5– 1+9 [Collecting like terms] = (7 – 3 + 5)x2 + (–4 + 2 – 1)x +(5 –1+ 9) [Adding like terms] 2 = 9x – 3x + 13
1 5 1 2 1 1 x + , – x + x – and 3 2 2 2 3 1 1 – 2x2 + x – . 5 6 Required sum Add : 5x2 –
5 1 2 1 1 2 1 = 5x – x + – 2 x 2 x – 3 3 2
Sol.
Ex.16
Sol. 1
1
2 + 2x 5 x – 6
= 5x2 –
1 2 1 1 1 5 x – 2x2 – x + x + x + 2 3 2 5 2 1 1 – – [Collecting like terms] 3 6
1 1 1 1 5 1 1 = 5 – – 2 x2 + – x + – – 2 3 2 5 2 3 6 [Adding like term]
10 – 1 – 4 2 10 15 6 15 – 2 – 1 x + x + = 30 2 6 =
Ex.17
5 2 11 x + x+2 2 30
Multiplication of Algebraic Expressions (i) The product of two factors with like signs is positive and the product of two factors with unlike signs is negative i.e., (a) (+) × (+) = + (b) (+) × (–) = – (c) (–) × (+) = – and, (d) (–) × (–) = + (ii) If a is any variable and m, n are positive integers, then am × an = am+n For example , a3 × a5 = a3+5 = a8, y4 × y = y4+1 = y5 etc.
Find the product of the following pairs of polynomials : (i) 4, 7x (ii) – 4a, 7a (iii) – 4x,7xy (iv) 4x3, – 3xy (v) 4x, 0 We have, (i) 4 × 7x = (4 × 7) × x = 28 × x = 28 x (ii) (–4a) × (7a) = (–4 × 7) × (a×a) = –28a2 (iii) (–4x) × (7xy) = (–4 × 7) × (x × xy) = –28x1+1y = –28x2y (iv) (4x3) × (–3xy) = (4×–3) × (x3 × xy) = – 12 (x3+1y) = –12x4y (v) 4x × 0 = (4 × 0) × x = 0 × x = 0 Find the areas of rectangles with the following pairs of monomials as their length and breadth respectively : (i) (x, y) (ii) (10x, 5y) (iii) (2x2, 5y2) (iv) (4a, 3a2) (v) (3mn, 4np) We know that the area of a rectangle is the product of its length and breadth. Length Breadth Length × Breadth = Area (i) x y x×y = xy (ii) 10 x 5y 10x × 5y = 50xy (iii) 2x2 5y2 2x2 × 5y2 = (2 × 5) × (x2 × y2) = 10x2y2 2 (iv) 4a 3a 4a × 3a2 = (4 × 3) × (a × a2) = 12 a3 (v) 3mn 4np 3mn × 4np = (3 × 4) × (m × n × n × p) = 12 mn2p Multiply : (i) 3ab2c3 by 5a3b2c 3 (ii) 4x2yz by – x2yz2 2 8 2 3 3 (iii) – x yz by – xy2z 5 4
3 2 7 4 x y by x y 14 2 2 (v) 2.1a bc by 4ab2 (i) We have, (3ab2c3) × (5a3b2c) = (3 × 5) × (a × a3 × b2 × b2 × c3 × c) = 15a1+3b2+2c3+1 = 15a4b4c4 (ii) We have, 3 2 2 (4x2yz) × – x yz 2 (iv)
Sol.
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(iii) We have, (3) × (15x) × (–23x2y) = (3 × 15 – 23) × (x × x2 × y) = –1035x1+2y = –1035x3y. (iv) We have,
3 = 4 × (x2 × x2 × y × y × z × z2) 2
= –6x2+2y1+1z1+2 = – 6x4y2z3 (iii) We have,
8 2 3 3 2 – x yz × – xy z 5 4
–3 xyz 2 (–x) × (x2yz) × 7
3 8 = – – × (x2 × x × y × y2 × z3 × z) 4 5
–3 × (x × x2 × x × y × y × z × z2) = – 1 7
6 21 1 2 31 6 3 3 4 x y z = x yz 5 5 (iv) We have, =
= Ex.19
7 4 3 2 x y × x y 2 14
3 7 = × (x2 × x4 × y × y) 14 2
3 2+4 1+1 3 x y = x6y2 4 4 (v) We have, (2.1a2bc) × (4ab2) = (2.1 × 4) × (a2 × a × b × b2 × c) = 8.4a2+1b1+2c = 8.4a3b3c =
Ex.18
Multiply :
6 5 12 + (a×a×b×b×b×c×c) 5 6 9 =
12 1+1 1+1+1 1+1 4 2 3 2 a b c = a b c 9 3
(iii) We have,
1 (– 6a2bc) × (2a2b) × – 4
1 = – 6 2 – × (a2 × a2 × b × b × c) 4 = 3a2+2b1+1c = 3a4b2c (ii) We have,
3 2 2 x yz × (0.5xy2z2) × (1.16x2yz3) × (2xyz) 4 3 0.5 1.16 2 × (x2 × x × x2 × x × y × y2 4
=
× y × y × z2 × z2 × z3 × z)
4 5 2 a b × (10a3b) × (6) 9
3 5 116 2 × (x2+1+2+1 × y1+2+1+1 4 10 100 × z2+2+3+1)
=
4 = 10 6 × (a5 × a3× b2 × b) 9
80 5+3 2+1 80 8 3 a b = a b 3 3
4 (v) (–3x2 y), (4xy2z), (–xy2z2) and z 5 (i) We have, (3xyz) × (5x) × 0 = (3×5×0)× (x×x×y×z) = 0 × x2yz = 0 12 6 5 ab × bc × abc 9 5 6
3 (iv) – x, x2yz and – xyz2 7 (i) We have,
=
Multiply each of the following mnomials : 6 5 12 (i) 3xyz, 5x, 0 (ii) ab, bc, abc 5 6 9 3 (iii) x2yz2, 0.5xy2z2, 1.16x2yz3, 2xyz 4 (vi) 20x10y20z30, (10xyz)2
(ii) We have,
1 (i) –6a2bc, 2a2b and – 4 4 5 2 (ii) a b , 10a3b and 6 9 (iii) 3.15x and – 23x2y
Sol.
Sol.
3 1+2+1 1+1 1+2 3 x y z = x4y2z3 7 7
=
87 6 5 8 x yz 100
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= 105 a6+1+2b2+3 = 105a9b5 1 Putting a = 1 and b = , we have 2
(iv)We have, (20x10y20z30) × (10xyz)2 = (20x10y20z30) × (10xyz) × (10xyz) = (20 × 10 × 10) × (x10 × x × x × y20 × y × y × z30 × z × z) 10+1+1 20+1+1 30+1+1 = 2000x y z = 2000x12y22z32
105a9b5
4 z 5
= 105 × 1×
(v) We have, (–3x2y) × (4xy2z) × (–xy2z2) ×
= – 3 4 –1
4 × (x2 × x × x × y × y2 × y2 5 × z×z2 ×z)
=
48 2+1+1 1+2+2 1+2+1 48 4 5 4 x y z = x y z 5 5
Ex.20
Express the following product as a monomial:
Sol.
1 2 (x3) × (7x5) × x × (–6x4) 5 Verify the product for x = 1 We have, 1 2 (x3) × (7x5) × x × (–6x4) 5 1 = 1 7 –6 × (x3 × x5 × x2 × x4 ) 5
Ex.23 Sol.
Multiply : (7xy + 5y) by 3xy We have, (7xy + 5y) × 3xy = 7xy × 3xy + 5y × 3xy = 21x1+1y1+1+15xy1+1 = 21x2y2 + 15xy2
Ex.24
Multiply : –
Sol.
We have,
= –
3 3 3 2 1+1 2 3 2+1 2 2 2 × a b + ab = – a b + ab 5 5 3 5 5
1 4 Multiply : 3x – y 2 x by xy.. 2 5 Horizontal method We have,
1 4 2 3x – y x × xy 2 5 1 4 1 = 3x × xy – y2x × xy 2 5 2
42 42 R.H.S. = – ×(1)14 = – 5 5 L.H.S. = R.H.S.
Sol.
Ex.25 Sol.
and,
Ex.21
2 3ab 2 × 2a – – 3ab × b 5 5 3
2
1 42 × – 6 = – 5 5
2a 3ab2 – b by 3 5
3ab 2 2a – × – b 5 3
1 2 L.H.S. = (x3) × (7x5) × x × (–6x4 ) 5
= 1 × 7 ×
1 105 = 32 32
Multiply : 2x by (3x + 5y) We have, 2x × (3x + 5y) = 2x × 3x + 2x × 5y = 6x2 + 10xy
= –
Find the value of (5a6) × (–10ab2) × (–2.1a2b3) 1 for a = 1 and b = . 2 We have, (5a6) × (–10ab2) × (–2.1a2b3) = (5 × –10 × – 2.1) × (a6 × a × a2 × b2 × b3)
21 = 5 –10 – × (a6 × a × a2 × b2 × b3) 10
5
Ex.22 Sol.
42 3+5+2+4 42 14 x =– x 5 5 Verification : For x = 1, we have
1 5
1 × 2
Multiplication of a Monomial & a Binomial
=–
= (1)3 × {7×(15)} × (1) × {–6 × (1)4}
= 105 ×
(1)9
1 = 3 ×x×x×y – 2
Column method We have, 3x –
×
1 xy 2
3 2 2 x y – x 2 y3 2 5
4 1 2 × y ×y×x×x 5 2 3 2 2 3 2 2 2 3 = x y – y3x2 = x y– x y 2 5 2 5
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4 2 y x 5
Ex.26
Sol.
Ex.27
Sol.
Determine each of the following products and find the value of each for x = 2, y = 1.15, z = 0.01. (i) 27x2 (1 – 3x) (ii) xz(x2 + y2) 2 (iii) z (x – y) (iv) (2z – 3x) × (– 4y) (i) We have, 27x2 (1 – 3x) = 27x2 × (1 – 3x) = 27x2 × 1 – 27x2 × 3x [Expanding the bracket] = 27x2 – 81x3 Putting x = 2, we have 27x2 (1 – 3x) = 27× (2)2 × (1 – 3×2) = 27 × 4 × (1 – 6) = 27 × 4 × – 5 = – 540 (ii) We have, xz(x2 + y2) = xz × (x2 + y2) = xz × x2 + xz × y2 = x3z + xy2z Putting x = 2, y = 1.15 and z = 0.01, we get xz (x2 + y2) = 2 × 0.01 × {(2)2 + (1.15)2} = 0.02 × (4 + 1.3225) = 0.02 × 5.3225 = 0.106450 (iii) We have, z2(x – y) = z2 × (x – y) = z2 × x – z2 × y = z2 x – z2 y Putting x = 2 y = 1.15 and z = 0.01, we get z2(x – y) = (0.01)2 × (2 – 1.15) = (0.0001) × (0.85) = 0.000085 (vi) We have, (2z – 3x) × (– 4y) = (2z) × (– 4y) – 3x × (– 4y) = – 8zy + 12xy Putting x = 2, y= 1.15 and z = 0.01, we have (2z – 3x) × – 4y = [(2 × 0.01) – (3 × 2)] × (–4 × 1.15) = (0.02 – 6) × (–4.6) = – 5.98 × –4.6 = 27.508 Simplify the expression and evaluate them as directed : (i) x(x – 3) + 2 for x = 1 (ii) 3y (2y – 7) – 3(y–4) – 63 for y = – 2 (i) We have, x(x – 3) + 2 = x2 – 3x + 2 For x = 1, we have x2 – 3x + 2 = (1)2 – 3 × 1 + 2 = 1 – 3 + 2 = 3 –3 = 0 (ii) We have, 3y(2y – 6) – 3 (y – 4) – 63 = (6y2 – 21y) – (3y – 4) – 63 = 6y2 – 21y – 3y + 12 – 63
= 6y2 – 24y – 51 For y = –2, we have 6y2 – 24y – 51 = 6 × (–2)2 – 24(–2) – 51 = 6 × 4 + 24 × 2 – 51 = 24 + 48 – 51 = 72 – 51= 21 Ex.28 Sol.
Subtract 3pq (p – q) from 2pq(p + q) (i) We have, 3pq (p – q) = 3p2q – 3pq2 and, 2pq (p + q) = 2p2q + 2pq2 Subtraction : 2p2q + 2pq2 3p2q – 3pq2 –
+
+
– p2q + 5pq2
Ex.29 Sol.
Add :
(i) p(p – q), q (q – r) and r(r – p) (ii) 2x (z – x – y) and 2y(z – y – x) (i) We have, p (p – q) + q (q – r) + r (r – p) = p2 – pq + q2 – qr + r2 – rp = p2 + q2 + r2 – pq – qr – rp (ii) We have, 2x(z – x – y) + 2y (z – y – x) = 2xz – 2x2 – 2xy + 2yz – 2y2 – 2xy = 2xz – 2x2 – 4xy + 2yz – 2y2
Ex.30
Sol.
Simplify each of the following expressions : (i) 15a2 – 6a (a – 2) + a(3 + 7a) (ii) x2(1 – 3y2) + x(xy2 – 2x) – 3y(y – 4x2y) (iii) 4st(s – t) – 6s2(t – t2) – 3t2(2s2 – s) + 2st(s– t) (i) We have, 15a2 – 6a(a – 2) + a(3 + 7a) = 15a2 – 6a2 + 12a + 3a + 7a2 = 15a2 – 6a2 + 7a2 + 12a + 3a = 16a2 + 15a (ii) We have, x2(1 – 3y2) + x(xy2 – 2x) – 3y(y – 4x2y) = x2 × 1 – 3y2 × x2 + x × xy2 – x × 2x – 3y × y + 3y ×4 x2y = x2 – 3x2y2 + x2y2 – 2x2 – 3y2 + 12x2y2 = (x2 – 2x2) + (–3x2y2 + x2y2 + 12x2y2) – 3y2 = –x2 + 10x2y2 – 3y2 (iii) 4st(s – t) – 6s2(t – t2) – 3t2(2s2 – s) + 2st(s – t) = 4st × s – 4st × t – 6s2 × t + 6s2 × t2 – 3t2 × 2s2 + 3t2 × s + 2st × s – 2st × t = 4s2t – 4st2 – 6s2t + 6s2t2 – 6s2t2 + 3st2 + 2s2t – 2st2 = (4s2t – 6s2t + 2s2t) + (–4st2 + 3st2 – 2st2) + (6s2t2 – 6s2t2) = – 3st2
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We have,
Multiplication of Two Binomials Ex.31 Sol.
Ex.32 Sol.
Multiply (3x + 2y) and (5x + 3y). We have, (3x + 2y) × (5x + 3y) = 3x × (5x + 3y) + 2y × (5x + 3y) = (3x × 5x + 3x × 3y) + (2y × 5x + 2y + 3y) = (15x2 + 9xy) + (10xy + 6y2) = 15x2 + 9xy + 10xy + 6y2 = 15x2 + 19xy + 6y2 Multiply (2x + 3y) and (4x – 5y) We have, (2x + 3y) × (4x – 5y) = 2x × (4x – 5y) + 3y × (4x – 5y) = (2x × 4x – 2x × 5y) + (3y × 4x – 3y × 5y) = (8x2 – 10xy) + (12xy – 15y2)
0.5x – y 0.5x y 0.25x 2 – 0.5xy 0.5xy – y 2
0.25x – y
Sol.
2a 3b
21ab 9b 2 14a 2 27ab 9b 2 Ex.34 Sol.
Multiplying 7a 3b by 2a Multiplying7a 3b by 3b Adding the like term
Multiply (7x–3y) by (4x–5y) by column method. We have,
7x – 3y 4x – 5y 28x 2 – 12xy – 35xy 15y 2 28x 2 – 47xy 15y 2 Ex.35 Sol.
Multiplying7x – 3y by 4x Multiplying7x – 3y by 5y Adding the like terms
Multiply (0.5x – y) by (0.5x + y) Horizontal Method: We have, (0.5x – y) × (0.5x + y) = 0.5x (0.5x + y) – y (0.5x + y) = 0.5x × 0.5x + 0.5x × y – y × 0.5x – y × y = 0.25x2 + 0.5xy – 0.5xy – y2 = 0.25x2 – y2 Column method:
Multiplying 0.5x – y by y Adding the like terms
3y 4y Multiplying 4 x and 3x – 5 5
Sol.
Horizontal Method :
3y 4y 4 x × 3x – 5 5
4y 4y 3y + = 4x × 3x – × 3x – 5 5 5 = 4x × 3x – 4x ×
Multiply (7a + 3b) and (2a + 3b) by column method. We have, 7a 3b
14a 2 6ab
2
Ex.36
= 8x2 – 10xy + 12xy – 15y2 = 8x2 + 2xy – 15y2
Ex.33
Multiplying 0.5x – y by 0.5x 2
= 12x2 –
4 y 3y 3y 4 y + × 3x – × 5 5 5 5
16 9 12 2 xy + xy – y 5 5 25
7 12 2 xy – y 5 25 Column method: We have, 3y 4x 5 4y 3x – 5 9 3y 12x 2 xy Multiplying 4 x by 3x. 5 5 16 12 2 3y 4y – xy – y Multiplying 4x by – . 5 25 5 5 7 12 2 12x 2 – xy – y Adding the like terms 5 25 = 12x2 –
Ex.37
Find the value of the following products: (i) (x + 2y) (x – 2y) at x = 1, y = 0 (ii) (3m – 2n) (2m – 3n) at m = 1, n = –1 (iii) (4a2 + 3b) (4a2 + 3b) at a = 1, b =2 Sol. (i) We have, (x + 2y) (x – 2y) = x(x – 2y) + 2y (x – 2y) = x × x – x × 2y + 2y × x – 2y × 2y = x2 – 2xy + 2yx – 4y2 = x2 – 4y2 When x = 1, y = 0, we get (x + 2y) (x – 2y) = x2 – 4y2 = (1)2 – 4 × (0)2 = 1 – 0 = 1.
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(ii) We have, (3m – 2n) (2m – 3n) = 3m (2m – 3n) – 2n (2m – 3n) = 3m × 2m – 3m × 3n – 2n × 2m + 2n × 3n = 6m2 – 9mn – 4mn + 6n2 = 6m2 – 13mn + 6n2 When m = 1, n = –1, we get (3m – 2n) (2m – 3n) = 6m2 – 13mn + 6n2 = 6 × (1)2 – 13×1× (–1) + 6× (–1)2 = 6 + 13 + 6 = 25 (iii) We have (4a2 + 3b) (4a2 + 3b) = 4a2 × (4a2 + 3b) + 3b × (4a2 + 3b) = 4a2 × 4a2 + 4a2 × 3b + 3b × 4a2 + 3b × 3b = 16a4 + 12a2b + 12a2b + 9b2 = 16a4 + 24a2b + 9b2 When, a = 1, b = 2, we get (4a2 + 3b) (4a2 + 3b) = 16a4 + 24a2b + 9b2 = 16 × (1)4 + 24 × (1)2 × 2 + 9 × (2)2 = 16 + 48 + 36 = 100 Ex.38
Simplify the following : (i) (2x + 5) (3x – 2) + (x + 2) (2x – 3) (ii) (3x + 2) (2x + 3) – (4x – 3) (2x – 1) (iii) (2x + 3y) (3x + 4y) – (7x + 3y) (x + 2y) Sol. (i) We have, (2x + 5) (3x – 2) + (x + 2) (2x – 3) = 2x(3x – 2) + 5(3x – 2) + x (2x – 3) + 2 (2x – 3) = 6x2 – 4x + 15x – 10 + 2x2 – 3x + 4x – 6 = (6x2 + 2x2) + (–4x + 15x – 3x + 4x) + (–10 – 6) = 8x2 + 12x – 16 (ii) We have, (3x + 2) (2x + 3) – (4x – 3) (2x – 1) = {3x(2x+3) + 2 (2x + 3)} – {4x(2x –1) – 3(2x –1)} = (6x2 + 9x + 4x + 6) – (8x2 – 4x – 6x + 3) = (6x2 + 13x + 6) – (8x2 – 10x + 3) 6x2 + 13x + 6 – 8x2 + 10x – 3 = – 2x2 + 23x + 3 (iii)We have, (2x + 3y) (3x + 4y) – (7x + 3y) (x + 2y) = {2x(3x + 4y) + 3y (3x + 4y) – 7x(x + 2y) + 3y (x + 2y)} 2 2 2 = (6x + 8xy + 9xy + 12y ) – (7x + 14xy + 3xy + 6y2) = (6x2 + 17xy + 12y2) – (7x2 + 17xy + 6y2) = 6x2 +17xy + 12y2 – 7x2 – 17xy – 6y2 = 6x2 – 7x2 + 17xy – 17xy + 12y2 – 6y2 = –x2 + 6y2.
Ex.39 Sol.
Multiply : (2x2 – 3x + 5) by (5x + 2). Horizontal method: We have, (2x3 – 3x + 5) × (5x + 2) = (2x2 – 3x + 5) × 5x + (2x2 – 3x + 5) × 2 = (10x3 – 15x2 + 25x) + (4x2 – 6x + 10) = 10x3 – 11x2 + 19x + 10 Column Method: We have,
2 x 2 – 3x 5
5x 2
10 x 3 – 15x 2 25x
Multiplying 2x 2 – 3x 5 by 5x
4x 2 – 6x 10
Multiplying 2x 2 – 3x 5 by 2
10 x 3 – 11x 2 19x 10
Adding the like terms
Ex.40
Simplify : (i) (3x – 2) (x – 1) (3x + 5) (ii) (5 – x) (3 – 2x) (4 – 3x) Sol. (i) We have, (3x – 2) (x – 1) (3x + 5) = {(3x – 2) (x – 1)} × (3x + 5) [By Associativity of Multiplication] = {3x(x –1) – 2 (x – 1)} × (3x + 5) = (3x2 – 3x – 2x + 2) × (3x + 5) = (3x2 – 5x + 2) × (3x + 5) = 3x2 ×(3x + 5) – 5x(3x + 5) + 2 × (3x + 5) = (9x3 + 15x2) + (– 15x2 – 25x) + (6x + 10) = 9x3 + 15x2 – 15x2 – 25x + 6x + 10 = 9x3 – 19x + 10 (ii) We have, (5 – x) (3 – 2x) (4 – 3x) = {(5 – x) (3 – 2x)} × (4 – 3x) = {5(3 – 2x) – x (3 – 2x)} × (4 – 3x) = (15 – 10x – 3x + 2x2) × (4 – 3x) = (2x2 – 13x + 15) + (4 – 3x) = 2x2 × (4 – 3x) – 13x × (4 – 3x) + 15 × (4 – 3x) = 8x2 – 6x3 – 52x + 39x2 + 60 – 45x = – 6x3 + 47x2 – 97x + 60
IDENTITIES Identity An indentity is an equality which is true for all values of the variables (s).
Standard Identities : Identity 1. Identity 2. Identity 3.
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(a + b2) = a2 + 2ab + b2 (a – b)2 = a2 – 2ab + b2 (a + b) (a – b) = a2 – b2
Ex.41
Evaluate : (i) (2x + 3y)2 (ii) (2x – 3y)2 (iii) (2x + 3y) (2x – 3y) Sol. (i) We have, (2x2 + 3y)2 = (2x)2 + 2×(2x) × (3y) + (3y)2 [Using: (a + b)2 = a2 + 2ab + b2] = 4x2 + 12xy + 9y2 (ii) We have, (2x – 3y)2 = (2x)2 – 2 × (2x) × (3y) + (3y)2 [Using: (a – b)2 = a2 – 2ab + b2] = 4x2 – 12xy + 9y2 (iii)We have, (2x + 3y) (2x – 3y) = (2x)2 – (3y)2 [Using : (a + b) (a – b) = a2 – b2] = 4x2 – 9y2. Ex.42
Sol.
4 2 4 2 x 3 x 3 3 3 4 2 = x 3 3
2
[ a.a =a2]
2
4 4 2 = x + 2 × x2 × 3 + (3)2 3 3 [Using : (a + b)2 = a2 + 2ab + b2]
16 4 x + 8x2 + 9 9 (ii) We have, =
2 2 2 2 2 2 x 5y x 5y 3 3
Write down the squares of each of the following binomials :
a 1 (i) x (ii) 5b – 2 2
(i) We have,
2 2 x 5y 2 3
2
=
y 2 y (iii) 2
[ a.a = a2]
2
2 2 2 = x + 2 × x2× 5y2 + (5y2)2 3 3
Sol. (i) We have,
[Using : (a + b)2 = a2 + 2ab + b2]
2
a a a x = x2 + 2 × x × + 2 2 2
2
=
4 4 20 2 2 x + x y + 25y4 9 3
[Using : (a + b)2 = a2 + 2ab + b2]
a2 = x2 + xa + 4 (ii) We have,
Ex.44
2
2
1 1 1 5b – = (5b)2 – 2 × 5b × + 2 2 2 2 [Using : (a –b) = a2 – 2ab + b2] 1 = 25b2 – 5b + 4 (iii)We have, 2 y y 2
2
= Ex.43
= y2
y2 +
2 y 2 y +2×y× + 2 2
y3
If x +
1 = 4, find the values of x
1 (i) x2 + 2 x Sol. (i) We have,
1 =4 x On squaring both sides, we get 2
1 x = 42 x 2
x2 + 2 × x ×
Find the product of the following binomials :
x2 + 2 +
4 2 4 2 (i) x 3 x 3 3 3
x2 +
2 2 2 2 2 2 (ii) x 5y x 5y 3 3
1 x4
x+
2
y4 + 4
(ii) x4 +
1 1 + = 16 x x
1 = 16 x2
1 = 16 – 2 [On transposing 2 on RHS] x2 1 x2 + 2 = 14 x
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(ii) We have,
2
1 1 x = x2 + 2 + 2 x x
1 x2 + 2 = 14 x On squaring both sides, we get
2
1 x = 27 + 2 x
2
2 1 x 2 = 142 x
2
1 x = 29 x
2
1 1 (x2)2 + 2 + 2 × x2 × 2 = 196 x x
x4
1 = 196 – 2 [On transposing 2 on RHS] x4
x4 +
Ex.45 Sol.
1 = ± 29 x [Taking square root of both sides] (ii) We have, x+
1 + 4 + 2 = 196 x
x4 +
2
1 1 1 x – = x2 – 2 × x × + x x x
1 = 194 x4
1 x – x
1 1 = 9, find the value of x2 + 2 . x x We have, 1 x– =9 x On squaring both sides, we get If x –
2 2 1 1 x – = 27 – 2 x 2 27(given ) x x 2
1 x – = 25 x 2
2
1 x – = 52 x
1 1 –2×x× + = 81 x x
1 x2 – 2 + 2 = 81 x
x2
x–
1 + 2 = 81 + 2 [On transposing – 2 on RHS] x
Ex.47 Sol.
1 x2 + 2 = 83 x 1 If x2 + 2 = 27, find the values of each of the x following : 1 1 (i) x + (ii) x – x x Sol. (i) We have, Ex.46
2
1 1 1 x = x2 + 2 × x + 2 x x x
1 x x
2
1 = x2 + 2 + 2 x
1 = x2 – 2 + 2 x
2
2
2
2
1 1 x – = x2 + 2 – 2 x x
1 x – = 81 x x2
2 1 x x 2 27(given )
Ex.48 Sol.
1 =±5 x
If 3x + 2y = 12 and xy = 6, find the value of 9x2 + 4y2. We have, (3x + 2y)2 = (3x)2 + (2y)2 + 2 × 3x × 2y (3x + 2y)2 = 9x2 + 4y2 + 12xy 122 = 9x2 + 4y2 + 12 × 6 [Putting 3x + 2y = 12 and xy = 6] 144 = 9x2 + 4y2 + 72 144 – 72 = 9x2 + 4y2 9x2 + 4y2 = 72 If 4x2 + y2 = 40 and xy = 6, find the value of 2x + y. We have, (2x + y)2 = (2x)2 + y2 + 2 × 2x × y (2x + y)2 = (4x2 + y2) + 4xy (2x + y)2 = 40 + 4 × 6 [Using 4x2 + y2 = 40 and xy = 6]
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(2xy + y)2 = 64
2 1 2 1 4 1 = x – 2 x 2 x 4 x x x
2x + y = ± 64 2x + y = ± 8 [Taking square root of both sides] Ex.49
2 2 1 2 1 4 = ( x ) – x 2 x 4 x
Find the continued product : (i) (x + 2) (x – 2) (x2 + 4) (ii) (2x + 3y) (2x – 3y) (4x2 + 9y2) (iii) (x – 1) (x + 1) (x2 + 1) (x4 + 1)
4 1 4 1 = x – 4 x 4 x x
1 1 2 1 4 1 (iv) x – x x 2 x 4 x x x x
=
y y (v) x – – 1 x 1 5 5
y y = x – 1 x 1 5 5
=
= ( x )
= x2 –
Ex.50
Sol.
– 12 (x4 + 1)
= (x4 – 1) (x4 + 1)
= ( x 4 ) 2 – 12
= x8 – 1 (iv) We have
Ex.51
1 1 2 1 4 1 x – x x 2 x 4 x x x x 1 1 2 1 4 1 = x – x x 2 x 4 x x x x
y + 1 5
b2]
= ( x 2 – 1)( x 2 – 1) + (x4 + 1) 2 2
x2
2
y 2 2y = x2 – 25 5 1
b2]
2
y y x – – 1 x 1 5 5
= (2 x ) 2 – (3y) 2 (4x2 + 9y2) [Using : (a + b) (a – b) = a2 – = (4x2 – 9y2) (4x2 + 9y2) = (4x2)2 – (9y2)2 [Using : (a + b) (a – b) = a2 – = 16x4 – 81y4. (iii) We have, (x – 1) (x + 1) (x2 + 1) (x4 + 1) = {(x – 1) (x + 1)} (x2 + 1) (x4 + 1) = (x2 – 1) (x2 + 1) (x4 + 1)
1 – 4 x
1 = x8 – 8 x (v) We have,
Sol. (i) We have, (x + 2) (x – 2) (x2 + 4) = {(x + 2)(x– 2)} (x2 + 4) [By associativity of multiplication] = (x2 – 22) (x2 + 4) [ (a + b) (a – b) = a2 – b2] = (x2 – 4) (x2 + 4) = (x2)2 – 42 [ (a + b) (a – b) = a2 – b2] 4 = x – 16 (ii) We have, (2x + 3y) (2x – 3y) (4x2 + 9y2) = {(2x + 3y) (2x – 3y)} (4x2 + 9y2) = {(2x + 3y) (2x – 3y)} (4x2 + 9y2)
(x4)2
Sol.
2y y2 – – 1 5 25
Prove that: 2a2 + 2b2 + 2c2 – 2ab – 2bc – 2ca = (a – b)2 + (b – c)2 + (c – a)2 We have, LHS = 2a2 + 2b2 + 2c2 – 2ab – 2bc – 2ca = (a2 – 2ab + b2) + (b2 – 2bc + c2) + (c2 – 2ca + a2) [Re-arranging the terms] = (a – b)2 + (b – c)2 + (c – a)2 = R.H.S. Hence , 2a2 + 2b2 + 2c2 – 2ab – 2bc – 2ca = (a – b)2 + (b – c)2 + (c – a)2 If a2 + b2 + c2 – ab – bc – ca = 0, prove that a = b = c. We have, a2 + b2 + c2 – ab – bc – ca = 0 2a2 + 2b2 + 2c2 – 2ab – 2bc – 2ca = 2 × 0 [Multiplying both sides by 2]
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(a2 – 2ab + b2) + (b2 – 2bc + c2) + (c2 – 2ac + a2) = 0 (a – b)2 + (b – c)2 + (c – a)2 = 0 a – b = 0, b – c = 0, c – a = 0
(iii)We have, 25x = 5362 – 1362 25x = (536 + 136) × (536 – 136) [Using : (a2 – b2) = (a + b) (a – b)] 25x = 672 × 400
Sum of positive quantities is zero if and only if each quantity is zero
25x 672 400 = 25 25 x = 672 × 16 x = 10752
a = b , b = c and c = a a = b = c. Ex.52
Using the fomulae for squaring a binomial, evaluate the following : (i) (101)2 (ii) (99)2 (iii) (93)2 We have, (i) (101)2 = (100 + 1)2 = (100)2 + 2 × 100 × 1+ (1)2 [Using : (a + b2) = a2 + 2ab + b2] = 10000 + 200 + 1 = 10201 (ii) (99)2 = (100 – 1)2 = (100)2 – 2 × 100 × 1 + (1)2 [Using : (a – b)2 = a2 – 2ab + b2] = 10000 – 200 + 1 = 9801 (iii) (93)2 = (90 + 3)2 = (90)2 + 2 × 90 × 3t (3)2 = 8100 + 540 + 9 = 8649
Sol.
Find the value of x, if (i) 6x = 232 – 172 (ii) 4x = 982 – 882 (iii) 25x = 5362 – 1362 Sol. (i) We have, 6x = 232 – 172 6x = (23 + 17) × (23 – 17) [Using : a2 – b2 = (a + b) (a – b)] 6x = 40 × 6
Ex.54 Sol.
Ex.55
Ex.53
6x 40 6 = [Dividing both sides by 6] 6 6 x = 40 (ii) We have, 4x = 982 – 882 4x = (98 + 88) × (98 – 88) [Using : a2 – b2 = (a + b) (a – b)] 4x = 186 × 10 4x 18610 = [Dividing both sides by 4] 4 4
1860 4 x = 465 x=
Sol.
[Dividing both sides by 25]
What must be added to 9x2 – 24x + 10 to make it a whole square ? We have, 9x2 – 24x + 10 = (3x)2 – 2 × 3x × 4 + 10 It is evident from the above expression that First term = 3x and , Second term = 4 To make the given expression a whole square, we must have (4)2 = 16 in place of 10. Hence, we must add 6 to it to make a perfect square. Adding and subtracting 6, we get 9x2 – 24x + 10 + 6 – 6 = 9x2 – 24x + 16 – 6 = (3x – 4)2 –6 Find the following products: (i) (x + 2) (x + 3) (ii) (x + 7) (x – 2) (iii) (y – 4) (y – 3) (iv) (y – 7) (y + 3) (v) (2x – 3) (2x + 5) (iv) (3x + 4) (3x – 5) Using the identitiy : (x + a) (x + b) = x2 + (a + b)x + ab, we have (i) (x + 2) (x + 3) = x2 + (2 + 3)x + 2 × 3 = x2 + 5x + 6 (ii) (x + 7) (x – 2) = (x + 7) {x + (– 2)} = x2 + {7 + (–2)}x + 7 × 2 = x2 + 5x – 14 (iii) (y – 4) (y – 3) = {y + (– 4)}{y + (–3)} = y2 + {(–4) + (–3)y + (–4) × (–3)} = y2 – 7y + 12 (iv) (y – 7) (y + 3) = {y + (– 7)} (y + 3) = y2 + {(–7) + 3} y + (–7) × 3 = y2 – 4y – 21 (v) (2x – 3) (2x + 5) = (y – 3) (y + 5), where y = 2x = {y + (–3)} (y + 5) = y2 + {(–3) + 5}y + (–3) × 5 = y2 + 2y – 15
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Ex.56
Evaluate the following: (i) 107 × 103 (ii) 56 × 48 (iii) 95 × 97 Sol. Using the identity : (x + a) (x + b) = x2 + (a + b)x + ab we have (i) 107 × 103 = (100 + 7) × (100 + 3) = (100)2 + (7 + 3) × 100 + 7 × 3 = 10000 + 10 × 100 + 21 = 10000 + 1000 + 21 = 11021 (ii) 56 × 48 = (50 + 6) × (50 – 2) = (50 + 6) × {50 + (–2)} = (50)2 + {6 + (–2)} × 50 + 6 × – 2 = 2500 + 4 × 50 – 12 = 2500 + 200 – 12 = 2700 – 12 = 2688 (iii) 95 × 97 = (100 – 5) × (100 – 3) = = = = =
Greatest Common Factor (GCF) Or Highest Common Factor (HCF) The greatest common factor of given monomials is the common factor having greastest coefficient and highest power of the variables. The following step-wise procedure will be helpful to find the GCF of two or more monomials. Step I Obtain the given monomials. Step II Find the numerical coefficient of the each monomial and their greatest common factor (GCF/HCF). Step III Find the common literals appearing in the given monomials. Step IV Find smallest power of each common literal. Step V Write a monomial of common literals with smallest powers obtained in step IV. Step VI The required GCF is the product of the coefficient obtained in step II and the monomial obtained in step V.
{100 + (–5)}× {100 + (–3)} (100)2 + {(–5) + (–3)} × 100 + (–5) × (–3) 10000 – 8 × 100 + 15 10000 – 800 + 15 9215
Ex.59
FACTORIZATION Sol. Factors : If an algebraic expression is written as the product of numbers or algebraic expressions, then each of these numbers and expressions are called the factors of the given algebraic expression and the algebraic expression is called the product of these exrpressions. Factorization : The process of writing a given algebraic expression as the product of two or more factors is called factorization. Ex.57 Sol.
Ex.58 Sol.
Write down all possible factors of 3x2y. We have, 3x2y = 1 × 3x2y = 3 × x2y = 3x × xy = 3xy × x = x2 × 3y = y × 3x2 Thus, the possible factors of 3x2y are 1, 3x2 y, 3, x2 y, 3x, xy, 3xy, x, x2, 3y, y, 3x2 Write down all possible factors of 12x2. We have, 12x2 = 1 × 12x2 = 12 × x2 = 3 × 4x2 = 4 × 3x2 = 2 × 6x2 = 6 × 2x2 = 3x × 4x = 6x × 2x = 2 × 3x × 2x = 3 × 2x × 2x Thus, the possible factors of 12x2 are 1, 12x2, 12, x2, 3, 4x2, 4, 3x2, 2, 6x2, 6, 2x2, 3x, 4x, 6x, 2x
Ex.60 Sol.
Find the greatest common factors of the monomials 21a3b7 and 35a5b5. The numerical coefficients of the given monomials are 21 and 35 The greatest common factor of 21 and 35 is 7 The common literals appearing in the given monomials are a and b The smallest power of ‘a’ in the two monomials =3 The smallest power of ‘b’ in the two monomials =5 The monomial of common literals with smallest powers = a3b5 The greatest common factor = 7a3b5 Find the greatest common factors of the monomials 14x2y3, 21x2y2, 35x4y5z. The numerical coefficients of the given monomials are 14, 21, and 35 The greatest common factor of 14, 21 and 35 is 7 The common literals appearing in the three monomials are x and y The smallest power of ‘x’ in the three monomials =2 The smallest power of ‘y’ in the three monomials =2 The monomial of common literals with smallest powers = x2y2 Hence, the greatest common factor = 7x2y2
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Factorization of Algebraic Expression when a common Monomial Factor Occurs in each Term
Sol.
We have, (i) 7(2x + 5) + 3 (2x + 5) = (7 + 3) (2x + 5) [Taking (2x + 5) common] = 10(2x + 5) (ii) (x + 2) y + (x + 2) x = (x + 2) (y + x) [Taking (x + 2) common] (iii) 5a (2x + 3y) – 2b (2x + 3y) = (2x + 3y) (5a – 2b) [Taking (2x + 3y) common] (iv) 8(5x + 9y)2 + 12(5x + 9y) = 4(5x + 9y) {2(5x + 9y) + 3} = 4(5x + 9y) (10x + 18y + 3)
Ex.63
Factorize : (i) (y – x) a + (x – y)b (ii) 9(a – 2b)2 + 6(2b – a) (iii) (x – 2y)2 – 4x + 8y (iv) 2a + 6b – 3 (a + 3b)2 We have, (i) (y – x)a + (x – y)b = – (x – y)a + (x – y)b [Taking (–1) common from (y – x)] = (x – y) (–a + b) [Taking (x – y) common] = (x – y) (b – a) [ –a + b = b – a] (ii) 9(a – 2b)2 + 6(2b – a) = 9(a – 2b)2 – 6(a – 2b) [ 2b – a = – (a – 2b)] = 3(a – 2b) {3 (a – 2b) – 2} [Taking 3(a – 2b) common] = 3(a – 2b) (3a – 6b – 2) (iii) (x – 2y)2 – 4x + 8y = (x – 2y)2 – 4(x –2y)
In order to factorize algebraic expressions consisting of a common monomial factors of each term we use the following step-wise procedure. Step I Obtain the algebraic expression. Step II Find the greatest common factor (GCF/HCF) of its terms. Step III Express each term of the given expression as the product of the GCF and the quotient when it is divided by the GCF. Step IV Use the distributive property of multiplication over addition to express the given algebraic expression as the product of the GCF and the quotient of the given expression by the GCF. Ex.61
Sol.
Ex.62
Sol.
Factorize each of the following algebraic expressions: (i) 3x + 15 (ii) 2x2 + 5x 2 2 (iii) 3x y – 6xy (iv) 6x3 + 8x2y (i) The greatest commo factor of the terms namely, 3x and 15 of the expression 3x + 15 is 3. Also, 3x = 3 × x and 15 = 3 × 5. 3x + 15 = 3 (x + 5) (ii) The greatest common factor of the terms 2x2 and 5x of the expression 2x2 + 5x is x. Also, 2x2 = 2x × x and 5x = 5 × x. 2x2 + 5x = 2x × x + 5 × x = (2x + 5)x (iii) Clearly, 3xy is the greatest common factor of the terms 3x2y and 6xy2 of the binomial 3x2y – 6xy2. Also, 3x2y = 3xy × x and 6xy2 = 3xy× 2y 3x2y – 6xy2 = 3xy × x – 3xy × 2y = 3xy(x – 2y) (iv) Clearly, 2x2 is the GCF of the terms 6x3 and 8x2y of the given binomial 6x3 + 8x2y. Also, 6x3 = 2x2 × 3x and 8x2y = 2x2 × 4y. 6x3 + 8x2y = 2x2 × 3x + 2x2 × 4y = 2x2(3x + 4y) Factorize : (i) 7(2x + 5) + 3 (2x + 5) (ii) (x + 2)y (x + 2)x (iii) 5a(2x + 3y) – 2b (2x + 3y) (iv) 8(5x + 9y)2 + 12 (5x + 9y)
Taking – 4 common from – 4 x 8 y
= (x – 2y) {(x – 2y) – 4} [Taking (x – 2y) common] = (x – 2y) (x – 2y – 4) (iv) 2a + 6b – 3(a + 3b)2 = 2(a + 3b) – 3 (a + 3b)2 [Taking 2 common from 2a + 6b] = (a + 3b) {2 – 3 (a + 3b)} [Taking (a – 3b) common] = (a + 3b) (2 – 3a – 9b)
Factorization by Grouping the Terms Ex.64
Factroize : (i) ax + bx + ay + by (ii) ax2 + by2 + bx2 + ay2 (iii) a2 + bc + ab + ac (iv) ax – ay + bx – by
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Sol.
We have, (i) ax + bx + ay + by = (ax + bx) + (ay + by) [Grouping the terms] = (a + b)x + (a + b)y = (a + b) (x + y) [Taking (a + b) common] (ii) ax2 + by2 + bx2 + ay2 = ax2 + bx2 + ay2 + by2 [Re-arranging the terms] = (a + b)x2 + (a + b)y2 = (a + b)(x2 + y2) [Taking (a + b) common] (iii) a2 + bc + ab + ac = (a2 + ab) + (ac + bc) [Re-grouping the terms] = a (a + b) + (a + b)c = (a + b) (a + c) [Taking (a + b) common] (iv) ax – ay + bx – by = a (x – y) + b (x – y)
(iv) (3a – b)2 – 9c2 = (3a – b)2 – (3c)2 = {(3a – b) + 3c}{(3a – b – 3c)} = (3a – b + 3c) (3a – b – 3c) Ex.67
Factorize : (i) 16a2 –
b 16a 2 (iii) 100(x + y)2 – 81 (a + b)2 (iv) (x – 1)2 – (x – 2)2 We have, (ii) 16a2b –
Sol.
(i) 16a2–
2 5 5 25 2 – 5 4a 4a – = (4a) 2 2 a 2 a 4a 2a
= (a + b) (x – y) [Taking (x – y) common]
Ex.65
Sol.
(ii) 16a2b –
Factorize each of the following expression: (i) a3x + a2 (x – y) – a(y + z) – z (ii) (x2 + 3x)2 – 5(x2 + 3x) – y (x2 + 3x) + 5y (i) We have, a3x + a2 (x – y) – a (y + z) – z = a3x + a2x – a2y – ay – az – z = (a3x + a2x) – (a2y + ay) – (az + z) = a2x (a + 1) – ay(a + 1) – z(a + 1) = (a + 1)(a2x – ay – z) (ii) (x2 + 3x)2 – 5 (x2 + 3x) – y (x2 + 3x) + 5y
Sol.
Factorize : (i) 9a2 – 16b2 (ii) 36a2 – (x – y)2 (iii) 80a2 – 45b2 (iv) (3a – b)2 – 9c2 We have, (i) 9a2 – 16b2 = (3a)2 – (4b)2 = (3a + 4b) (3a – 4b) [Using : (a2 – b2) = (a + b) (a – b)] 2 (ii) 36a – (x – y)2 = (6a)2 – (x – y)2 = {6a + (x – y)} {6a – (x – y)} [Using : a2 – b2 = (a + b) (a – b)] = (6a + x – y) (6a – x + y) (iii) 80a2 – 45b2 = 5(16a2 – 9b2) = 5{(4a)2 – (3b)2} = 5 (4a + 3b) (4a – 3b) [Using : a2 – b2 = (a + b) (a – b)]
1 – 4a
2
1 1 = b 4a 4a – 4a 4a
(iii) 100(x + y)2 – 81 (a + b)2 = {10 (x + y)}2 – {9(a + b)}2 = {10(x + y) + 9 (a + b)} {10 (x + y) – 9 (a + b)} = (10x + 10y + 9a + 9b) (10x + 10y – 9a – 9b) (iv) (x – 1)2 – (x – 2)2 = {(x – 1) + (x – 2)}{(x –1) – (x – 2)} = (2x – 3) (x – 1 – x +2) = (2x – 3) × 1 = 2x – 3
= (x2 + 3x – 5) (x2 + 3x – y)
Ex.66
1 b 2 = b 16a – 16a 2 16a 2
2 = b ( 4 a )
= (x2 + 3x) ( x 2 3x ) – 5 – y ( x 2 3x ) – 5
Factorization of Binomial Expressions expressible as the difference of two squares
25 4a 2
Ex.68
Sol.
Factorize each of the following algebraic expression: (i) x4 – 81y4 (ii) 2x5 – 2x 4 (iii) 3x – 243 (iv) 2 – 50x2 (v) x8 – y8 (vi) a12x4 – a4x12 4 4 2 2 (i) x – 81y = (x ) – (9y2)2 = (x2 – 9y2) (x2 + 9y2)
= x 2 – (3y) 2 (x2 + 9y2) = (x – 3y) (x + 3y) (x2 – 9y2) (ii) 2x5 – 2x = 2x (x4 – 1)]
= 2x ( x 2 ) 2 – 12
= 2x (x2 – 1) (x2 + 1) = 2x (x – 1) (x + 1) (x2 + 1) (iii) 3x4 – 243 = 3 (x4 – 81)
= 3 ( x 2 ) 2 – 9 2 = 3(x2 – 9) (x2 + 9) 3(x2
= – (x2 + 9) = 3(x + 3) (x – 3) (x2 + 9)
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3 2)
(iv) 2 – 50x2 = 2{1 – 25x2} = 2{12 – (5x)2} = 2(1 – 5x) (1 + 5x) (v) x8 – y8 = {(x4)2 – (y4)2} = (x4 – y4) (x4 + y4) = {(x2)2– (y2)2}(x4 + y4) = (x2 – y2) (x2 + y2) (x4 + y4) = (x – y) (x + y) (x2 + y2) (x4 + y4) = (x – y) (x + y) (x2 + y2) {(x2)2 + (y2)2 + 2x2y2 – 2x2y2}
2 2 2 = (x – y) (x + y) (x2 + y2) ( x y ) –
(vi)
a12x4
–
a4x12
=
a 4 x4
(a )
=
a 4 x4
(a4
=
4 2
a 4 x4 4 2
a
8
–x
8
– (x ) + x4) (a4 – x4)
= a4x4 (a4 + x4) (a 2 ) 2 – ( x 2 ) 2 a 4 x4
(a4
x4)
(a2
x2 )
2
Ex.71
Sol.
2
2 xy
= (x – y) (x + y) (x2 + y2) x 2 y 2 – 2 xy
x
= (x2 – 5y2)2 = (x2 – 5y2) (x2 – 5y2) (iii) a4 – 2a2b2 + b4 = (a2)2 – 2 × a2 × b2 + (b2)2 = (a2 – b2)2 = {(a – b) (a + b)}2 = (a – b)2 (a +b)2
y 2 2 xy
(a2
= + + – x2 ) 4 4 4 4 2 2 = a x (a + x ) (a + x ) (a + x) (a – x)
Factorization of Algebraic Expressions expressible as a perfect square
Ex.73
(i) a2 + 2ab + b2 = (a + b)2 = (a + b) (a + b) (ii) a2 – 2ab + b2 = (a – b)2 = (a – b) (a – b) Ex.69 Sol.
Ex.70
Sol.
Sol.
Factorize : (i) x2 + 8x + 16 (ii) 4a2 – 4a + 1 We have, (i) x2 + 8x + 16 = x2 + 2 × x × 4 + 42 = (x + 4)2 [Using : a2 + 2ab + b2 = (a + b)2] = (x + 4) (x + 4) (ii) 4a2 – 4a + 1 = (2a)2 – 2 × 2a × 1 + (1)2 = (2a – 1)2 [Using : a2 – 2ab + b2 = (a – b)2] = (2a – 1) (2a – 1) Factorize : (i) 4x2 + 12xy + 9y2 (ii) x4 – 10x2y2 + 25y4 (iii) a4 – 2a2b2 + b4 We have, (i) 4x2 + 12xy + 9y2 = (2x)2 + 2 × 2x × 3y + (3y)2 = (2x + 3y)2 = (2x + 3y) (2x + 3y) (ii) x4 – 10x2y2 + 25y4 = (x2)2 – 2 × x2 × 5y2 + (5y2)2
Ex.74 Sol.
Factorize each of the following expressions: (i) x2 – 2xy + y2 – x + y (ii) 4a2 + 12ab + 9b2 – 8a – 12b (iii) a2 + b2 –2 (ab – ac + bc) (i) x2 – 2xy + y2 – x + y = (x2 – 2xy + y2) = (x – y)2 – (x – y) = (x – y) {(x – y) – 1} = (x – y) (x – y – 1) (ii) 4a2 + 12ab + 9b2 – 8a – 12b = (2a)2 + 2 × 2a × 3b + (3b)2 – 4(2a + 3b) = (2a + 3b)2 – 4(2a + 3b) = (2a + 3b) (2a + 3b – 4) (iii) a2 + b2 – 2 (ab – ac + bc) = a2 + b2 –2ab + 2ac – 2bc = (a – b)2 + 2c (a – b) = (a – b) {(a – b) + 2c} = (a – b) (a – b + 2c) Factorize each of the following expressions: (i) x2 + 2xy + y2 – a2 + 2ab – b2 (ii) 25x2 – 10x + 1 – 36y2 (iii) 1 – 2ab –(a2 + b2) (i) x2 + 2xy + y2 – a2 + 2ab – b2 = (x2 + 2xy + y2) – (a2 – 2ab + b2) 2 = (x + y) – (a – b)2 ={(x + y) + (a – b)} {(x + y) – (a – b)} = (x + y + a – b) (x + y – a + b) (ii) 25x2 – 10x + 1 – 36y2 = (5x)2 – 2× 5x × 1 + 12 – (6y)2 2 = (5x – 1) – (6y)2 = (5x – 1 + 62) (5x – 1 – 6y) (iii) 1 – 2ab – (a2 + b2) = 1 – (2ab + a2 +b2) = 1– (a + b)2 = {1 + (a + b)} {1– (a + b)} = (1 + a + b) (1 – a – b) Factorize: (i) x2 + 8x + 15 (ii) x4 + x2 + 1 (iii) x4 + 4 We have, (i) x2 + 8x + 15 = (x2 + 8x + 16) – 1 [Replacing 15 by 16 – 1] = {(x)2 + 2 × x × 4 + 42} – 1 = (x + 4)2 – 12 = {x + 4 + 1} {(x + 4) – 1} = (x + 5) (x + 3)
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x2 + 4x – 21 = x2 + 7x – 3x – 21 = (x2 + 7x) – (3x + 21) = x(x + 7) – 3 (x + 7) = (x + 7) (x – 3)
(ii) x4 + x2 + 1 = x4 + 2x2 + 1 – x2 [Adding and subtracting x2] = (x4 + 2x2 + 1) – x2
= ( x 2 ) 2 2 x 2 1 12 – x2 = (x2 + 1)2 – x2 = {(x2 + 1) + x} {(x2 + 1) – x} = (x2 + x + 1) (x2 – x + 1) (iii) x4 + 4 = x4 + 4x2 + 4 – 4x2 [Adding and subtracting 4x2] = {(x2)2 + 2 × x2 × 2 + 22} – 4x2 = (x2 + 2)2 – (2x)2 = {(x2 + 2) + 2x} {(x2 + 2) – 2x} = (x2 + 2x + 2) (x2 – 2x + 2)
Factorization of Quadratic Polynomials in one varibale Algorithm : Step I Step II Step III Step IV Step V
Ex.75
Obtain the quadratic polynomial x2 + px + q. Obtain p = coefficient of x and, q = constant term. Find the two numbers a and b such that a + b = p and ab = q. Split up the middle term as the sum of two terms ax and bx. Factorize the expression obtained in step IV by grouping the term.
Factorize each of the following expressions: (i) x2 + 6x + 8 (ii) x2 + 4x – 21 (iii) x2 – 7x + 12 Sol. (i) In order to factorize x2 + 6x + 8, we find two numbers p and q such that p + q = 6 and pq = 8. Clearly, 2 + 4 = 6 and 2 × 4 = 8 We now split the middle term 6x in the given quadratic as 2x + 4x. x2 + 6x + 8 = x2 + 2x + 4x + 8 = (x2 + 2x) + (4x + 8) = x (x + 2) + 4 (x + 2) = (x + 2) (x + 4) (ii) In order to factorize x2 + 4x – 21, we have to find two numbers p and q such that p + q = 4 and pq = – 21 Clearly, 7 + (–3) = 4 and 7 × –3 = 21. We now split the middle term 4x of x2 + 4x – 21 as 7x – 3x.
(iii) In order to factorize x2 – 7x + 12 we have to find two numbers p and q such that p + q = –7 and pq = 12. Clearly, – 3 – 4 = – 7 and – 3 × – 4 = 12. We now split the middle term – 7x of the given quadratic as – 3x – 4x. x2 – 7x + 12 = x2 – 3x – 4x + 12 = (x2 – 3x) – (4x – 12) = x(x – 3) – 4 (x – 3) = (x – 3) (x – 4) Ex.76
Factorize each of the following quadratic polynomials : (i) x2 – 23x + 132 (ii) x2 – 21x + 108 (iii) x2 + 5x – 36 Sol. (i) In order to factorize x2 – 23x + 132, we have to find numbers p and q such that p + q = –23 and pq = 132. Clealry, – 12 – 11 = –23 and –12 × –11 = 132. We now split the middle term – 23x of x2 – 23x + 132 as – 12x – 11x x2 – 23x + 132 = x2 – 12x – 11x + 132 = (x2 – 12x) – (11x – 132) = x (x – 12) – 11 (x – 12) = (x – 12) (x – 11) (ii) In order to factorize x2 – 21x + 108, we have to find two numbers such that their sum is – 21 and the product 108. Clearly, –21 = –12 – 9 and – 12 × – 9 = 108 So, we split the middle term – 21x as – 12x – 9x x2 – 21x + 108 = x2 – 12x – 9x + 108 = (x2 – 12x) – (9x – 108) = x(x – 12) – 9(x – 12) = (x – 12) (x – 9) (iii) In order to factorize x2 + 5x – 36, we have to find two numbers p and q such that p + q = 5 and pq = – 36. Clearly, 9 + (–4) = 5 and 9 × –4 = –36. So, we write the middle term 5x of x2 + 5x – 36 as 9x – 4x. x2 – 5x – 36 = x2 + 9x – 4x – 36 = (x2 + 9x) – (4x + 36) = x (x + 9) – 4 (x + 9) = (x + 9) (x – 4)
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(iii) The given expression is 6x2 – 13x + 6. Here, coefficient of x2 = 6, coefficient of x = – 13, and constant term = 6. We shall now split up the coefficient of x i.e. – 13 into two parts whose sum is –13 and product equal to product of the coefficient of x2 and constant term i.e, 6× 6 = 36. Clearly, –4, –9 = – 13 and –4 × –9 = 36. So, we write the middle term – 13x as – 4x – 9x Thus, we have 6x2 – 13x + 6 = 6x2 – 4x – 9x + 6 = 2x (3x – 2) – 3 (3x – 2) = (3x – 2) (2x – 3)
Factorization of Quadratic Polynomials of Theorem ax2 + bx + c, a 1 Procedure: Step I Obtain the quadratic trinomial ax2 + bx + c Step II Obtain a = coefficient of x2, b = coefficient of x and c = constant terms. Step III Find the product of the coefficient of x2 and the constant term i.e. ac. Step IV Split up the coefficient of x i.e. b into two parts whose sum is b and product ac and write the middle term as the sum of two terms. Step V Factorize the expression obtained in step IV by grouping the term. Factors so obtained will be the required factors of the given quadratic trinomial. Factorize: (i) 2x2 + 5x + 3 (ii) 6x2 + 5x – 6 (iii) 6x2 – 13x + 6 (iv) –2x2 – 3x + 2 Sol. (i) The given expression is 2x2 + 5x + 3 Here, coefficient of x2 =2, coeffcient of x = 5, and constant term = 3. We shall now split up the coefficient of the middle term i.e. 5 into two parts such that their sum is 5 and product equal to the product of coefficient of x2 and constant term i.e. 2 × 3 = 6. Clearly, 2 + 3 = 5 and 2 × 3 = 6. So, we replace the middle term 5x by 2x + 3x. Thus, we have 2x2 + 5x + 3 = 2x2 + 2x + 3x + 3 = (2x2 + 2x) + (3x + 3) = 2x(x + 1) + 3(x + 1) = (x + 1) (2x + 3)
(iv) The given expression is – 2x2 – 3x + 2. Here, coefficient of x2 = – 2, coefficient of x = –3 and constant term = 2. We shall now split up the coefficient of the middle term i.e. – 3 into two parts such that their sum is – 2 and the product is equal to the product of the coefficient of x2 and constant term i.e. – 2 × 2 = –4. Clearly, – 4 + 1 = – 3 and – 4 × 1 = –4. So, we write the middle term – 3x as – 4x + x. Thus, we have –2x2 – 3x + 2 = –2x2 – 4x + x + 2 = –2x (x + 2) + 1 (x + 2) = (x + 2) (–2x + 1)
Ex.77
(ii) The given expression is 6x2 + 5x – 6 Here, coefficient of x2 = 6, coefficient of x = 5, constant term = – 6 We shall now split up the coefficient of x i.e., 5 into two parts such that their sum is equal to coefficient of x i.e., 5 and product equal to the product of coefficient of x2 and constant term i.e., 6 × – 6 = – 36. Clearly, 9 + (–4) = 5 and 9 × – 4 = – 36. So, we replace the middle term 5x by 9x – 4x. Thus, we have 6x2 + 5x – 6 = 6x2 + 9x – 4x – 6 = 3x(2x + 3) – 2 (2x + 3) = (2x + 3) (3x – 2)
Ex.78
Factorize: (i) 12x2 – 23xy + 10y2 (ii) 12x2 + 7xy – 10y2 Sol. (i) The given expression is 12x2 – 23xy + 10y2 Here, coefficient of x 2 = 12, coefficient of x = –23y, and constant term = 10y2. Now, we split up the coefficient of the middle term i.e., –23y into two parts whose sum is –23y and product equal to the product of the coefficient of x 2 and constant term i.e., 12 × 10y2 = 120y2. Clearly, –15y – 8y = – 23y and –15y × –8y = 120y2 So,we replace the middle term –23xy by –15xy –8xy. Thus, we have 12x2 – 23xy + 10y2 = 12x2 – 15xy – 8xy + 10y2 = 3x (4x – 5y) – 2y (4x – 5y)= (4x – 5y) (3x – 2y) (ii) The given expression is 12x2 + 7xy – 10y2 Here, coefficient of x2 = 12, coefficient of x = 7y and constant term = – 10y2. We shall now split up the coefficient of the middle term i.e. 7y into two parts whose sum is 7y and product equal to the product of the coefficient of x 2 and constant term i.e. 12 × –10y2 = –120y2.
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Clearly, 15y – 8y = 7y and 15y × –8y = – 120y2 So, we replace the middle term 7xy by 15xy – 8xy Thus, we have 12x2 + 7xy – 10y2 = 12x2 + 15xy – 8xy – 10y2 = 3x (4x + 5y) – 2y (4x + 5y) = (4x + 5y) (3x – 2y)
Factorization of Quadratic Polynomials by using the method of completing the perfect square
= (y2 + 6y + 32) –1 = (y + 3)2 – 12 [By completing the square] = {(y + 3) – 1} {(y + 3) + 1} [Using : a2 – b2 = (a – b) (a + b)] = (y + 2) (y + 4) Ex.80 Sol.
3 2 = 4 y – 2 y [Making coefficient of y2 as 1] 4
Procedure: Step I Obtain the quadratic polynomial. Let the polynomial be ax2 + bx + c, where a 0. Step II Make the coefficient of x2 unity by dividing and multiplying throughout by it, if it is not unity i.e., write
3 2 2 2 = 4 y – 2 y 1 – 1 4 2 1 Adding and subtractin g Coeff . of y 2 2 i.e.,1
c 2 b ax2 + bx + c = a x x a a
1 2 2 = 4 ( y – 2 y 1 ) – 4
Step III Add and subtract square of half of the coefficient of x i.e., write
c 2 b ax2 + bx + c = a x x a a
2 2 1 ( y – 1 ) – [Completing the square] =4 2
2 2 2 c b b b x 2 x – =a a 2a 2a 2a
1 1 = 4 ( y – 1) – ( y – 1) 2 2
[Using a2 – b2 = (a – b) (a + b)]
Step IV Write first three terms as the square of a binomial and simplify last two terms i.e., write ax2 + bx + c
1 1 = 4 y –1 – y –1 2 2 3 1 2y – 3 2 y – 1 = 4y – y – = 4 2 2 2 2
2 2 2 c b b b x 2 x – = a a 2a 2a 2a 2 b x =a 2a
b 2 – 4ac – 2 4a
Sol.
= (2y – 3) (2y – 1) Ex.81 Sol.
Step V Factorize last step obtained in step IV by using a2 – b2 = (a – b) (a + b) to get desired factors. Ex.79
Factorize: 4y2 – 8y + 3 We have, 4y2 – 8y + 3
Factorize : 6 – x – 2x2 We have, 6 – x – 2x2 = – 2x2 – x + 6
2 1 = – 2 x x – 3 2 Dividing and multiplying by – 2 2 i.e., the coeff . of x
y2
Factorize + 6y + 8 by using the method of completing the square. Here, coefficient of y2 is unity. So, we add and subtract the square of the half of coefficient of y. y2 + 6y + 8 = y2 + 6y + 32 – 32 + 8 6 Adding and subtracting 32 2 2
2 2 2 1 1 1 x x – – 3 =–2 2 4 4 2 2 1 1 Adding and subtracting Coeff . of x i.e., 2 a
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2 1 1 1 2 x 2 x – 3 = – 2 4 4 16
2 2 3 2 x – x + x –5 is a polynomial in variables 3 2
Ex.82
1 3 x – 3x2 + 5x1/2 + x – 1 is not a 2 polynomial, because it contains a term 5x1/2 which x whereas
2 1 49 x – = –2 4 16
1 as the power of variable x, which is 2 not a non-negative integer. contains
2 2 1 7 x – =–2 4 4
Ex.83 1 7 1 7 = – 2 x – x 4 4 4 4
3 – 2x2 + 4x2y + 8y –
5 xy2 is a polynomial in 3
two variables x and y. Ex.84 (i) 2x + 3 is a polynomial in x of degree 1. 7 (ii) 2x2 – 3x + is polynomial in x of degree 2. 5
1 7 1 7 = – 2 x – x 4 4 4 4 3 = – 2 x – (x + 2) 2
(iii)
2 2 7 2 a – a + 4 is a polynomial in a dgree 3. 3 2
= (– 2x + 3) (x +2) Ex.85
3x4 – 2x3y2 + 7xy3 – 9x + 5y + 4 is a polynomial
POLYNOMIALS in x and y of degree 5, whereas
Polynomials : An algebraic expression in which the variables involved have only non-negative integral powers, is called a polynomial. Degree of a polynomial in one variable: In a polynomial in one variable, the highest power of the variable is called degree. Degree of a polynomial in two variable: In a polynomial in more than one variable the sum of the powers of the variables in each term is computed and the highest sum so obtained is called the degree of the polynomial. Constant Polynomial : A polynomial consisting of a constant term only is called a constant polynomial. The degree of a constant polynomial is zero. Linear Polynomial : A polynomial of degree 1 is called a linear polynomial. Quadratic Polynomial : A polynomial of degree 2 is called a quadratic polynomial. Cubic Polynomial : A polynomial of degree 3 is called a cubic polynomial. Biquadratic Polynomials : A polynomial of degree 4 is called a biquadratic polynomial.
1 – 3x + 7x2y – 2
3 2 2 x y is a polynomial of degree 4 in x and y.. 4 Ex.86 Ex.87
Ex.88
2–
3 1 3 x, + y, 2 + 3a etc. are linear polynomials. 4 2 5
3 1 2x2 – 3x + 4, 2 – x + x2 , 2y2 – y+ are 2 4 quadratic polynomials. x3 – 7x + 2x – 3, 2 +
1 3 y – y2 + 4y3 are cubic 2 2
polynomial. Ex.89
2 2 3 3x4 – 7x3 + x2 – x + 9, 4 – x + x4 are 3 5 biquadratic polynomials.
Division of a monomial by a monomial While dividing a monomial by a monomial, we follow the following two rules: Rule-1 Coefficient of the quotient of two monomial is equal to the quotient of their coefficients. Rule-2 The variable part in the quotient of two monomials is equal to the quotient of the variables in the given monomials.
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Ex.90
Divide : (i) 12x3y3 by 3x2y Sol. (i) We have,
(ii) –15a2bc3 by 3ab
12 x x x y y 12x 3 y 2 = = 4 × x × y = 4xy 2 3 x x y 3x y (ii) We have,
–15 a a b c c c – 15a 2 bc3 = = – 5ac3 3 a b 3ab
Division of a Polynomial by a Monomial Obtain the polynomial (dividend) and the monomial (divisor). Step II Arrange the terms of the dividend in descending order of their degrees. For example, write 7x2 + 4x –3 + 5x3 as 5x2 + 7x2 + 4x – 3. Step III Divided each term of the plynomial by the given monomial by using the rules of division of a monomial by a monomial.
Divide 6 + x – 4x2 + x3 by x – 3. We go through the following steps to perform the division: Step I We write the terms of the dividend as well as of divisor in descending order of their degress. Thus, we write = 6 + x – 4x2 + x3 as x3 – 4x2 + x + 6 and x – 3 as x – 3 Step II We divide the first term x3 of the dividend by the Ex.92 Sol.
first term x of the divisor and obtain
Step I
Ex.91
Divide : (i) 9m5 + 12m4 – 6m2 by 3m2 (ii) 24x3y + 20x2y2 – 4xy by 2xy Sol. (i) We have,
12m 4 9m 5 12m 4 – 6m 2 9m 5 6m 2 = + – 3m 2 3m 2 3m 2 3m 2 3 = 3m + 4m2 – 2 (ii) We have, 24 x 3 y 20 x 2 y 2 – 4 xy 2 xy 4 xy 24x 3 y 20 x 2 y 2 = + – = 12x2 + 10xy – 2 2 xy 2xy 2 xy
Division of a Polynomial by a Binomial by using long division Step I Step II Step III Step IV Step V
Arrange the terms of the dividend and divisor in descending order of their degrees. Divide the first term of the dividend by the first term of the divisor to otbain the first term of the quotient. Multiply the divisor by the first term of the quotient and subtract the result from the dividend to obtain the remainder. Consider the remainder (if any) as dividend and repeat step II to obtain the second term of the quotient. Repeat the above process till we obtain a remainder which is either zero or a polynomial of degree less than that of the divisor.
x3 = x2 as x
the first term of the quotient. Step III We multiply the divisor x – 3 by the first term x of the quotient and subtract the result from the dividend x3 – 4x2 + x + 6. We obtain – x2 + x + 6 as the remainder. x2 – x – 2 3 2 x – 4x + x + 6 x–3 3 2 x – 3x – + 2 –x –x+6 – x2 + 3x + – – 2x + 6 – 2x + 6 + – 0 Step IV We take –x2 + x + 6 as the new dividend and
x2 repeat step II to obtain the second term x – x of the quotient. Step V We multiply the divisor x – 3 by the second term – x of the quotient and subtract the result –x2 + 3x from the new dividend. We obtain – 2x + 6 as the remainder. Step VI Now we treat – 2x + 6 as the new dividend and divide its first term – 2x by the first term x of the –2x divisor to obtain = –2 as the third term of x the quotient. Step VII We multiply the divisor x – 3 and the third term –2 of the quotient and subtract the result – 2x + 6 from the the new dividend. We obtain 0 as the remainder. Thus, we can say that (6 + x – 4x2 + x3) (x – 3) = x2 – x – 2
6 x – 4x x 3 = x2 – x – 2 x –3 The above procedure is displaced on the right side of the above step. or,
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Note : In the above example, the remainder is zero. So, we can say that (x – 3) is a factor of 6 + x – 4x2 + x3. Ex.93 Sol.
Sol.
Divide : x3 – 6x2 + 11x – 6 by x2 – 4x + 3 On dividing, we get x–2 3 2 2 x – 4x + 3 x – 6x + 11x – 6 3 x – 4x2 + 3x – + – – 2x22 + 8x – 6 – 2x + 8x – 6 + – + 0
4
Sol.
Using division show that 3y2 + 5 is factor of 6y5 + 15y4 + 16y3 + 4y2 + 10y – 35. On dividing 6y5 + 15y4 + 16y3 + 4y2 + 10y – 35 by 3y2 + 5, we obtain
2y3 + 5y2 + 2y – 7 2
3y + 5
6y5 + 15y4 + 16y3 + 4y2 + 10y – 35 5 3 6y + 10y – – 15y4 + 6y3 + 4y2 + 10y – 35 4 2 15y + 25y – – 3 2 6y – 21y + 10y – 35 3 6y + 10y – – – 21y2 – 35 2 – 21y – 35 + + 0
Ex.96 Sol.
Find the values of a and b so that x4 + x3 + 8x2 + ax + b is divisible by x2 + 1. If x4 + x3 + 8x2 + ax + b is exactly divisible by x2 + 1, then the remainder should be zero. On dividing, we get x2 + x + 7 4 3 2 x2 + 1 x + x + 8x + ax + b x4 + x2 – – x3 + 7x2 + ax + b x3 +x – – 7x2 + x(a – 1) + b 7x2 +7 – – x (a – 1) + b – 7
Division Aglorithm:
Ex.95
What must be subtracted from 8x4 + 14x3 – 2x2 + 7x – 8 so that the resulting polynomial is exactly divisible by 4x2 + 3x – 2.
2
Quotient = 2x2 + 2x – 1 and, Remainder = 14x – 10 Thus, if we subtract the remainder 14x – 10 from 8x4 + 14x3 – 2x2 + 7x – 8, it will be divisible by 4x2 + 3x – 2
Since the remainder is zero. Therefore, 3y2 + 5 is a factor of 6y5 + 15y4 + 16y3 + 4y2 + 10y – 35.
We know that if a number is divided by another number, then Dividend = Divisor × Quotient + Remainder Similarly, if a polynomial is divided by another polynomial, then Dividend = Divisor × Quotient + Remainder This is generally known as the division algorithm.
3
2 4x + 3x – 2 8x + 14x – 2x + 7x – 8 4 3 2 8x + 6x – 4x – – + 3 2 8x + 2x + 7x – 8 3 2 8x + 6x – 4x – – + 2 – 4x + 11x – 8 2 – 4x – 3x + 2 + + – 14x – 10
x3 – 6x2 + 11x – 6 = (x –2) (x2 – 4x + 3) Ex.94
We know that Dividend = Quotient × Divisor + Remainder Dividend – Remainder = Quotient × Divisor Clearly, R.H.S of the above result is divisble by the divisor. Therefore, L.H.S. is also divisible by the divisor. Thus, if we subtract remainder from the dividend, then it will be exactly divisible by the divisior. Dividing 8x4 + 14x3 – 2x2 + 7x – 8 by 4x2 + 3x – 2, we get 2x2 + 2x – 1
Quotient = x2 + x + 7 and, Remainder = x (a – 1) + b –7 Now, Remainder = 0
x (a – 1) + (b – 7) = 0 x (a – 1) + (b – 7) = 0x + 0 a – 1 = 0 and b – 7 =0 [Comparing coefficients of x and constant terms] a = 1 and b = 7
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Ex.97 Sol.
Ex.98 Sol.
Ex.99 Sol.
Divide x4 – x3 + x2 + 5 by (x + 1) and write the quotient and remainder. We have, x4 – x3 + x2 + 5 = x3 (x + 1) – 2x2(x + 1) + 3x (x + 1) – 3 (x + 1) + 8 = (x + 1) (x3 – 2x3 + 3x – 3) + 8 Hence, Quotient = x3 – 2x2 + 3x – 3 and, Remainder = 8. Divide 12x3 – 8x2 – 6x + 10 by (3x – 2). Also, write the quotient and the remainder. We have, 12x3 – 8x2 – 6x + 10 = 4x2 (3x – 2) – 2 (3x – 2) + 6 = {4x2 (3x – 2) – 2 (3x – 2)} + 6 = (3x – 2) (4x2 – 2) + 6 Hence, Quotient = 4x2 – 2 and, Remainder = 6. Divide 6x3 – x2 – 10x – 3 by (2x – 3). We have, 6x3 – x2 – 10x – 3 = 3x2 (2x – 3) + 4x (2x – 3) – 1(2x – 3) – 6 = {3x2 (2x – 3)+ 4x (2x – 3) – 1 (2x – 3)} – 6 = (2x – 3) (3x2 + 4x – 1) – 6 Hence, Quotient = 3x2 + 4x – 1 and, Remainder = – 6.
ax 2 (b ac) x bc ( x c) (ax b)( x c) = ( x c)
=
= ax + b
a12 a 6 b 6 b12 a 6 – a 3b3 b 6
(a 6 – a 3b 3 b 6 )(a 6 a 3b 3 b 6 ) = (a 6 – a 3 b 3 b 6 ) = a 6 + a 3b 3 + b 6 [Cancelling a6 – a3b3 + b6 from Nr and Dr]
Ex.100 Divide: (i) 35a2 + 32a – 99 by 7a – 9 (ii) ax2 + (b + ac) x + bc by x + c Sol. (i) We have, 35a2 + 32a – 99 = 35a2 + 77a – 45a – 99 = 7a (5a + 11) – 9 (5a + 11) = (5a + 11) (7a – 9) ...(i) (35a2 + 32a – 99) (7a – 9)
35a 2 32a – 99 7a – 9
=
(5a 11)(7a – 9) = 5a + 11 (7a – 9)
[Using (i)]
Just as numbers, we cancel common factor (7a – 9) in numerator and deno min ator
(ii) We have, ax2 + (b + ac)x + bc = (ax2 + bx) + (acx + bc) = x (ax + b) + c (ax + b) = (ax + b) (x + c) ...(i)
Cancelling common factor ( x c) in numerator and deno min ator
Ex.101 Divide: a12 + a6b6 + b12 by a6 – a3b3 + b6 Sol. We have, a12 + a6b6 + b12 = a12 + 2a6b6 + b12 – a6b6 [Adding and subtracting a6b6] 6 6 2 = (a + b ) – (a3b3)2 = (a6 + b6 – a3b3) (a6 + b6 + a3b3) = (a6 – a3b3 + b6) (a6 – a3b3 + b6) ...(i)
Division of Polynomials by using Factorization
=
[Using (i)]
ax 2 (b ac) x bc (x + c)
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EXERCISE # 1 Q.1
Add the following algebraic expressions: 2,
Q.3
Q.5
Q.6
Q.7
Q.11
(i) Subtract: 3l (l – 4m + 5n) from 4l (10n – 3m + 2l) (ii) Subtract : 3a (a + b + c)–2b (a + b + c) from 4c(–a + b + c)
1 2 Subtract : – 2 y y – 3 from 7y2 – 2y + 10. 2
Subtract:
3 2 x y + 2
4 y – 5
1 2 x yz from 3
12 2 3 2 x yz – xyz + x2y.. 5 5 3 Q.4
Add: (i) 5m(3 – m) and 6m2 – 13m (ii) 4y(3y2 + 5y – 7) and 2(y3 – 4y2 + 5)
2y 4 y 5y 2 5y 3 2y 2 – + ,– + – , 3 3 2 3 2 3 6 5y 3 + 3y2 + 3y + 5 3
Q.2
Q.10
Find the volume of the rectangular boxes with following length, breadth and height : Length Breadth Height (i) 2ax 3by 5cz (ii) m2n n2 p p2m 2 (iii) 2q 4q 8q3 Find each of the following products: (i) (–2x2) × (7a2x7) × (6a5x5) (ii) (4s2t) × (3s3t3) × (2st4) × (–2)
4 3 6 2 xy by x y and verify your 3 7 result for x = 2 and y = 1. Multiply –
Find the product of – 5x2y , –
2 2 xy z, 3
Q.12
1 1 Multiply x – y and (5x2 – 4y2) 4 5
Q.13
Multiply (3x2 + y2) by (x2 + 2y2).
Q.14
Multiply: {2m + (–n)} by {–3m + (–5)}
Q.15
2 2 Find the product of y y and (7y – y2) and 7 verify the result for y = 3.
Q.16
Simplify the following: 1 (i) (6x2 + 15y2) (6x2 – 15y2) 3 (ii) 9x4(2x3 – 5x4) × 5x6 (x4 – 3x2)
Q.17
Multiply: (2x2 – 4x + 5) by (x2 + 3x – 7)
Q.18
Find the product of the following binomials: (i) (6x2 – 7y2) (6x2 – 7y2)
1 1 (ii) x – y 2 5 Q.19
8 1 xyz2 and – . Verify the result when x = 1, 15 4 y = 2 and y = q. Q.8
Q.9
3 3 (ii) 2a 2a – b b (iii) (a2 + b2) (–a2 + b2) (iv) (–a + c) (–a – c) Q.20
b (ii) 121.5ab × ac 10 (iii) 0.1a × (0.01a × 0.001b)
Find the product of the following binomials:
5 3 5 3 (i) x y x – y 6 4 6 4
7 2 Find the product of s t and s + t. Verify the 2 1 result for s = and t = 5. 2 Find the following products: (i) 100x × (0.01x4 – 0.01x2)
1 1 x – y 2 5
If x+ x–
Q.21
1 1 = 9 and x2 + 2 = 53, find the value of x x
1 . x
If x + y = 12 and xy = 14, find the value of x2 + y2.
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Q.22
Simplify the following products: (i) (x2 + x + 1) (x2 – x + 1) (ii) (x2 + 2x + 2) (x2 – 2x + 2)
Q.23
Simplify the following by using: (a + b) (a – b) = a2 – b2 . (i) 68 × 72 (ii) 101 × 99 (iii) 67 × 73 (iv) 1282 – 772
Q.33
Factorize : (i) 4x2 – 4xy + y2 – 9z2 (ii) 16 – x2 – 2xy – y2 (iii) x4 – (x – z)4
Q.34
Factorize : (i) 4(x + y)2 – 28y (x + y) + 49y2 (ii) (2a + 3b)2 + 2(2a + 3b) (2a – 3b) + (2a – 3b)2
Q.35
Factorize each of the following expressions: (i) 9x2– 4y2 (ii) 36x2 – 12x + 1 – 25y2 (iii) a2 – 1 + 2x – x2
Q.24
Find the greatest common factor of the monomials 6x3a2b2c, 8x2ab3c3 and 12a3b2c2.
Q.25
Factorize: (i) 12x3y4 + 16x2y5 – 4x5y2 (ii) 18a3b2 + 36ab4 – 24a2b3
Q.36
Factorize: (i) (x + y)(2x + 3y) – (2x + 3y) – (x + y) (x + 1) (ii) (x + y) (2a + b) – (3x – 2y) (2a + b)
Factorize: (i) 9 – a6 + 2a3b3 – b6 (ii) x16 – y16 + x8 + y8
Q.37
Factorzie: (2x + 3y)2 – 5(2x + 3y) – 14
Q.38
Factorize: 3m2 + 24m + 36
Q.39
Divide: (i) 6x4yz – 3xy3z + 8x2yz4 by 2xyz
Q.26
Q.27
Q.28
Q.29
Q.30
Q.31
Q.32
Factorize : (i) x2 + xy + 8x + 8y (ii) 15xy – 6x + 10y – 4 (iii) n – 7 + 7lm – lmn Factorize: (i) a2 + 2a + ab + 2b (ii) x2 – xz + xy – xz
(ii)
Factorize each of the following expressions: (i) a2 – b + ab – a (ii) xy – ab + bx – ay (iii) 6ab – b2 + 12ac – 2bc (iv) a(a + b – c) – bc (v) a2x2 + (ax2 + 1) x + a (vi) 3ax – 6ay – 8by + 4bx Factorize: (i) x3 – 2x2y + 3xy2 – 6y3 (ii) 6ab – b2 + 12ac – 2bc Factorize : (i) x4 – y4 (iii) x4 – (y + z)4 (v) 3a4 – 48b4
(ii) 16x4 – 81 (iv) 2x – 32x5 (vi) 81x4 – 121x2
2 2 2 2 4 2 3 1 1 a b c + ab c – ab3c2 by abc 3 3 5 2
Q.40
Divide the polynomial 2x4 + 8x3 + 7x2 + 4x + 3 by x + 3.
Q.41
Divide 10x 4 + 17x 3 – 62x 2 + 30x – 3 by 2x2 + 7x – 1
Q.42
Divide 3y5 + 6y4 + 6y3 + 7y2 + 8y + 9 by 3y3 + 1 and verify that Dividend = Divisor × Quotient + Remainder
Q.43
Divide 16x4 + 12x3 – 10x2 + 8x + 20 by 4x – 3. Also, write the quotient and remainder.
Q.44
Divide 8y3 – 6y2 + 4y – 1 by 4y + 2. Also, write the quotient and the remainder.
Q.45
Divide: a4 – b4 by a – b
Q.46
Divide: x4a + x2ay2b + 4y4b by x2a + xayb + y2b
Factorize each of the following algebraic expressions: (i) 16(2x – 1)2 – 25z2 (ii) 4a2 – 9b2 – 2b – 3b (iii) x2 – 4x + 4y – y2 (iv) 3 – 12 (a – b)2 (v) x (x + z) – y(y + z) (vi) a2 – b2 – a – b
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ANSWERS (Ex. # 1)
Sol.1
28 19 25 3 + y + 2y2 + y 15 6 6
Sol.2
9y2 –
Sol.3
41 2 5 2 3 4 x yz – x y – xyz – y 15 6 5 5
5 y + 13 2
(i) 30 abcxyz
(ii) m3 n3 p 3
(iii) 64q6
Sol.5
(i) – 84x14a7
(ii) – 48s6t8
(iii) 1000x14y11
Sol.6
–
Sol.8 Sol.9
(ii) (–2x + 3y) (2a + b)
Sol.28 (i) (a + 2) (a + b)
(ii) (x + y) (x – z)
Sol.29 (i) (a + b) (a – 1) (iii) (b + 2c) (6a – b) (v) (x + a) (ax2 + 1)
(ii) (y + b) (x – a) (iv) (a + b) (a – c) (vi) (3a + 4b) (x – 2y)
Sol.30 (i) (x – 2y) (x2 + 3y2)
Sol.4
Sol.7
Sol.26 (i) (x + y) (x + 3y – 1)
8 3 4 x y 7
(x2
(ii) (6a – b)(b + 2c) y2)
Sol.31 (i) (x – y)(x + y) + (ii) (2x – 3) (2x + 3) (4x2 + 9) (iii) (x – y – z) (x + y + z) {x2 + (y + z)2} (iv) 2x (1 + 4x2) (1 – 2x) (1 + 2x) (v) 3(a – 2b) (a + 2b) (a2 + 4b2) (vi) x2 (9x – 11) (9x + 11)
7 3 7 22 s t+ s t 2 2
Sol.32 (i) (8x – 5z – 4) (8x + 5z – 4) (ii) (2a + 3b) (2a – 3b – 1) (iii) (x – y) (x + y – 4) (iv) 3(1 + 2a – 2b) (1 – 2a + 2b) (v) (x – y) (x + y + z) (vi) (a + b) {(a – b) – 1}
(i) x5 – x3 (ii) 121.5a2bc + 12.15ab2 (iii) 0.001a2 + 0.0001ab
Sol.33 (i) (2x – y + 3z) (2x – y – 3z) (ii) (4 + x + y) (4 – x – y) (iii) (2x2 – 2xz + z2) (2x – z)z
4 – x 4 y4z 4 9
Sol.10 (i) 2m + m2
Sol.34 (i) (2x – 5y)2
Sol.11 (i) 25ln + 5l2 (ii) –7ac + 6bc + 4c2 – 3a2 – ab – 2b2
Sol.35 (i) (3x + 2y) (3x – 2y) (ii) (6x – 5y – 1) (6x + 5y – 1) (iii) (a – 1 + x) (a + 1 – x)
Sol.12 x3 –
4 2 5 xy – x 2 y + y3 5 4
Sol.36 (i) (a3 – b3 + 3) (–a3 + b3 + 3) (ii) (x8 + y8) (x8 – y8 + 1) (iii) (p + q – a + b) (p + q + a – b + 1)
Sol.13 3x4 + 7x2y2 + 2y4 2 4 y 7 Sol.16 (i) 12x4 – 75y4 (ii) –225x18 + 90x17 + 675x16 – 270x15
Sol.15 7y2 + y3 –
Sol.17 2x4 + 2x3 – 21x2 + 43x – 35 1 4 xy 1 2 x y 4 5 25 9 (ii) 4a2 – 2 b (iv) a2 – c2
Sol.18 (i) 36x4 – 842y2 + 49y4 (ii)
9 2 25 2 x – y 16 36 (iii) b4 – a4
Sol.19 (i)
Sol.20 ± 5 Sol.21 116 Sol.23 (i) 4896
Sol.22 (i) x4 + x2 + 1, (ii) x4 – 2x2 + 4 (ii) 9999
(ii) 16a2
(iii) 4891
(iv) 10455
Sol.37 (2x + 3y – 7) (2x + 3y + 2) Sol.38 3(m + 2) (m + 6)
3 2 y + 4xz3 2 4 8 2 (ii) abc + bc – b c 3 3 5
Sol.39 (i) 3x3 –
Sol.40 (x + 3) (2x3 + 2x2 + x + 1) Sol.41 (2x2 + 7x – 1) (5x2 – 9x + 3)
9 2 5 11 Sol.44 (4y + 2) 2 y – y – 2 4 2 Sol.45 (a + b)(a2 + b2) Sol.46 x2a – xayb + y2b
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EXERCISE # 2 1 If x = 3, then x 10 82 (A) (B) 3 9
2 1 x x2
Q.2
1 1 If x = , then x 2 (A) 7 (B) –7
2 4 4 x is equal to x2 (C) 9 (D) –9
Q.3
1 If x = 4, then x (A) 196 (B) 194
Q.4
1 If x = x is -
Q.1
Q.5
Q.6
Q.7
(C) 7
4 1 x x4 (C) 192
is equal to
Q.11
4 1 3 1 If x 4 = 727, the value of x 3 is x x (A) 125 (B) 140 (C) 155 (D) 170
Q.12
3 If 2 x = 5, the value of x (A) 25 (B) 30 (C) 35
Q.13
1 6 1 If x = 3, the value of x 6 is x x (A) 927 (B) 414 (C) 364 (D) 322
(D) 11
is equal to (D) 190
3 1 3 , then the value of x 3 x
Q.14
2 9 4 x is x2 (D) 49
1 If t2 – 4t + 1 = 0, then the value of t 3 ist3 (A) 44 (B) 48 (C) 52 (D) 64
(A) 0
(B) 3 3
If x + y = 7 and xy = 12, the value of (x2 + y2) is -
(C) 3( 3 – 1)
(D) 3( 3 + 1)
(A) 25
Q.15
1 3 1 If x = 2, then x 3 is equal to x x (A) 64 (B) 14 (C) 8 (D) 2 2 1 1 If x 2 = 102, the value of x is x x (A) 8 (B) 10 (C) 12 (D) 13 4 1 If x 4 x
1 = 322, the value of x is x
(A) 4
(B) 6
(C) 8
Q.16
Q.19
(D) 3 2
(D) 49
(B) 133
(C) 217
(D) 343
If x + y = 5 and xy = 6, the value of (x3 – y3) is(A) –19
Q.18
(C) 37
If x + y = 5 and xy = 6, the value of (x3 + y3) is (A) 91
Q.17
(B) 29
(B) 19
(C) –63
(D) 63
If x1/3 + y1/3 + z1/3 = 0, then (A) x + y + z = 0
(B) (x + y + z)3 = 27 xyz
(C) x + y + z = 3xyz
(D) x3 + y3 + z3 = 0
If a + b + c = 0, then (a3 + b3 + c3) is equal to (A) 0
(B) abc
(C) 3abc
(D) (ab + bc + ca)
Q.8
3 1 1 If x 3 = 52, the value of x is x x (A) 4 (B) 3 (C) 6 (D) 13
Q.20
Q.9
3 1 1 If x 3 = 14, the value of x is x x (A) 5 (B) 4 (C) 3 (D) 2
If a + b + c = 0, then the value of a 2 b2 c2 bc ca ab is (A) 1 (B) 0 (C) –1 (D) 3
Q.21
If x + y + z = 9 & xy + yz + zx = 23, then the value of (x3 + y3 + z3 – 3xyz) is (A) 108
Q.10
17 1 If x is an integer such that x = , x 4 1 then the value of x is x 13 15 1 (A) 4 (B) (C) (D) 4 4 4
(B) 207
(C) 669
(D) 729
x
5 = 1, then x is equal to 125 (A) 5 (B) 2 (C) 0 (D) 3
Q.22
If
Q.23
If 3x – 3x–1 = 18, then the value of xx is (A) 3
manishkumarphysics.in
(B) 8
(C) 27
(D) 216
Q.24
If 2x – 2x–1 = 16, then the value of x2 is (A) 4
(B) 9
(C) 16
Q.33
(D) 25
(x29 – x25 + x13 – 1) is divisible by (A) both (x – 1) & (x + 1) (B) (x – 1) but not by (x + 1)
Q.25
If x and y are non-zero rational unequal numbers,
( x y) ( x y) 2
then
1 (A) xy
Q.26
2
(D) neither (x – 1) nor (x + 1)
is equal to -
x y xy 1 4 2 (B) (C) (D) xy xy xy 2
(C) (x + 1) but not by (x – 1)
2
x y If = (b c)(b c 2a ) (c a )(c a 2b) z , the value of (x + y + z) is (a b)(a b 2c) (A) a + b + c (B) a2 + b2 + c2
Q.34
(A) –1
Q.27
ab a ab (C) a
Q.28
(B)
Q.37
The factors of (x3 – 5x2 + 8x – 4) are (A) (x + 2) (x – 2) (x – 1) (B) (x + 1) (x + 2) (x – 2) (C) (x – 2)2 (x – 1) (D) (x – 2)2 (x + 1)
Q.38
The factors of (x4 + 4) are (A) (x2 + 2)2 (B) (x2 + 2) (x2 – 2) (C) (x2 + 2x + 2) (x2 – 2x + 2) (D) None of these
Q.39
(x + y)3 – (x – y)3 can be factorized as -
(D) None of these
(A) 2 Q.29
(B) –2
(B) –7 x3
Q.32
(C) –8
(D) –10
6x2
Let f(x) = – + 11x – 6. Then, which one of the following is not factor of f(x) ? (A) x – 1
Q.31
(D) –1
If x3 + 6x2 + 4x + k is exactly divisible by (x + 2), then the value of k is (A) –6
Q.30
(C) 1
(B) x – 2 (C) x + 3
(D) x – 3
Q.40
(B) x – 2
(C) x + 2
(D) None of these
Which of the following statements are correct ?
(A) 2, 3, 4
(B) 1, 3, 4
(C) 1, 2, 4
(D) 1, 2, 3
(C) 2y (3y2 + x2)
(D) 2x (x2 + 3y2)
The factors of (x3 + 8y3) are -
(C) (x + 2y) (x – 2y)2 (D) (x + 2y)3 Q.41
2. x + 2 is a factor of x3 + 2x2 – x – 2 4. x – 2 is a factor of 2x3 – 3x + 4
(B) 2x (3x2 + y2)
(B) (x + 2y) (x2 + 2xy + 4y2)
1. x + 3 is a factor of x3 + 2x2 + 3x + 18 3. x + 1 is a factor of x3 + x2 – 4x – 4
(A) 2y(3x2 + y2)
(A) (x + 2y) (x2 – 2xy + 4y2)
The polynomial (x4 – 5x3 + 5x2 – 10x + 24) has a factor as (A) x + 4
(D) –8
27y3)
The factors of (x3 + y3 + 2x2 – 2y2) are (A) (x + y) (x2 + y2 + xy + 2x – 2y) (B) (x + y) (x2 + y2 – xy + 2x –2y) (C) (x + y) (x2 + y2 + 2x – 2y) (D) None of these
ab b
If (x – 2) is a factor of (x2 + 3qx – 2q), then the value of q is -
(C) 8
Q.36
Let
(A)
(8x3
The factors of – are 2 2 (A) (2x – 3y) (4x + 9y – 6xy) (B) (2x – 3y) (4x2 + 9y2 + 6xy) (C) (2x – 3y) (4x2 – 9y2 – 6xy) (D) (2x – 3y) (4x2 – 9y2 + 6xy)
(D) indeterminate
a b a b – = x : y. If (x – y) = , then b a b a x is equal to -
(B) 1
Q.35
=
(C) 0
Value of k for which (x – 1) is a factor of (x3 – k) is -
Q.42
If (x + 2) and (x – 1) are the factors of (x3 + 10x2 + mx + n), the values of m and n are (A) m = 5, n = – 3
(B) m =17, n = – 8
(C) m = 7, n = –18
(D) m = 23, n = – 19
On dividing (x3 – 6x + 7) by (x + 1), the remainder is (A) 2
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(B) 12
(C) 0
(D) 7
Q.43
If (x5 – 9x2 + 12x – 14) is divided by (x – 3), the remainder is (A) 184
Q.44
(B) 56
(C) 2
(D) 1
When (x4 – 3x3 + 2x2 – 5x + 7) is divided by (x – 2), the remainder is (A) 3 (B) –3 (C) 2 (D) 0 5x3
5x2
Q.45
If + – 6x + 9 is divided by (x + 3), the remainder is (A) 135 (B) –135 (C) 63 (D) –63
Q.46
If (x11 + 1) is divided by (x + 1), the remainder is (A) 0 (B) 2 (C) 11 (D) 12
Q.47
The value of expression (16x2 + 24x + 9) for 3 x=– is 4 (A) 2 (B) 1 (C) 0 (D) –1
Q.48
If 2x3 + 5x2 – 4x – 6 is divided by 2x + 1, the remainder is (A) –
Q.49
Q.50
Q.51
Q.55
13 2
(B) 3
(C) –3
When (x 3 – 2x 2 + px – q) is divided by (x2 – 2x – 3), the remainder is (x – 6). The values of p and q are (A) p = – 2, q = –6 (B) p = 2, q = – 6 (C) p = – 2, q = 6 (D) p = 2, q = 6 The sum of (x 2 + 1) and the reciprocal of (x2 – 1) is x4 (A) 2x2 (B) 2 x 1 2 x 2x 2 2x 2 (C) (D) 4 x2 1 x 1 11x2y2
y4)
Q.52
For making – + a perfect square, the expression to be added is (A) 5x2y2 (B) 9x2y2 (C) –5x2y2 (D) 7x2y2
Q.53
(x4 + 5x3 + 6x2) is equal to (A) x(x +3) (x2 + 2) (B) x2 (x + 3) (x + 2) (C) x2 (x – 2) (x – 3) (D) x(x2 + 3) (x + 2)
Q.54
If (x3/2 – xy1/2 + x1/2 y – y3/2) is divided by (x1/2 – y1/2), then the quotient is (A) x + y (B) x – y (C) x1/2 + y1/2 (D) x2 – y2
4 x4 are The factors of 4 81 x (A) not possible 2 x2 2 x2 (B) 2 x2 9 9 x 2 2 x 2 2 x 2 2 (C) 2 9 3 x 2 9 3 x 2 x2 2 (D) 2 x 9 3
2 x2 2 x2 9 3
Q.57
The factors of (x2 – 1 – 2a – a2) are (A) (x – a + 1) (x – a – 1) (B) (x + a – 1) (x – a + 1) (C) (x + a + 1) (x – a – 1) (D) None of these
Q.58
The factors of (x2 – 8x – 20) are (A) (x + 10) (x – 2) (B) (x – 10) (x + 2) (C) (x – 5) (x + 4) (D) (x + 5) (x – 4)
Q.59
The factors of (x2 – xy – 72y2) are (A) (x – 8y) (x + 9y) (B) (x – 9y) (x + 8y) (C) (x – y) (x + 72y) (D) (x – 6y) (x + 12y)
Q.60
The factors of (x2 – 11xy – 60y2) are (A) (x + 15y) (x – 4y) (B) (x – 15y) (x + 4y) (C) (15x + y) (4x – y) (D) None of these
Q.61
The factors of (x4 + x2 + 25) are (A) (x2 + 5 + 3x) (x2 + 5 – 3x) (B) (x2 + 3x + 5) (x2 + 3x – 5) (C) (x2 + x + 5) (x2 – x + 5) (D) None of these
Q.62
The factors of (x4 – 7x2y2 + y4) are (A) (x2 + y2 – 3xy) (x2 + y2 + 3xy) (B) (x2 – y2 – 3xy) (x2 – y2 + 3xy) (C) (x2 – 3xy + y2) (x2 – 3xy – y2) (D) None of these
(D) 6
If x3 + 5x2 + 10 k leaves remainder –2x when divided by x2 + 2, then the value of k is (A) –2 (B) –1 (C) 1 (D) 2
(x4
Q.56
The factors of (x4 + 625) are (A) (x2 – 25), (x2 + 25) (B) (x2 + 25), (x2 + 25) (C) (x2 – 10x + 25), (x2 + 5x + 25) (D) do not exist
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Q.63
Q.64
The factors of (x2 + 4y2 + 4y – 4xy – 2x – 8) are(A) (x – 2y – 4) (x – 2y + 2) (B) (x – y + 2) (x – 4y – 4) (C) (x + 2y – 4) (x + 2y + 2) (D) None of these
Q.65
The factors of (x3 – x2y – xy2 + y3) are (A) (x + y) (x2 + y2 – xy) (B) (x + y) (x2 + y2 + xy) (C) (x + y)2 (x – y) (D) (x – y)2 . (x + y)
Q.66
The factors of (216x3 – 64y3) are (A) 8(3x – 2y) (9x2 + 4y2 – 6xy) (B) 8(3x – 2y) (9x2 – 4y2 – 6xy) (C) 8(3x – 2y) (9x2 + 4y2) (D) 8(3x – 2y) (9x2 + 4y2 + 6xy)
The factors of (x2 + xy –2y2) are (A) (x – 2y) (x + y) (B) (x + 2y) (x – y) (C) (x – 2y) (x – y) (D) (x + 2y) (x + y)
ANSWER KEY
Q.No
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
Ans .
C
C
B
A
D
B
A
A
D
C
B
C
D
C
A
C
B
B
C
D
Q.No
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
Ans .
A
D
C
D
C
C
D
D
C
C
B
D
B
B
B
B
C
C
A
A
Q.No
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
Ans .
C
B
A
B
D
A
C
C
C
C
B
B
B
A
D
D
C
B
B
B
Q.No
61
62
63
64
65
66
Ans .
A
A
A
B
D
D
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Hints & Solution # 1 Sol.1
Required sum =
2y 4 y 2y 2 5y 2 5y 3 =2+ – + – + – 3 3 2 3 3 2 +
=2–
12 1 2 = x2yz + – 5 3 3
6 5y 3 + 3y2 + 3y + 5 3
4 6 y 2y 5y 2 2y 2 + + – + 3y – + 3 3 5 2 3 3 + 3y2 +
5y 3 5y 3 + 2 3
=
Sol.5
(i) We have, (–2x2) × (7a2x7) × (6a5x5) = (–2 × 7 × 6) × (x2 × x7 × x5 × a2 × a5) = –84x2+7+5a2+5 = –84x14a7 (ii) We have, (4s2t) × (3s3t3) × (2st4) × (–2) = (4 × 3 × 2 × – 2)×(s2× s3 × s × t × t3 × t4) = – 48s2+3+1t1+3+4 = – 48s6t8 (iii) We have, (5x6) × (–10xy4) × (–2x6y6) × (10xy) = (5 × –10 × –2 × 10) × (x6 × x × x6 × x × y4 × y6 × y) = 1000x6+1+6+1y4+6+1 = 1000x14y11
Sol.6
We have,
30 – 20 18 4 – 3 18 –529 2 + y + y = 15 6 3
Sol.2
28 19 25 3 + y + 2y2 + y 15 6 6
The required difference is given by 1 2 (7y2 – 2y + 10) – – 2 y y – 3 2 1 = 7y2 – 2y + 10 + 2y2 – y+3 2 1 = 7y2 + 2y2 – 2y – y + 10 + 3 2 [Grouping like terms]
1 = (7 + 2)y2 + – 2 – y + 13 2 5 = 9y2 – y + 13 2 Sol.3
4 3 6 2 4 6 – xy × x y = – ×(x× x2× y3 × y) 3 7 3 7
8 1 2 31 8 x y = – x 3y4 7 7 Verification: For x = 2 and y = 1, we have =–
We have,
3 2 2 3 2 4 1 2 12 2 x yz – xyz x y – x y y – x yz 5 3 5 3 5 2 =
41 2 5 2 3 4 x yz – x y – xyz – y 15 6 5 5
We know that the volume of a rectangular box is given by Volume = Length × Breadth × Height (i) Volume = 2ax × 3by × 5 cz = (2 × 3 × 5) × (ax × by × cz) = 30 abcxyz (ii) Volume = m2n × n2p × p2m = m2+1n1+2p1+2 = m3n3p 3 (iii) Volume = 2q × 4q2 × 8q3 = (2 × 4 × 8)q1+2+3 = 64q6
5 5 + y3 2 2
=
3 3 4 x2y – xyz – y 2 5 5
Sol.4
4 6 2 1 5 2 = 2 – + – 3 y + – 3 y2 3 5 3 2 3 3
15 10 3 y + 6
12 2 1 2 3 3 4 x yz + x2yz + x2y – x2y – xyz – y 5 3 3 2 5 5 [Grouping like terms]
12 2 3 2 3 4 x yz – xyz + x2y – x2y – y 5 5 3 2 5 +
1 2 x yz 3
4 3 6 2 L.H.S. = – xy × x y 3 7 4 8 24 64 3 6 2 = – 2 (1) × ( 2) 1 =– × =– 3 7 3 7 7 and,
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8 3 4 8 64 x y = – × 23 × (1)4 = – 7 7 7 Hence, for x = 2 and y = 1, we have L.H.S. = R.H.S. R.H.S. = –
Sol.7
2 2 8 2 1 × – xy z × xy z × – z 3 15 4
2 8 1 = – 5 – – × (x2 × x × x × y × y2 3 15 4 × z × z2 ) 4 4 4 4 =– x y z 9 Verification : For x = 1, y = 2 and z = 3, we have 2 L.H.S. = (– 5 × (1)2 × 2) × – 1 (2) 3 3
8 2 1 × 1 2 (3) × – 3 15 4 2 8 3 = (–5×1×2)× – 1 4 3 × 1 2 9 × – 3 15 4 48 3 = (–10) × (–8) × × – = –576 5 4
4 4 × (1)4 × 24 × 34 = × 1 × 16 × 81 9 9 = –576 L.H.S. = R.H.S. and , R.H.S. =
Sol.8
We have,
7 7 7 7 7s 2 t × (s + t) = s2t×s+ s2t×t = s3t+ s2t2 2 2 2 2 2 Verification: When s = L.H.S. =
1 and t = 5, we have 2
7 2 s t × (s + t) 2
1 1 x5 – 100 x3 = x5 – x3 = 100 100 100 (ii) We have,
b 121.5ab× ac 10 = 121.5ab × ac + 121.5ab ×
b 10
= 121.5a2bc + 12.15ab2 (iii)We have, 0.1a×(0.01a + 0.001b) = 0.1a×0.01a+0.1a×0.001b = 0.001a2 + 0.0001ab Sol.10 (i) We have, 5m (3 – m) = 5m × 3 – 5m × m = 15m – 5m2 5m (3 – m) + 6m2 – 13m = 15m – 5m2 + 6m2 – 13m = (15m – 13m) + (6m2 – 5m2) = 2m + m2 (ii) We have, 4y(3y2 + 5y – 7) = 4y × 3y2 + 4y × 5y – 4y × 7 = 12y3 + 20y2 – 28y 2(y3 – 4y2 + 5) = 2 × y3 + 2 (–4y2) + 2 × 5 = 2y3 – 8y2 + 10 Adding the two expressions
2
10
14 y 12 y – 28y 10 3
7 3 7 s t + s2t 2 2 2 3
35 350 385 = 16 16
Sol.9 (i) We have, 100x × (0.01x4 – 0.01x2) = 100x × 0.01x4 – 100x × 0.01x2 = (100×0.01)x5 – (100 × 0.01)x3
2 y 3 – 8y 2
7 1 11 35 11 385 = 5 × = × = 8 2 16 2 4 2
=
=
12 y 3 20 y 2 – 28y
7 1 2 1 = 2 2 5 × 5 2
R.H.S. =
7 1 7 1 35 175 × ×5+ × × 25 = + 2 8 2 4 16 8
L.H.S. = R.H.S.
We have, (–5x2y) 5
=
7 1 7 1 × × 5 + × × (5)2 2 2 2 2
2
Sol.11 (i) We have, Required difference = 4l (10n – 3m + 2l) – 3l (l – 4m + 5n) = (40ln – 12lm + 8l2) – (3l2 –12lm + 15ln) = 40ln – 12lm + 8l2 – 3l2 + 12lm – 15ln = 25ln + 5l2
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(ii) We have, Required difference = 4c (–a + b + c) – {3a(a + b + c) – 2b (a – b + c)} = –4ac + 4bc + 4c2 – {3a2 + 3ab + 3ac – 2ab + 2b2 – 2bc} 2 2 = –4ac + 4bc + 4c – {3a + ab + 3ac + 2b2 –2bc} = –4ac + 4bc + 4c2 – 3a2 – ab – 3ac – 2b2 + 2bc = –7ac + 6bc + 4c2 – 3a2 – ab – 2b2
1 1 x – y × (5x2 – 4y2) 5 4
1 1 x × (5x2 – 4y2) – y × (5x2 – 4y2) 5 4
=
1 1 1 1 x × 5x2 – x × 4y2 – y ×5x2 + y×4y2 5 5 4 4
4 2 5 2 = x3 – xy – x y + y3 5 4
Multiplying 2m – n by – 3m
– 10m 5n
Multiplying 2m – n by – 5
– 6m 2 3mn – 10m 5n
Adding the like terms
Sol.15 We have,
2 2 2 2 = y× 7y – y × y2 + y × 7y – y × y2 7 7 2 4 = 7y2 – y3 + 2y3 – y 7 2 4 y 7
Verification: When y = 3, we have
2 2 L.H.S. = y y × (7y – y2) 7
3x 2 y 2 2
3x 4 x 2 y 2 6x y 2 y
– 6m 2 3mn
= 7y2 + y3 –
Sol.13 Column method: We have,
2 2
– 3m – 5
2 2 = y × (7y – y2) + y × (7y – y2) 7
=
x 2y
2m – n
2 2 y y × (7y – y2) 7
Sol.12 We have,
2
Multiplying 3x 2 y 2 by x 2 4
Multiplying 3x y by 2 y 2
2
2
3x 4 7 x 2 y 2 2 y 4 Adding the like terms Horizontal Method: We have, (3x2 + y2) (x2 + 2y2) = 3x2 × (x2 + 2y2) + y2 (x2 + 2y2) = 3x2 × x2 + 3x2 × 2y2 + y2 × x2 + y2 × 2y2 = 3x4 + 6x2 y2 + x2y2 + 2y4 = 3x4 + 7x2y2 + 2y4 Sol.14 Horizontal Method: We have, {2m + (–n)} × {–3m + (–5)} = (2m – n) × (–3m – 5) = 2m × (–3m – 5) – n × (–3m – 5) = –6m2 – 10m + 3mn + 5n Column method: We have, 2m + (–n) = 2m – n and – 3m + (–5) = –3m – 5
2 2 = 3 (3) × (7 × 3 – (3)2) 7
2 = 3 9 × (21– 9) 7 18 21 18 × 12 = 3 × 12 = 7 7 =
39 468 ×12 = 7 7
2 4 R.H.S. = 7y2 + y3 – y 7 = 7 × (3)2 + (3)2 – = 7 × 9 + 27 – = 63 + 27 –
2 ×(3)4 7
2 × 81 7
162 162 630 – 162 = 90 – = 7 7 7 =
L.H.S. = R.H.S.
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468 7
Column Method: We have,
Sol.16 (i) We have,
1 (6x2 + 15y2) (6x2 – 15y2) 3
2 x 2 – 4x 5
1 2 2 = (6x 15y ) × (6x2 – 15y2) 3 By using associativity of multiplication 1 1 = 6 x 2 15 y 2 × (6x2 – 15y2) 3 3 By using distributi vity of multiplication over addtion
(2x2
5y2)
(6x2
15y2)
= + × – 2 2 2 = 2x × (6x – 15y ) + 5y2 × (6x2 – 15y2) = 2x2 × 6x2 – 2x2 × 5y2 + 15y2 × 6x2 – 5y2 × 15y2 = 12x4 – 30x2y2 + 30x2y2 – 75y4 = 12x4 – 75y4
(2x3
5x4)
5x6
(x4
2 x 4 – 4 x 3 5x 2
Multiplying 2x 2 – 4x 5 by x 2
6x – 12x 15x 3
Multiplying 2x 2 – 4x 5 by – 3x
2
– 14x 2 28x – 35 Multiplying 2x 2 – 4x 5 by – 7 2x 4 2x – 21x 2 43x – 35 Adding the like terms
Sol.18 (i) We have, (6x2 – 7y2) (6y2 –7y2) = (6x2 – 7y2)2 [ a.a = a2] 2 2 2 2 2 2 = (6x ) – 2×6x × 7y + (7y ) = 36x4 – 842y2 + 49y4
1 1 1 1 (ii) x – y x – y 5 2 5 2 2
2
1 1 1 1 x – y = x – 2 x 5 2 2 2 =
(ii) We have, 9x4(2x3 – 5x4) × 5x6 (x4 – 3x2) 9x4
x 2 3x – 7
1 1 y + y 5 5
2
1 4 xy 1 2 x y 4 5 25
Sol.19 (i) We have, 3x2)
= × – × × – 4 3 4 6 = {9x × (2x – 5x )} × {5x ×(x4 – 3x2)} By using associativity of multiplication
= (9x4 × 2x3 – 9x4 × 5x4) × (5x6 × x4 – 5x6 × 3x2) = (18x7 – 45x8) × (5x10 – 15x8) = 18x7 × (5x10 – 15x8) – 45x8(5x10 – 15x8) = 18x7 × 5x10 – 18x7 × 15x8 – 45x8 × 5x10 + 45x8 × 15x8 = 90x17 – 270x15 – 225x18 + 675x16 = –225x18 + 90x17 + 675x16 – 270x15
5 3 5 3 x y x – y 6 4 6 4 2
3 5 = x – y 4 6
2
[Using : (a + b) (a – b) = a2 – b2]
9 2 25 2 x – y 16 36 (ii) We have, =
3 3 2a 2a – b b 2
3 = (2a)2 – [Using: (a + b) (a – b) = a2 – b2] b
Sol.17 Horizontal method: We have, (2x2 – 4x + 5) × (x2 + 3x – 7) = 2x2 (x2 + 3x – 7) – 4x (x2 + 3x – 7) =
(2x4
+
6x3
–
14x2)
+
(–4x3
+ 5(x2 + 3x –7) – 12x2 + 28x) + (5x2 + 15x – 35)
= 2x4 + 6x3 – 4x3 – 14x2 – 12x2 + 5x2 + 28x + 15x – 35 = 2x4 + 2x3 – 21x2 + 43x – 35
9 b2 (iii)We have, (a2 + b2) (–a2 + b2) = (a2 + b2) {–(a2 –b2)} = –(a2 + b2) (a2 – b2) = –{(a2)2 – (b2)2} = – (a4 – b4) = – a4 + b4 = b4 – a4 = 4a2 –
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(iv) We have, (–a + c) (–a – c) = {–(a – c)} {–(a + c)} = (a – c) (a + c) = a2 – c2 Sol.20 We have,
1 1 1 1 x + x – = x2 + 2 × x × + x x x x
2
2
2
1 1 + x2 – 2 × x × + x x
2
2
2
1 1 1 1 x + x – = x2+2+ 2 + x 2 – 2 + 2 x x x x 2
2
2 1 1 x + x – = 2x2 + 2 x x x 2 2 2 1 1 1 x + x – = 2 x 2 x x x
Putting the values of x +
1 92 + x – x
1 1 and x2 + 2 ,we get x x
2
= 2 × 53 2
1 81 + x – = 106 x 2
1 x – x x–
Sol.23 (i) We have,
68 72 = 70 2 So, we express 68 and 72 in terms of 70. 68 × 72 = (70 – 2) × (70 + 2) = (70)2 – 22 = 4900 – 4 = 4896 (ii) We have, 101 99 = 100 2 So, we express 101 and 99 in terms of 100. 101 × 99 = (100 + 1) × (100 – 1) = (100)2 – 12 = 10000 – 1 = 9999 (iii) We have, 67 73 = 70 2 So, we express 67 and 73 in terms of 70. 67 × 73 = (70 – 3) × (70 + 3) = (70)2 – (3)2 = 4900–9 = 4891 (iv) We have, 1282 – 772 = (128 + 77) × (128 – 77) = 205 × 51 = 10455
1 x – = 106 – 81 x 1 x – x
Sol.22 (i) We have, (x2 + x + 1) (x2 – x + 1) {(x2 + 1) + x} {(x2 + 1) – x} = (x2 + 1)2 – x2 = x4 + 2x2 + 1 – x2 = x4 + x2 + 1 (ii) We have, (x2 + 2x + 2) (x2 – 2x + 2) = {(x2 + 2) + 2x} {(x2 + 2) – 2x} = (x2 + 2)2 – (2x)2 = x4 + 2x2 + 4 – 4x2 = x4 – 2x2 + 4
2
= 25
2
= 52
1 =±5 x
Sol.21 We have, (x + y)2 = x2 + y2 + 2xy Putting the values of x + y and xy, we otbain 122 = x2 + y2 + 2 × 14 144 = x2 + y2 + 28 144 – 28 = x2 + y2 x2 + y2 = 116
Sol.24 The numerical coefficients of the given monomials are 6, 8 and 12. The greatest common factor of 6, 8 and 12 is 2. The common literals appearing in the given monomials are a, b and c. The smallest power of ‘a’ in the three monomials = 1 The smallest power of ‘b’ in the three monomials = 2 The smallest power of ‘c’ in the three monomials = 1 The monomial of common literals with smallest powers = a1b2c1 = ab2c Hence, the greatest common factor = 2ab2c.
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Sol.25 (i) The greatest common factor of the terms 12x3 y4 , 16x2 y5 and 4x5 y2 of the expression 12x3y4 + 16x2y5 – 4x5y2 is 4x2y2. Also, we can write 12x3y4 = 4x2y2 × 3xy2, 16x2y5 = 4x2y2 × 4y3 and, 4x5y2 = 4x2y2 × x3 12x3y4 + 16x2y5 – 4x5y2 = 4x2y2 × 3xy2 + 4x2y2 × 4x2y2 × 4y3 – 4x2y2 × x3 4x2y2 (3xy2 + 4y3 – x3) (ii) We have, 18a3b2 + 36ab4 – 24a2b3 The greatest common factor of the terms 18a3b2, 36ab4 and 24a2b2 is 6ab2. Also, we can write 18a3b2 = 6ab2 × 3a2, 36ab4 = 6ab2 × 6b2 and,24a2b2 = 6ab2 × 4ab 18a3b2 + 36ab4 – 24a2b3 = 6ab2 × 3a2 + 6ab2 × 6b2 – 6ab2 × 4ab = 6ab2(3a2 + 6b2 – 4ab) Sol.26 We have, (i) (x + y) (2x + 3y) – (x + y) (x + 1) = (x + y) (2x + 3y) – (x + 1) [Taking (x + y) common] = (x + y) (2x + 3y – x – 1) = (x + y) (x + 3y – 1) = (x + y) (x + 3y – 1) (ii) (x + y) (2a + b) – (3x – 2y) (2a + b) = {(x + y) – (3x – 2y)} (2a + b) [Taking (2a + b) common] = (x + y – 3x + 2y) (2a + b) = (–2x + 3y) (2a + b) Sol.27 (i) We observe that there is no common factor among all terms. Also, there are four terms. So, let us think of grouping the terms in pairs in such a way that there are some factors common to them and after taking factors common from each pair same binomial is left inside thetwo brackets. We observe that first two terms have x as a common factor . Taking x common from them, we have, x2 + xy = x(x + y) Also, 8 is a common factor from the last two terms. Taking 8 common from the last two terms, we have 8x + 8y = 8(x + y) Clearly, x + y is common from the two groups. Thus, we group the terms as follows: x2 + xy + 8x + 8y = (x2 + xy) + (8x + 8y) (ii) We have, 15xy – 6x + 10y – 4 Clearly, there is no common factor among all the
terms. Also, there are four terms. So, let us think of grouping the terms in pairs in such a way that there are some factors common to the terms in each pair and after taking factors common from each pair same binomial is left inside the two brackets. We observe that first two terms have 3x as a common factor. Taking 3x common from them, we have 15xy – 6x = 3x (5y – 2) Last two terms have 2 as the common factor. Taking 2 common from these two, we have 10y – 4 = 2 (5y – 2) Clearly, (5y – 2) is the binomial common from these two groups. Thus, we group the terms as follows: 15xy – 6x + 10y – 4 = 3x(5y – 2) + 2 (5y – 2) = (3x + 2) (5y – 2) (iii) We have, n – 7 + 7lm – lmn = (n – 7) + (7lm – ln) = (n – 7) + (7 – n) lm = (n – 7) – (n – 7)lm = (n – 7) (1 – lm) Sol.28 We have, (i) a2 + 2a + ab + 2b = (a2 + 2a) + (ab + 2b) [Grouping the terms] = a(a + 2) + (a + 2)b = (a + 2) (a + b) = (a + 2) (a + b) [Taking (a + 2) common] 2 (ii) x – xz + xy – yz = (x2 – xz) + (xy – yz) [Grouping the terms] = x(x – z) + y (x – z) = (x + y) (x – z) [Taking (x – z) common] Sol.29 (i) We have, a2 – b + ab – a = a2 + ab – b – a = (a2 + ab) – (b + a) = a(a + b) – (a + b) = (a + b) (a – 1) (ii) xy – ab + bx – ay = xy + bx – ab – ay = x(y + b) – a (b + y) = x (y + b) – a(y + b) = (y + b) (x – a) (iii)6ab – b2 + 12ac – 2bc = 6ab + 12ac – b2 – 2bc = 6a(b + 2c) – b (b + 2c) = (b + 2c) (6a – b) (iv) a(a + b – c) – bc = a2 + ab – ac – bc = (a2 + ab) – (ac + bc) = a(a + b) – c(a + b) = (a + b) (a – c)
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(v) a2x2 + (ax2 + 1)x + a = a2x2 + ax3 + x + a = ax2 (x + a) + (x + a) = (x + a) (ax2 + 1) (vi) 3ax – 6ay – 8by + 4bx = 3ax + 4bx – 6ay – 8by = (3ax + 4bx) – (6ay + 8by) = x(3a + 4b) – 2y (3a + 4b) = (3a + 4b) (x – 2y) Sol.30 (i) We have, x3 – 2x2y + 3xy2 – 6y3 = (x3 – 2x2y) + (3xy2 – 6y3) = x2 (x – 2y) + 3y2 (x – 2y) = (x – 2y) (x2 + 3y2) (ii) We have, 6ab – b2 + 12ac – 2bc = b(6a – b) + 2c(6a – b) = (6a – b)(b + 2c) Sol.31 We have, (i) x4 – y4 = (x2)2 – (y2)2 = (x2 – y2) (x2 + y2) [Using : a2 – b2 = (a – b) (a + b)] = (x – y)(x + y) (x2 + y2) [Using : a2 – b2 = (a – b) (a + b)] 4 (ii) 16x – 81 = (4x2)2 – (9)2 = (4x2 – 9) (4x2 + 9) [Using : a2 – b2 = (a – b) (a + b)] = {(2x)2 – (3)2}+ (4x2 + 9) = (2x – 3) (2x + 3) (4x2 + 9) [Using : a2 – b2 = (a – b) (a + b)] (iii) x4 – (y + z)4 = (x2)2 – {(y + z)2}2 = {x2 – (y + z)2} {x2 + (y + z)2} = {x – (y + z)}{x + (y + z)} {x2 + (y + z)2} = (x – y – z) (x + y + z) {x2 + (y + z)2} (iv) 2x – 32x5 = 2x (1 – 16x4) = 2x {12 – (4x2)2} = 2x (1 + 4x2) (1 – 4x2) = 2x (1 + 4x2) {1 – (2x)2} = 2x (1 + 4x2) (1 – 2x) (1 + 2x) (v) 3a4 – 48b4 = 3(a4 – 16b4) = 3{(a2)2 – (4b2)2} = 3(a2 – 4b2) (a2 + 4b2) = 3{a2 – (2b)2} (a2 + 4b2) = 3(a – 2b) (a + 2b) (a2 + 4b2) (vi) 81x4 – 121x2 = x2 (81x2 – 121) = x2 {(9x)2 – (11)2} = x2 (9x – 11) (9x + 11) Sol.32 (i) 16(2x – 1)2 – 25z2 = {4 (2x – 1)}2 – (5z)2 = {4 (2x – 1) – 5z} {4 (2x – 1) + 5z} = (8x – 4 – 5z) (8x – 4 + 5z) = (8x – 5z – 4) (8x + 5z – 4)
(ii) 4a2 – 9b2 – 2a – 3b = {(2a)2 – (3b)2} – (2a + 3b) = (2a – 3b) (2a + 3b) – (2a + 3b) = (2a + 3b) {(2a – 3b) – 1} = (2a + 3b) (2a – 3b – 1) (iii)x2 – 4x + 4y – y2 = (x2 – y2) – (4x – 4y) = (x – y) (x + y) – 4 (x – y) = (x – y) {(x + y – 4)} = (x – y) (x + y – 4) (vi) 3 – 12 (a – b)2 = 3 {1 – 4 (a – b)2} = 3[12 – {2(a – b)}2] = 3[{1 + 2 (a – b)}{1 – 2 (a – b)}] = 3 (1 + 2a – 2b) (1 – 2a + 2b) (v) x(x + z) – y (y + z) = x2 + xz– y2 – yz = (x2 – y2) + (xz – yz) = (x – y) (x + y) + z (x – y) = (x – y) {(x + y) + z} = (x – y) (x + y + z) (vi) a2 – b2 – a – b = (a2 – b2) – (a + b) = (a – b) (a + b) – (a + b) = (a + b) {(a – b) – 1} Sol.33 We have, (i) 4x2 – 4xy + y2 – 9z2 = (4x2 – 4xy + y2) – 9z2 = {(2x2) – 2 × 2x × y + y2} – (3z)2 = (2x – y)2 – (3z)2 = (2x – y + 3z) (2x – y – 3z) (ii) 16 – x2 – 2xy – y2 = 16 – (x2 + 2xy + y2) = 42 – (x + y)2 = {4 + (x + y)}{4 – (x + y)} = (4 + x + y) (4 – x – y) (iii)x4 – (x – z)4 = (x2)2 – {(x – z)2}2 = {x2 + (x – z)2}{x2 – (x – z)2} = (x2 + x2 – 2xz + z2) [{x + (x – z)}{x – (x – z)}] = (2x2 – 2xz + z2) (x + x – z) (x – x + z) = (2x2 – 2xz + z2) (2x – z)z Sol.34 (i) 4(x + y)2 – 28y (x + y) + 49y2 = {2(x + y)}2 – 2 × 2 (x + y) × 7y + (7y)2 = {2(x + y) – 7y}2 = (2x + 2y – 7y)2 = (2x – 5y)2 (ii) (2a + 3b)2 + 2 (2a + 3b) (2a – 3b) + (2a – 3b)2 = [(2a + 3b) + (2a – 3b)]2 = (4a)2 = 16a2 Sol.35 (i) 9x2 – 4y2 = (3x)2 – (2y)2 = (3x + 2y) (3x – 2y) (ii) 36x2 – 12x + 1 – 25y2 = (6x)2 – 2 × 6x × 1 + 12 – (5y)2 = (6x – 1)2 – (5y)2 = {(6x – 1) – 5y}{(6x – 1) + 5y} = (6x – 1 – 5y) (6x – 1 + 5y) = (6x – 5y – 1) (6x + 5y – 1)
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(iii) a2 – 1 + 2x – x2 = a2 – (1 – 2x + x2) = a2 – (12 – 2 × 1 × x + (x)2) = a2 – (1 – x)2 = {a – (1 – x)} {a + (1 – x)} = (a – 1 + x) (a + 1 – x)
(ii) We have,
Sol.36 We have, (i) 9 – a6 + 2a3b3 – b6 = 9 – (a6 – 2a3b3 + b6) = 32 – {(a3)2 – 2 × a3 × b3 + (b3)2}2 = {3 + (a3 – b3)} {3 – (a3 – b3)} = (3 + a3 – b3) (3 – a3 + b3) = (a3 – b3 + 3) (–a3 + b3 + 3) 16 16 8 (ii) x – y + x + y8 = {(x8)2 – (y8)2} +(x8 + y8) = (x8 – y8) (x8 + y8) + (x8 + y8) = (x8 + y8) (x8 – y8 + 1) (iii)(p + q)2 – (a – b)2 + p + q – a + b = {(p + q)2 – (a – b)2} + (p + q) – (a – b) = {(p + q) + (a – b)} {(p + q) – (a – b)} + {(p + q) – (a – b)} = (p + q + a – b) (p + q – a + b) + (p + q – a + b) = (p + q – a + b) (p + q + a – b + 1) Sol.37 The given expression is (2x + 3y)2 – 5(2x + 3y) – 14 Let 2x + 3y = a. Then, (2x + 3y)2 – 5 (2x + 3y) – 14 = a2 – 5a – 14 = a2 – 7a + 2a – 14 = a (a – 7) + 2 (a – 7) = (a – 7) (a + 2) = (2x + 3y – 7) (2x + 3y + 2) Sol.38 We have, 3m2 + 24m + 36 = 3(m2 + 8m + 12) [Making coefficient of m2 as 1] 2 = 3{m + 8m + 42 – 42 + 12} 2 8 Adding and subtracting 4 2 2
= = = = =
3 {m2 + 2 × m × 4 + 42 – 4} 3{(m + 4)2 – 22} [Completing the square] 3{(m + 4) – 2}{(m + 4) + 2} 3(m + 4 – 2) (m + 4 + 2) 3(m + 2) (m + 6)
Sol.39 (i) We have,
6 x yz – 3xy z 8x yz 2xyz 4
3
2
4
4
=
3
6 x yz 3xy z – 2 xyz 2 xyz
2 2 2 2 4 2 3 1 3 2 a b c ab c – ab c 3 3 5 1 abc 2 2 2 2 2 4 2 3 1 3 2 a b c ab c ab c 3 3 = + – 5 1 1 1 abc abc abc 2 2 2 4 8 2 = abc + bc – b c 3 3 5 Sol.40 We have 2x3 + 2x2 + x + 1 x+3
4
3
2
2x + 8x + 7x + 4x + 3 2x4 + 6x3 – – 3 2 2x + 7x + 4x + 3 3 2 2x + 6x – – 2 x + 4x + 3 2 x + 3x – – x +3 x +3 – – 0
4 3 First term of quotient 2x = 2x x
New dividend is 2x 3 + 7x2 + 4x + 3 3 2x 2 Second term of quotient = = 2x x 2 New dividend is x + 4x + 3 2 Third term of quotient = x = x x New dividend is x + 3 x Fourth term of quotient = x = 1
As the remainder is zero, we can say that x + 3 is a factor of the given polynomial 2x4 + 8x3 + 7x2 + 4x + 3. 2x4 + 8x3 + 7x2 + 4x + 3 = (x + 3) (2x3 + 2x2 + x + 1) Sol.41 On dividing, we get
5x2 –9x + 3 4
3 2 2x2 + 7x – 1 10x + 17x – 62x + 30x – 3 4 3 2 10x + 35x – 5x – – + 3 2 – 18x – 57x + 30x – 3 3 2 –18x – 63x + 9x + + –
6x2 + 21x – 3 6x2 + 21x – 3 – – + 0 10x4 + 17x3 – 62x2 + 30x – 3 = (2x2 + 7x – 1) (5x2 – 9x + 3)
8x 2 yz 4 3 2 + = 3x3 – y + 4xz3 2 xyz 2
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Sol.42 On dividing, we get
9 2 5 11 = (4y + 2) 2 y – y – 2 4 2
y2 + 2y + 2 5 4 3 2 3 3y + 1 3y + 6y + 6y + 7y + 8y + 9 5 2 3y +y – – 4 3 2 6y + 6y + 6y + 8y + 9 4 6y + 2y – –
Hence, Quotient = 2y 2 – Remainder = –
6y3 + 6y2 + 6y + 9 6y3 +2 – – 2 6y + 6y + 7 Clearly, the degree of the remainder 6y2 + 6y + 7 is 2, which is less than the degree of the divisor 3y3 + 1. So, division process stops at this and we have, Quotient = y2 + 2y + 2 and, Remainder = 6y2 + 6y + 7 Verification: We have, Divisor × Quotient + Remainder = (3y3 + 1) (y2 + 2y + 2) + 6y2 + 6y + 7 = 3y3(y2 + 2y + 2) + 1 (y2 + 2y + 2) + 6y2 + 6y +7 = 3y5 + 6y4 + 6y3 + y2 + 2y + 2 + 6y2 + 6y + 7 = 3y5 + 6y4 + 6y3 + 7y2 + 8y + 9 = Dividend Sol.43 We have, 16x4 + 12x3 – 10x2 + 8x + 20 = 4x3 (4x – 3) + 6x2 (4x – 3) + 2x (4x – 3)
7 61 + (4x – 3) + 2 2 7 3 61 2 = (4x – 3) 4x 6x 2 x + 2 2 Hence, Quotient = 4x3 + 6x2 + 2x + Remainder =
11 . 2
Sol.45 We have, a4 – b4 = (a2)2 – (b2)2 a4 – b4 = (a2 – b2) (a2 + b2) [Using: x2 – y2 = (x + y) (x – y)] 4 4 a – b = (a – b) (a + b) (a2 + b2) ...(i) (a4 – b4) (a – b) =
a 4 – b4 a–b
=
(a – b)(a b)(a 2 b 2 ) (a – b)
Sol.46 We have, x4a + x2ay2b + y4b = (x2a)2 + x2ay2b + (y2b)2 =(x2a)2 + 2x2ay2b + (y2b)2 – x2ay2b = (x2a + y2b)2 – (xayb)2 = (x2a + y2b + xayb) (x2a + y2b – xayb) = (x2a + xaya + y2b) (x2a – xayb + y2b)
x 4a x 2a y 2 b y 4 b x 2a x a y b y 2 b ( x 2a x a y b y 2 b )( x 2 a – x a y b y 2 b )
=
7 and, 2
[Using (i)]
= (a + b)(a2 + b2) [Cancelling common factor (a – b) in Nr and Dr]
( x 2a x a y b y 2 b )
= x2a – xayb + y2b
61 2
Sol.44 We have, 8y3 – 6y2 + 4y – 1 = 2y2 (4y + 2) –
5 9 y + and, 2 4
5 9 11 y (4y + 2) + (4y + 2) – 2 4 2
5 9 2 11 = 2 y (4 y 2) – y(4 y 2) (4 y 2) – 2 4 2
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Hints & Solution # 2 18.[B] Using a + b+ c = 0 a3 + b3 + c3 = 3abc, we get :
x1/3
+
y1/3
+
z1/3
29.[C] Since (x + 2) is a factor, so x = –2 will make it zero. (–2)3 + 6 (–2)2 + 4 (–2) + k = 0
=0
x + y + z = 3x1/3 y1/3 z1/3
or k = – 8.
(x + y + z)3 = 27xyz . 43.[A] Let f(x) = x5 – 9x2 + 12x – 14. This when divided by (x – 3) gives remainder
19.[C] a + b + c = 0 a3 + b3 + c3 = 3abc.
f(3) = 35 – 9 × 32 + 12 × 3 – 14 = 184. 20.[D] a + b + c = 0 a3 + b3 + c3 = 3abc;
a3 b3 c3 + + =3 abc abc abc
or
a2 b2 c2 + + = 3. bc ac ab
44.[B] Remainder is f(2) = 24 – 3 × 23 + 2 × 22 – 5 × 2 + 7 = – 3. 49.[C] On actually dividing x3 + 5x2 + 10k by (x2 + 2), the remainder obtained is – 2x + 10k – 10. –2x + 10k – 10 = – 2x
21.[A] (x3 + y3 + z3 – 3xyz)
or k = 1.
= (x + y + z) (x2 + y2 + z2 – xy – yz – zx) 2
= (x + y + z) [(x + y + z)2 – 3 (xy + yz + zx)] 56.[D]
= 9 × (81 – 3 × 23) = (2 × 12) = 108.
2 2 2 x4 2 x +2× 2 × x – 4 + = + 9 9 81 x 2 9 x4 x2
4
2
x a b y = ab 2
27.[D]
a b ab xy = 2 xy a b 2 ab 2
2
Also, (x – y) =
2
a 2 b2 ab
....(i)
(a b 2 ab) (a 2 b 2 ) . ab (a 2 b 2 ab) Adding (i) and (ii) we get : x+y=
x=
=
1 2
2 2 x2 2 = 2 – 9 3 x 2 x2 2 2 x2 2 = 2 9 3 2 9 3 x x
....(ii)
a 2 b 2 a 2 b 2 ab 1 ab . a 2 b 2 ab (a 4 b 4 )
(a b 2 ab)
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