1 4
Number and algebra
Polynomials Optional Stage 5.3 topic We have analysed and graphed linear equations y ¼ mx þ b , quadratic equations y ¼ ax þ bx þ c and simple cubic equations y ¼ ax þ c . In this topic we will look at 2
3
x, their graphs and the equations involving higher powers of x, methods for factorising them and sketching their graphs.
NEW CENTURY MATHS ADVANCED for the
A u s tr a l i an C u r ri c u l um
10 10A
þ
d a u h a s / m o c . k c o t s r e t t u h S
n Chapter outline
n Wordbank
14-01 Polynomials* 14-02 Adding and subtracting polynomials* 14-03 Multiplying polynomials* 14-04 Dividing polynomials* 14-05 The remainder theorem* 14-06 The factor theorem* 14-07 The cubic curve y a (x r )( )(x s )( )(x t )* )* 14-08 Graphing polynomials* 14-09 Transforming graphs of polynomials* ¼
*STAGE 5.3
9780170194662
Proficiency strands U R C U F R C U U U U U
F F F F F
R R R R R
C C C C C
U U
F PS R F PS R
C C
degree of a polynomial The highest power in a polynomial. For example, the degree of 8 x 3 4 x 7 is 3.
þ
polynomial An algebraic expression involving powers x that are positive integers. For example, of x P ( x) 8 x 3 4 x 7.
¼
þ
quotient The The ‘whole’ part of the answer when a polynomial is divided by another polynomial remainder The ‘left-over’ part of the answer when a polynomial is divided by another polynomial root of an equation A value of x x that is a solution to the P ( x) 0 is the same as a zero equation P ( x) 0. A root of P P ( x). of P
¼
¼
zero of a polynomial A value of x that makes the value of the polynomial P ( x) equal to 0
Chapter 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 Polynomials
n In this chapter you will: • (STAGE (STAGE 5.3) investigate investigate the concept concept of a polynomia polynomiall and apply the factor factor and remainder remainder theorems to solve problems • (STAG (STAGE E 5.3) apply understan understanding ding of polynomia polynomials ls to sketch a range of curves curves and describe describe the features of these curves from their equation • (STAG (STAGE E 5.3) recognis recognisee a polynomial polynomial and and use polynomi polynomial al notation notation • (STAG (STAGE E 5.3) 5.3) add add and and subtract subtract poly polynomia nomials ls • (STAG (STAGE E 5.3) multiply multiply and and divide divide a polynomial polynomial by a linear linear polynomi polynomial al • (STAG (STAGE E 5.3) understand understand and and apply apply the remainder remainder and factor factor theorems theorems • (STAG (STAGE E 5.3) factorise factorise polynomi polynomials als and solve solve polynomia polynomiall equations equations • (STAGE 5.3) graph cubic equations equations of the form y a ( x r )( )( x s )( x t ) • (STAG (STAGE E 5.3) graph graph quadratic, quadratic, cubic cubic and and quartic quartic polynomial polynomialss • (STAG (STAGE E 5.3) graph graph transfo transformation rmationss of a basic basic polynom polynomial ial y P ( x)
¼
¼
SkillCheck 1
If x x a
2
¼ 2, evaluate each expression. x þ 2 x x 1 b 2 x þ 3 x 11 3
2
b x3 f x 2
16 2 x 15
Solve each equation.
b x2 e x2
þ ¼ þ þ ¼
Worksheet StartUp assignment 14 (Advanced) MAT10NAWK10228
c 3 x 2 g 2 x 2
16 x þ 5 x 24
a (2 x 5)( x 2) 0 d x 2 6 x 5 0
Stage 5.3
c x4
Factorise each expression. a x2 e x2
3
2
d 3 x 3 27 x h x 3 3 x 2 70 x
27 þ x 10
10 x ¼ 0 2 x 120 ¼ 0
3
x þ 2 x
c 5 x 2 f 2 x 2
3 x ¼ 0 þ 7 x þ 6 ¼ 0
14-01 Polynomials A polynomial is an algebraic expression involving powers of x that are positive integers, for polynomial is 3 example, 8 x 4 x 7. It is written using the notation P ( x), meaning ‘a polynomial using the variable x ’.
þ
Summary A polynomial has the general form polynomial has P ( x)
1
2
2
¼ a x þ a x þ a x þ … þ a x þ a x þ a n
n
n
1
n
n
n
2
2
1
0
x are positive integers, n where the powers of x and a , a 1, a 2, …, a 1, a 0 are called coefficients called coefficients.. n
538
n
n
9780170194662
NEW CENTURY MATHS ADVANCED for the
A u s tr a l i an C u r ri c u l um
10 10A
þ
The leading term of The leading term of a polynomial is the term with the highest power. For example, the leading term of the polynomial P ( x) 2 x 5 8 x 3 7 is 2 x 5. The leading The of a polynomial is the coefficient of the leading term. For P ( x) 2 x 5 8 x 3 7, leading coefficient of the leading coefficient is 2. The highest power of the polynomial is the degree of the polynomial. For P ( x) 2 x 5 8 x 3 7, degree of the degree is 5. A monic polynomial is polynomial is a polynomial that has a leading coefficient of 1. For example, 4 7 x 2 x 8 is monic. P ( x) x The constant The constant term of term of a polynomial is the term ‘at the end’ of the polynomial that is independent of x x . For P ( x) x 4 7 x 2 x 8, the constant term is 8.
¼
þ
¼
¼
¼
þ
þ
Stage 5.3
þ þ
¼
þ þ
Names of polynomials Linear polynomials have polynomials have degree 1 and their graph is a straight line, for example, P ( x) 2 x 7. Quadratic polynomials have polynomials have degree 2 and their graph is a parabola, for example, P ( x) 3 x 2 2 x 5. Cubic polynomials have polynomials have degree 3, for example P ( x) 6 x 3 7 x 2 9 x 10. 2 x 4 x 3 8 x 2 4 x 13. Quartic polynomials have polynomials have degree 4, for example P ( x)
¼ þ ¼ þ þ
Example
¼ ¼ þ
1
Determine which of these expressions represents a polynomial. 3 A x C x 2x4 7 x 5 2 x x3 7 2
ð Þ ¼ p ffiffiffi ffi B( x) ¼ 5 x þ 2
ð Þ ¼ p ffiffiffi ffi þ þ D( x) ¼ ( x 2)( x þ 5)
þ x
2
Solution
ð Þ ¼ p ffiffiffi ffi
Since A ( x) can be written as A x 2x4 7 x 5 3 x2 , it is not a a polynomial because 2 powers of x has a negative power. x must be positive integers and 3 x • B( x) 5 x 2 is a polynomial. 1 1 • Si Sinc nce, e, C x 2 x2 x3 7, it is not a a polynomial because 2 x2 has a fractional power. x2 2 x 5 x 3 5 x2 2 x 10, which is a polynomial. • D x •
¼ þ ðÞ¼ þ þ ð Þ ¼ ð Þð þ Þ ¼ þ
Example
þ
2
For the polynomial P ( x)
6
3
¼ 7 x þ 5 x þ 9 x 2, state:
a the degree constant term d the constant
leading term leading coefficient b the leading c the leading e whether the polynomial is monic.
Solution a b c d e
The degree degree is 6. The leading leading term term is 7 x 6. The leading leading coefficient coefficient is 7. The constant term is 2. Since the leading leading coefficient is not 1, the polynomial is not monic.
9780170194662
The highest power of the polynomial is 6. The term with the highest power The term independent of x. x.
539
Chapter 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 Polynomials
Stage 5.3
P ( x )
notation
To find the value of a polynomial P ( x) for x x in the polynomial.
Example
¼ c, we write P (c). This means that c is substituted for
3 4
For the polynomial P ( x) (3) a P (3)
b
2
¼ 3 x 6 x þ 4, find: (0) P (1) c P (0) P (2)
Solution
ð Þ ¼ 3ð34Þ 6ð32 Þ þ 4 ¼ 193 P ð0Þ ¼ 3 ð04 Þ 6ð02 Þ þ 4 ¼ 4 ð1Þ ¼ 3ð1Þ4 6ð1Þ2 þ 4 P ð ¼ 1 ð1Þ ¼ 4 1 P ð0Þ P ð ¼ 3
ð Þ ¼ 3ð2Þ4 6ð2Þ2 þ 4 ð ¼ 28
a P 3 c
b P 2
)
Exercise 14-01 Polynomials Example 1 See Example
1
Determine whether each each expression is a polynomial. polynomial. If it is a polynomial, state whether whether it is monic. 5 2 x 7 a 9 x 2 5 x b 2 c 3 x 5 x 6 3 d g j
Example 2 See Example
2
x 10 p ffiffi 5ffi ffi x2 2 p ffiffiffiffiffi 10x4 þ x3 x 3 þ x3 3 4
6 x
2
5
i
2
b 7 x 4
l
5
e
2
4
3
þ p ffiffiffi ffi x þ x 11 (20 x)(20 x þ x 9 x3 þ x2 þ x 3
2
2
)
x
iii the constant term. c 11 x 2
6 x þ 3 1 8 x x þ 7 x þ 3 2
þ p ffiffi2ffi ffi x 11 þ x h 22 54 x P ( x) ¼ 3 x 2 x þ x 1, evaluate: If P (0) (2) a P (0) b P (2) c P (1) If P P ( x) ¼ x x and Q ( x) ¼ 1 x , find: (0) þ Q (0) (2) þ Q (1) a P (0) b P (2) 1 1 p ffiffiffi ffi p ffiffiffi ffi þ Q e P f P 2 3 þ Q 3
5
10
f 9
6
i
x3
2
x þ þ x 3 2
2
2
2
540
3
2 x þ 5 x þ 1
g 4 x 3
4
k
4
f
For each polynomial, state: leading coefficient i the degree ii the leading
d
3
h
x
a 9 x 5
Example 3 See Example
e
þ x þ x p ffiffiffi ffi 4 x þ 2 x p ffiffiffi ffi 3 x 2 x þ x ( x 2) ( x þ 3)
p ffiffi2ffi ffi
1
d P
e P
2
3
4
(4) c P (4)
þ Q(1)
g P [Q(2)]
d P ( 2)
Q(2)
(2)] h Q[ P (2)] 9780170194662
NEW CENTURY MATHS ADVANCED for the
A u s tr a l i an C u r ri c u l um
10 10A
14-02 Adding and subtracting polynomials Example If P P ( x) a
þ
Stage 5.3
4 3
2
2
¼ x þ 3 x 2 x 5 and Q( x) ¼ 3 x 5 x þ 7, simplify each expression. b P ( x) Q ( x) P ( x) þ Q ( x)
Solution
ð Þ þ Qð xÞ ¼ x 3 þ 3 x2 2 x 5 þ ð3 x2 5 x þ 7Þ ¼ x3 þ 6 x2 7 x þ 2 P ð xÞ Qð xÞ ¼ x 3 þ 3 x2 2 x 5 ð3 x2 5 x þ 7Þ ¼ x3 þ 3 x 12
a P x
b
Exercise 14-02 Adding and subtracting subtracting polynomial polynomialss 1
Example 4 See Example
Simplify each expression. (9 x 3 x 2 x ) (7 x 2 5 x 2) (3 x 2) ( x 3 x ) (8 x 2) (11 x 4 x 2) 2( x 4 x 2 2) 3( x 3 x 2 x 4) x 4 ( x 3 x 2 5 x) f 2(5 x 4 x 3 2 x) 2(11 2 x 2 x 5) g (7 x 6 x 5) ( x 2 x 3) (2 3 x) h 8 x 4 x 3 6 x 2 5 x ( x 6) ( x 3 9 x 2 1) i x 4 8 x 3 2 x 2 7 x 2 (6 x 3 x 2 5 x 4) j 6 x 3 2 x 2 x (2 x 3) (3 x 2) a b c d e
2 3
4
þ þ þ þ þ þ þ þ þ þ þ þ þ þ þ þ þ þ þ þ þ þ þ þ þ P ( x) ¼ 4 x 3 and Q ( x) ¼ x þ 7 x þ 2, simplify each expression. If P ( x) a P ( x) þ Q ( x) b P ( x) Q ( x) c Q( x) P ( d 3 P ( x) þ 2 Q( x) If A A ( x) ¼ x 6, B ( x) ¼ 11 x 2 x and C ( x) ¼ 9 4 x, simplify each expression. a C ( x) B ( x) b A( x) C ( x) c B( x) A ( x) d A( x) þ B ( x) þ C ( x) e A( x) B ( x) þ C ( x) f A( x) [ B( x) þ C ( x)] R ( x) ¼ x 6 x þ 5, P ( x) ¼ x 1 and Q ( x) ¼ R ( x) P ( x), simplify each expression. If R p ffiffiffi ffi 2
2
2
2
a Q( x) x if Q Q ( x) c the values of x
9780170194662
b Q 3 2
¼ 0
541
Chapter 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 Polynomials
Stage 5.3
14-03 Multiplying polynomials Example If P P ( x)
5 2
¼ 3 x þ 1 and Q( x) ¼ x 5 x 2, find P ( x) Æ Q( x).
Solution
ð Þ ð Þ ¼ ð3 x þ 1Þð x2 5 x 2Þ ¼ 3 xð x2 5 x 2Þ þ 1ð x2 5 x 2Þ ¼ 3 x3 15 x2 6 x þ x2 5 x 2 ¼ 3 x3 14 x2 11 x 2
P x Q x
Multiplying g polynomials Exercise 14-03 Multiplyin Example 5 See Example
Expand each product.
1
3
a (3 x
2
þ 2)( x x) b (11 x þ 4 x )(8 x 2) )(7 7 x þ x þ x þ x ) d (6 x 2 x x 3)(3 x þ 2) e (2 3 x)( P ( x) ¼ 4 x 3 and Q ( x) ¼ x þ 7 x þ 2, find P ( x) Æ Q ( x). If P If A A ( x) ¼ x 6, B ( x) ¼ 11 x 2 x and C ( x) ¼ 9 4 x, find: 3
2
6
2
3
2
f
4
3
2
2
a A( x) Æ B ( x)
Finding the quotient
3
2
þ x þ x)(5 x 2) ( x x þ 6)(8 x x þ 6 x þ 5 x 2)
2
2
Puzzle sheet
5
c (9 x 3
b A( x) Æ C ( x)
c B( x) Æ C ( x)
14-04 Dividing polynomials
MAT10NAPS00052
Long division The long division process for 9947 4 65 is shown on the right. • 65 int into o 99 is 1, rem remain ainder der 34 • Bri Bring ng dow down n the the 4 from from 994 9947 7 • 65 into into 344 goe goess 5, rem remain ainder der 19 • Bri Bring ng dow down n the the 7 from from 994 9947 7 • 65 into into 197 goe goess 3, rem remain ainder der 2 )
9947 4 65
)
153 65 9947 – 65 344 – 325
¼ 153 remainder 2 ¼ 153 652 9947 ¼ 65 153 þ 2 ;
197 195 2
3
9947 is the dividend the dividend,, 65 is the divisor the divisor,, 153 is the quotient the quotient , 2 is the remainder the remainder..
542
9780170194662
NEW CENTURY MATHS ADVANCED for the
A u s tr a l i an C u r ri c u l um
The long division process can also be used to divide polynomials. For example, ( x 3 5 x 2 6 x 4) 4 ( x 3) is
þ
þ
( x 3) into ( x 3 5 x 2) is x 2, remainder 2 x 2 Brin Br ingg do down wn th thee ( 6 x) from the dividend ( x 3) into (2 x 2 6 x) goes 2 x, remainder 12 x Brin Br ingg do down wn th thee ( 4) from the dividend ( x 3) into ( 12 x 4) goes ( 12), remainder 32
• • • • •
3
þ
10 10A
þ
x2 + 2x – 12 x + 3 x3 + 5x2 – 6x – 4 x3 + 3x2 x 2 2x2 – 6x 2x2 + 6x 2 x – 12x – 4 – 12x – 36
Stage 5.3
þ x x ¼ þ x ¼ 2 x þ 12 x x ¼ 12 32 ( x þ 5 x 6 x 4) ( x þ 3) ¼ x þ 2 x 12, remainder 32 ( x þ 5 x 6 x 4) ¼ ( x þ 3)( x þ 2 x 12) þ 32 ( x þ 5 x 6 x 4) is the dividend the dividend,, ( x þ 3) is the divisor the divisor,, ( x þ 2 x 12) is the quotient the quotient , 32 is [ [
3
2
3
2
3
4
2
4
4
2
4
2
2
2
the remainder.. the remainder Note that we can express the dividend P ( x) as the product of its factors plus the remainder: P ( x) divisor 3 quotient remainder.
¼
þ
Example
6
Divide P ( x) x 3 x 4 by A form m P ( x) A( x) Æ Q( x) R( x), A( x) x 3. Then write P ( x) in the for where Q( x) is the quotient quotient and and R( x) is the remain remainder. der.
¼ þ
¼
¼
þ
Solution x
3Þ x3 þ x 3
x2 0 x2 3 x2 3 x2 3 x2
3 x þ 8 x þ 4 x 9 x 8 x þ 4 8 x 24
Write 0 x2 as there is no x2 term
28
Example
)
x3
x þ 4 ¼ ð x 3Þð x2 þ 3 x þ 8Þ þ 28
7 3
Show that (2 x
2
þ 5) is a factor of 2 x þ 5 x 2 x 30.
Solution If 2 x 3
2
þ 5 x 12 x 30 ¼ (2 x þ 5) Æ Q( x) with no remainder, then (2 x þ 5) is a factor. 6 x2 2 x þ 5 Þ 2 x3 þ 5 x2 12 x 30 2 x3 þ 5 x2 þ 0 x2 12 x 30 12 x 30 0
Since the remainder is 0, this means that 2 x 3 5 x 2 [ (2 x 5) is a factor of 2 x 3 5 x 2 2 x 30.
þ
9780170194662
þ
þ
2
12 x 30 ¼ (2 x þ 5)( x 6) 543
Chapter 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 Polynomials
Stage 5.3 Example 6 See Example
Exercise 14-04 Dividing polynomial polynomialss 1
Perform the following following divisions, then write the first polynomial in the form: dividend divisor 3 quotient remainder.
¼
a ( x 2
þ
4
2
2
3
5
4
3
4
2
4
2
4
3
2
2
þ 7 x þ 4) ( x þ 2) b ( x 6 x þ 2) ( x 3) c (4 x þ 3 x þ 10) ( x 1) d (8 x þ 9 x þ 11) (2 x þ 1) e ( x þ 6 x þ 5 x 4) ( x 3) f (4 x þ 2 x þ x ) ( x þ 4) g (2 x x þ 5 x þ 3) ( x þ 6) h (11 x þ 3 x ) ( x þ 2) i ( x x þ 8 x þ 2 x x 1) ( x þ 1) j ( x x 10) ( x þ 3) If P P ( x) ¼ 3 x 7 x þ 5, W ( x) ¼ x þ 5 x and T ( x) ¼ x 2, then find the quotient and 2
2
4
4
3
2
4
4
2
4
3
2
4
4
2
remainder in each expression.
Example 7 See Example
3
a P ( x) 4 T ( x)
b W ( x) 4 T ( x)
c [W ( x) 3 P ( x)] 4 T ( x)
d [ P ( x)
T ( x)
1) is a factor of each polynomial. Express P ( x) as a product of the two
a P ( x)
2
c
2
g
The remainder theorem
4
Show that (2 x factors.
e
Puzzle sheet
þ W ( x)]
¼ 6 x P ( x) ¼ 8 x P ( x) ¼ 2 x P ( x) ¼ 6 x
4 3
þ x 2 þ 10 x 7 7 x 5 x þ 8 x 2 3 x þ 2 x 1 3
2
2
b P ( x)
3
2
d
3
2
4
3
f h
¼ 2 x þ x P ( x) ¼ 6 x þ x P ( x) ¼ 2 x x P ( x) ¼ 11 x þ x
5
þ x 1 1 2 x þ 7 x 3 6 x 2 x 4 2
2
6
14-05 The remainder theorem
MAT10NAPS00053
Summary The remainder theorem If a polynomial P ( x) is divided by the linear expression ( x
a), then the remainder is P (a).
Proof: Since ( x a ) is a polynomial of degree 1: P ( x) ( x a ) Æ Q ( x) R , where the remainder R is a constant. Substituting x a , gives:
¼
þ
¼ P ðaÞ ¼ ða aÞ QðaÞ þ R ¼ 0 þ R ¼ R :
544
9780170194662
NEW CENTURY MATHS ADVANCED for the
Example
A u s tr a l i an C u r ri c u l um
10 10A
þ
Stage 5.3
8
Find the remainder when P ( x) a ( x
2)
4
3
¼ 6 x 3 x þ 2 x þ 5 is divided by each linear expression. b ( x þ 3)
Solution a Dividing by ( x
b
(2). 2) will give the remainder P (2). 4 3 P ð2Þ ¼ 6 ð2Þ 3ð2Þ þ 2ð2Þ þ 5 ¼ 81 Dividing by ( x þ 3) will give the remainder P (3). ð3Þ ¼ 6ð3Þ4 3ð3Þ3 þ 2ð3Þ þ 5 P ð ¼ 566
Exercise 14-05 The remainder theorem 1
2
Find the remainder when P ( x) 2 x 3 a ( x 2) b ( x 4) e ( x 1) f ( x 10)
þ
þ
¼
2
3 x þ x 1 is divided by each linear expression. c ( x 1) d ( x 5) g ( x þ 3) h ( x 3)
Example 8 See Example
Determine the remainder remainder when the first polynomial polynomial is divided by the second polynomial. polynomial. a x 2 11 x 6, x 4 b x 2 2 x 3, x 1 c 3 x 2 x 4, x 2 d x 3 x 2 x , x 2 e 3 x 3 2 x 2 11, x 1 f 2 x 2 x 16, x 10 g 5 x 3 x 2 6 x, x 1 h x 6 3 x 4 x 3 2, x 2 i 2 x 4 3 x 2 6 x 2, x 3
þ þ þ þ þ þ þ
þ
þ þ þ þ
Puzzle sheet
14-06 The factor theorem If a poly polynomia nomiall P ( x) is divided divided by by ( x a) and the the remai remainde nderr is zer zero, o, the then n ( x Furthermore, by the remainder theorem, P (a) 0.
¼
Factorising cubic functions
P ( x). or of of P a) is a fafactctor
MAT10NAPS00051
Summary The factor theorem If ( x
a) is a factor of P P ( x), then P (a) ¼ 0.
9780170194662
545
Chapter 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 Polynomials
Stage 5.3 Video tutorial
Factorising polynomials Example
9
Factorising polynomials MAT10NAVT10018
P ( x) 2 x 3 7 x 2 a Show that ( x 2) is a factor of P b Hence express P ( x) as a product of its factors.
¼
3 x þ 18.
Solution
ð Þ ¼ 2ð23Þ 7ð22 Þ 3ð2Þ þ 18 ¼ 0 P ( x). ( x 2) is a factor of P
a P 2 [
factors. b Use long division to find the other factors. x
[
3 x 9 3 x þ 18
2 Þ 2 x2 4 x2 3 x2 3 x 3 x2 þ 6 x 9 x þ 18 9 x þ 18 2 x3
0
ð Þ ¼ ð x 2Þð2 x2 3 x 9Þ ¼ ð x x 2Þð2 x2 6 x þ 3 x 9Þ ¼ ð x x 2Þ½2 xð x 3Þ þ 3ð x 3Þ ¼ ð x x 2Þð x 3Þð2 x þ 3Þ 2 x 7 x 3 x þ 18 ¼ ( x 2)( x 3)(2 x þ 3)
)
2 x2 7 x2
P x
3
Factorising 2 x 2
3 x 9
2
Zeroes of a polynomial If ( x a ) is a factor of P a zero of of the polynomial P ( x). P ( x), x a is called a zero A zero of a polynomial P ( x) is a value of x x that makes P ( x) equal to zero. It is a solution of P ( x) 0. zero of In Example 9 Example 9 above, above, x 2 is a zero a zero of of the polynomial P ( x) 2 x 3 7 x 2 3 x 18 because P (2) (2) 0.
¼
¼
546
¼
þ
¼ ¼
9780170194662
NEW CENTURY MATHS ADVANCED for the
A u s tr a l i an C u r ri c u l um
10 10A
Solving polynomial equations Example
þ
Stage 5.3
10
a Factorise 2 x 3 3 x 2 29 x 30. b Hence solve the equation 2 x 3 3 x 2
29 x 30 ¼ 0.
Solution a Let P ( x) 2 x 3 3 x 2 29 x 30 Use ‘guess and check’ to find a zero of P ( x). P ( x) must be a factor of the constant term of P P ( x), which is ( 30), because if Any zero of P expressed as a product of its factors, factors, the constant constant terms of each facto factorr must multiply multiply P ( x) is expressed together to make ( 30). Factors of 30 are 1, 1, 2, 2, 3, 3, 5, 5, 6, 6, 10, 10, 15, 15, 30 and 30. Guessing and checking:
¼
P 1
ð Þ ¼ 2 3 29 30 ¼ 60 6¼ 0 ð x 1Þ is not a factor ð1Þ ¼ 2 3 þ 29 30 P ð ¼ 6 6¼ 0 ð x þ 1Þ is not a factor ð2Þ ¼ 16 12 þ 58 30 P ð ¼ 0 ð x þ 2Þ isis a a factor )
)
)
Now, by long division:
x
2 x2 7 x 15 3 x2 29 x 30
þ 2Þ 2 x3 2 x3 þ 4 x2 7 x2 29 x 7 x2 14 x 15 x 30 15 x 30
0
ð Þ ¼ ð x þ 2Þð2 x2 7 x 15Þ ¼ ð x þ 2Þð2 x2 10 x þ 3 x 15Þ ¼ ð x þ 2Þ½2 xð x 5Þ þ 3ð x 5Þ ¼ ð x þ 2Þð x 5Þð2 x þ 3Þ 2 x 3 x 29 x 30 ¼ ( x þ 2)( x 5)(2 x þ 3)
)
[
P x
3
9780170194662
Factorising 2 x 2
7 x 15
2
547
Chapter 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 Polynomials
b 2 x 3 3 x 2 29 x 30 ( x 2)( x 5)(2 x 3)
Stage 5.3
þ x þ 2 ¼ 0
¼ 0 þ ¼ 0 or x 5 ¼ 0
¼ 2
x
x
or
¼ 5
2 x
or
þ 3 ¼ 0 2 x ¼ 3 3 x ¼ 2 x ¼ 1 12
Exercise 14-06 The factor theorem Example 9 See Example
1
P ( x). Determine which linear polynomial, polynomial, A, B and/or C, is a factor of P P ( x )
a b c d e 2
2
A
8 x þ 7 þ 2 x x 2 þ x þ 3 5 x 22 x 16 þ 2 x 13 x þ 10
x x3 2 x 3 x3 x3
( x ( x ( x ( x ( x
2
2 2
3
4
( x ( x ( x ( x ( x
7) þ 1) 2) þ 2) 2)
( x ( x ( x ( x ( x
1) 1) 3) 8) 3)
þ þ þ þ þ þ þ þ þ þ 4)
Factorise each polynomial. a x 3 6 x 2 8 x c x 3 2 x 2 x 2 e x 3 6 x 2 11 x 6 g 3 x 3 16 x 2 13 x 6 i 2 x 4 5 x 3 2 x 2
b d f h
þ þ þ þ þ þ
x 3 x 2 2 x 2 x 3 3 x 2 18 x 8 x 3 x 2 46 x 80 6 x 3 13 x 2 x 2
þ þ þ þ þ þ
Solve each equation. a c e g i k
548
2) þ 3) þ 1) þ 1) þ 5)
C
Show that the second polynomial polynomial is a factor of the first polynomial. a x 2 10 x 24, x 2 b x 3 5 x 2 11 x 10, x 2 c x 3 3 x 2 x 3, x 1 d x 4 x 3 2 x 2 x 1, x 1 e x 3 x 2 12 x, x 3 f x 5 4 x 4 3 x 3 x 2 4, x 2 1 g 8 x 4 2 x 2 1, x h 2 x 4 7 x 3 56 x 2 37 x 84, ( x 2
þ þ þ þ
Example 10 See Example
B
(2 x 1)( x 3)( x 4) 0 (2 x 5)( x 2 x 6) 0 ( x 2)( x 3)( x 2 16) 0 x 3 2 x 2 9 x 18 0 x 3 21 x 20 0 12 x 3 23 x 2 13 x 2 0
þ
b d f h j l
þ ¼ ¼ þ ¼ ¼ ¼ þ ¼
( x 4)( x 2 1) 0 ( x 2 25)( x 2 3) 0 x 3 5 x 2 14 x 0 x 3 x 2 10 x 8 0 2 x 3 7 x 2 2 x 3 0 x 3 x 24 0
þ þ ¼ þ ¼ þ ¼ ¼ þ þ ¼ ¼
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NEW CENTURY MATHS ADVANCED for the
A u s tr a l i an C u r ri c u l um
10 þ10A
14-07 The cubic curve y ¼ a (x r )( )(x s )( )(x t ) In Chapter 11, when graphing parabolas of the form y ¼ ax 2 þ bx þ c, we found its x -intercepts by solving the equation ax 2 þ bx þ c ¼ 0. Now we can graph cubic curves of the form y ¼ ax 3 þ bx 2 þ cx þ d by by factorising the RHS as y ¼ a( x r )( )( x s)( x t ) to find its x -intercepts at r , s and t .
Example
Stage 5.3 NSW Worksheet Graphing cubics 2 MAT10NAWK10230
11
Sketch the graph of the cubic equation y ¼ x( x þ 5)( x 1).
Solution Substitute y ¼ 0 to find the x -intercepts. 0 ¼ x( x þ 5)( x 1) x ¼ 0 or x þ 5 ¼ 0 or x 1 ¼ 0 [ x ¼ 0, 5 and 1 The x -intercepts are 5, 0 and 1. Substitute x ¼ 0 to find the y -intercept. y ¼ 0ð0 þ 5Þð0 1Þ
¼0 The y -intercept is 0. Possible graphs are:
A
B
y
–5
0 1
x
y
0 1
–5
x
To determine which graph is correct, we look at the leading coefficient of the cubic equation. y ¼ x( x þ 5)( x 1), the coefficient of x x 3 is 1, which is positive, so the In the expansion of y correct graph is B , an increasing cubic curve. x 3 is negative, the correct graph is A , a decreasing cubic curve) (If the coefficient of x We can check this by substituting a value of x x , say x x ¼ 3, into the equation to find a point on the curve: y ¼ 3ð3 þ 5Þð3 1Þ
¼ 24 This means (3, 24) lies on the curve. So B must be the correct graph.
(–3, 24)
–5
y y = x(x +
0 1
5)(x − 1)
x
Use GeoGebra or other graphing technology to check the shape of this graph.
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549
Chapter 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 Polynomials
Stage 5.3
Example 11 See Example
)(x Exercise 14-07 The cubic curve y a (x r )(
¼
¼ þ þ ¼ þ ¼ þ
¼ þ ¼ þ ¼ þ þ ¼ ¼ þ ¼ þ What are the x- intercepts of the graph of y y ¼ 2 x( x þ 1)( x 3)? Select the correct answer
or D A, B B,, C C or D.. A x 2, 1, 3 3
4
¼ 0, 1, 3 D x ¼ 2, 1, 3 Which cubic cubic equation equation has has a graph with x- intercepts 2, 5 and 6? Select A Select A,, B or D B,, C C or D.. A y ¼ ( x þ 2)( x þ 5)( x 6) B y ¼ ( x 2)( x 5)( x þ 6) C y ¼ ( x þ 2)( x 5)( x þ 6) D y ¼ ( x 2)( x þ 5)( x þ 6) ¼
B x
¼ 0, 1, 3
C x
Sketch the graph of each cubic cubic equation. ( x 3)( x 1)( x 1) a y x ( x 3)( x 1) b y d y (2 x 3)( x 1)( x 2) e y ( x 1)( x 1)( x 1)
¼ ¼
Polynomials review
s)(x t )
GeoGebra or other graphing technology can be used to check the shapes of the graphs in this exercise. 1 For each cubic equation, find the x - and y-intercepts and sketch its graph. a y ( x 2)( x 2)( x 3) b y x ( x 2)( x 1) c y ( x 1)( x 3)( x 1) d y (4 x )( x 1)( x 5) e y (1 x )(2 x )( x 3) f y 2 x( x 6)( x 3) ( x 3)( x 2)( x 5) h y 2( x 1)( x 2)( x 3) i y ( x 2)( x 1)( x 2) g y 2
Worksheet
þ
¼ þ ¼ þ þ þ
þ
¼ 2 x( x 1)( x þ 6) ¼ ( x 3) ( x þ 2)
c y f y
2
14-08 Graphing polynomials
MAT10NAWK10229
Summary To graph the polynomial y • • •
¼ P ( x):
substi subs titu tute te y 0 to find the x -intercepts subs su bsti titu tute te x 0 to find the y -intercepts use the sign sign of the leadi leading ng coefficient coefficient to to sketch the the shape shape of the curve curve
¼ ¼
Example Sketch y
12 3
2
¼ x x 10 x 8.
Solution Substitute y 0 to find the x -intercepts. 0 x 3 x 2 10 x 8 Use the factor theorem to factorise the RHS. Let P ( x) x 3 x 2 10 x 8 Test factors of 8.
¼ ¼
¼ P ð1Þ ¼ 1 1 10 8 ¼ 18 6¼ 0 550
9780170194662
NEW CENTURY MATHS ADVANCED for the
A u s tr a l i an C u r ri c u l um
10 10A
þ
ð1Þ ¼ ð1Þ3 ð1Þ2 10ð1Þ 8 ð ¼ 0 ( x þ 1) is a factor. x2 2 x 8 x þ 1 Þ x3 x2 10 x 8 x3 þ x 2 2 x2 10 x 2 x2 2 x 8 x 8 8 x 8
P
Stage 5.3
[
0
ð Þ ¼ ð x þ 1Þð x2 2 x 8Þ ¼ ð x þ 1Þð x 4Þð x þ 2Þ P ( x) ¼ 0: ( x þ 1)( x 4)( x þ 2) ¼ 0 If P The x -intercepts are 2, 1 and 4. Substitute x ¼ 0 to find the y -intercept. y ¼ 0 3 02 10ð0Þ 8 ¼ 8 The y -intercept -intercept is is 8. )
P x
Factorising x 2
2 x 8
[
y
[
y = x3 − x2 − 10x − 8
The leading coefficient is 1, which is positive, so the cubic curve is increasing. –2
–1
0
4
x
–8
Use GeoGebra or other graphing technology to check the shape of this graph.
Single, double and triple roots of a polynomial equation If x a root of of the equation. x a is a solution of the polynomial equation P ( x) 0, then x a is called a root 2 P ( x) 0. If ( x a ) is a factor of a polynomial P ( x), then x a is a double root of of P 2 2 For example, if P P ( x) ( x 3) ( x x 1), then x 3 is a double root of P ( x) 0. 3 P ( x) 0. If ( x a ) is a factor of a polynomial P ( x), then x a is a triple root of of P 3 For example, if P 1 is a triple root of P P ( x) ( x 1) ( x 2), then x P ( x) 0. If ( x a ) is a factor of a polynomial P ( x), then x a is a single root of of P P ( x) 0. For example, if P 2, x 3 are all single roots of P P ( x) ( x 1)( x x 2)( x x 3), then x 1, x P ( x) 0.
¼
9780170194662
¼
¼ þ
¼
þ
¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼
¼
¼ ¼ ¼
¼
¼
551
Chapter 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 Polynomials
Stage 5.3
Example
13
Sketch P ( x)
3
¼ ( x 3) ( x þ 1).
Solution Solve P ( x) 0 to find the x -intercepts. 1 x 3 and x 1 is a single root, so the graph will cross the x -axis at x x 3 is a triple root and x (and take a shape similar to y x 3 there) and at x 1. Substitute x 0 to find the y -intercept.
¼ ¼
¼ ¼
¼ 3 y ¼ ð0 3Þ ð0 þ 1Þ ¼ 27
¼ ¼
¼
The polynomial is a quartic (degree 4) and the leading term is x 3 3 x negative, so the quartic curve will decrease as x increases.
¼ 3
4
¼ x , which is
P (x) 27
y = –(x − 3)3(x + 1)
–1 0
3
x
Use GeoGebra or other graphing technology to check the shape of this graph.
Summary P ( x) 0 has a single If P a single root at at x a , then the graph of the polynomial crosses the x -axis at crosses the x a . If P a double root at at x a , then the graph of the polynomial touches the x -axis touches the P ( x) 0 has a double at x a with a flat gradient, taking the shape of a parabola there. P ( x) 0 has a triple If P a triple root at at x a , then the graph of the polynomial crosses the x -axis at crosses the x a with a flat gradient, taking the shape of a cubic curve there.
¼
¼
¼
¼
¼
¼
¼
¼
¼
Exercise 14-08 Graphing polynomials Example 12 See Example
GeoGeb GeoG ebra ra or ot othe herr gr grap aphi hing ng te tech chno nolo logy gy ca can n be us used ed to ch chec eck k th thee sh shap apes es of th thee gr grap aphs hs in th this is ex exer erci cise se.. the graph of each polynomial. 1 Sketch the a P ( x) x 3 6 x 2 8 x b P ( x) x 3 2 x 2 x 2 c e g
552
¼ þ þ P ( x) ¼ 2 x þ 3 x 18 x þ 8 P ( x) ¼ 2 x þ 17 x þ 31 x 20 P ( x) ¼ x þ 3 x þ 6 x 8 x 3 3
4
2
2
3
2
d f h
¼ þ P ( x) ¼ 2 x 5 x þ 2 x P ( x) ¼ x þ 5 x þ 8 x 12 P ( x) ¼ x þ 13 x 36 4
3 4
3
2
2
2
9780170194662
NEW CENTURY MATHS ADVANCED for the
2
Sketch the graph of each polynomial. polynomial. a y ( x 2)( x 4) 2 c e g i k
¼ þ y ¼ ( x þ 1) ( x 2) y ¼ ( x 1)( x 2)( x þ 2) y ¼ ( x 2) ( x þ 1) y ¼ ( x 4)( x 4) P ( x) ¼ x 5 x þ 4 2
2
2
2
h
2
4
¼ ( x þ 2)( x 3) y ¼ x ( x þ 1)( x 3) y ¼ x( x 4) y ¼ ( x 4) ( x þ 1) y ¼ x x þ 5 x 3 y ¼ x 3 x 12 x 44 x 48 2
3
4
l
Example 13 See Example
2
3
j
2
þ
2
b y f
2
10 10A Stage 5.3
d 2
A u s tr a l i an C u r ri c u l um
2
3
2
Technology Transforming graphs graphing technology to graph P ( x) 1 Use GeoGebra or other graphing 2 Write the equations of each polynomial. a
P ( x)
b P ( x)
c P ( x)
polynomials shown in question question 2 3 Graph the polynomials 2.. of the polynomials y 4 Describe how the graphs of can be drawn from the graph of y P ( x).
¼
3
2
¼ x 3 x þ 2. d 2 P ( x)
¼ P ( x), y ¼ P ( x), y ¼ P ( x) 3, y ¼ 2 P ( x)
14-09 Transforming graphs of polynomials
Worksheet Advanced graphs MAT10NAWK10231
Summary If the graph of the polynomial y • • • •
¼ P ( x) is drawn: thee gr th grap aph h of y y ¼ P ( x) is a reflection of y y ¼ P ( x) in the x -axis thee gr th grap aph h of y y ¼ P ( x) is a reflection of y y ¼ P ( x) in the y -axis thee gr th grap aph h of y y ¼ P ( x) þ c is a vertical translation of the graph y ¼ P ( x) y ¼ aP ( x) is y ¼ P ( x) either ‘stretched’ or ‘compressed’ vertically, with the thee gr th grap aph h of y
same x -intercepts
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Chapter 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 Polynomials
Stage 5.3
Example
14
The graph of y y
y
¼ P ( x) is shown.
3
Draw the graphs of each polynomial. a y c y
¼ P ( x) ¼ P ( x) þ 2
b y d y
y = P (x)
¼ P ( x) ¼ 3 P ( x)
–1 0
1
x
2
Solution a y
¼ P ( x) is a reflection of y y ¼ P ( x) in the
x-axis.
y b y P ( x) is a reflection of y the y -axis.
¼
y
y
y = –P (x)
y = P (x)
3
3
(–x) y = P (–
y = P (x)
–1 0
–1
¼ P ( x) in
1
0
–2
x
2
1 2
x
–3
c y P ( x) 2 is y 2 units up.
¼
þ
¼ P ( x) vertically translated
d y 3 P ( x) is y P ( x) stretched verticall verti callyy by a factor of 3.
¼
¼
y 5
y
9
y = P (x) + 2 y = 3P (x)
3
3
y = P (x) 0
x
–1
0
1
2
x
y = P (x)
554
9780170194662
NEW CENTURY MATHS ADVANCED for the
A u s tr a l i an C u r ri c u l um
10 10A
þ
Stage 5.3
Exercise 14-09 Transformi Transforming ng graphs graphs of polynomials polynomials 1
The graph graph of the polynomial polynomial y each polynomial below.
¼ 2 P ( x) y ¼ P ( x) 3
¼ P ( x) is shown. Sketch the graph of
¼ P ( x) þ 2 y ¼ P ( x)
a y
b y
c y
d
e
f
1 P x 2 P ( x)
¼ ð Þ y¼
y
y –3 –2
Example 14 See Example
3
–1 0
=
P (x)
1
x
–4
2
y Use the graph of y
a d
¼ P ( x) shown to sketch each polynomial. b y ¼ P ( x) 2 c y ¼ 2 P ( x) y ¼ P ( x) þ 1 y ¼ P ( x) e y ¼ P ( x) f y ¼ 3 P ( x)
y
y
=
P (x) 2
–1 0
1
x
3 Graph P ( x) ( x 1)( x 3) 2 and use it to sketch the graph of each polynomial. a y b y P ( x) c y P ( x) 3 d y 2 P ( x) P ( x) 4
¼
¼
¼
¼
¼
Describe the transformation needed to graph each each cubic equation equation using the graph of y y d g
3
¼ x 2 y ¼ x þ 3 y ¼ 2 x 5
a y
9780170194662
3
3
e h
3
¼ x þ 1 y ¼ 3 x y ¼ 4 3 x
b y
3
f 3
i
3
¼ 2 x y ¼ 2 x 1 y ¼ x3 þ 4 2
c y
¼ x
3
.
3
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