AQA Core 1 Polynomials Section 2: The factor and remainder theorems Notes and Examples These notes contain subsections on The factor theorem Dividing polynomials The remainder theorem
The factor theorem You already know that you can solve some quadratics by factorising them. e.g.
to solve the quadratic equation you factorise: and deduce the solutions
x² + 3 x – 10 x² 10 = 0 ( x x + + 5)( x 5)( x – 2) 2) = 0 x = x = -5 and x and x = = 2
f( x x)) = x = x²² + 3 x – 10 10, f(-5) = 0 and f(2) = 0. Clearly, for f( ( x x + + 5) is a factor of f ( x x)) f(-5) = 0 ( x – 2) 2) is a factor of f( x x)) f(2) = 0
This idea can be extended to other polynomials such as cubics. For example, for the cubic function g( x) ( x 1)( x 2)( x 3) , g(1) = 0, g(-2) = 0 and g(3) = 0. ( x – 1) 1) is a factor of g( x x)) g(1) = 0 ( x x + + 2) is a factor of g( x x)) g(-2) = 0 ( x – 3) 3) is a factor of g( x x)) g(3) = 0 In general, the factor theorem states theorem states that: f( x), ), then f(a f(a) = 0 and x = x = a is a root of the equation f( x f( x)) = 0. is a factor of f( x Conversely, if f(a f(a) = 0, then is a factor of f( x f( x). ). If
x a
x
a
You can use the factor theorem t heorem to solve cubic and higher order equations. These are the steps you need to take t ake to solve a cubic equation of the form f( x f( x)) = 0, where f( f( x x)) is a cubic function: First, work out f( f( x x)) for different values of x until you find one for which f( x x)) = 0. Using the factor factor theorem, you now know know one one factor factor of the function. function. Divide the function function by this linear linear factor. You can now express the function as the product of the linear factor and a quadratic factor. Factorise the quadratic factor, if possible. If the quadratic factor does does factorise, factorise, you now have have three linear factors, factors, from which you can deduce the three solutions to the equation. equation.
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AQA C1 Polynomials 2 Notes and Examples
If the quadratic factor does does not not factorise, you can use use the quadratic formula to find the two further solutions, if they exist.
For higher order equations, you will have to find more than one factor by trial and error, and you will have to divide more than once. The following example shows how this method works.
Example 1 (i) Solve the equation x equation x³³ + 2 x 2 x²² – 5 5 x – 6 6 = 0 (ii) Sketch the graph graph of y of y = = x x³³ + 2 x 2 x²² – 5 5 x – 6 6
Solution (i)
The first step is to find one solution by trial and error. If there is an integer solution x x = = a, then by the factor theorem ( x – a) must be a factor of x x³³ + 2 x x²² – 5 5 x – 6 6. So a must be a factor of 6. a could therefore be 1, -1, 2, -2, 3, -3, 6 or – –6. 6.
Let f( x f( x)) = x = x³³ + 2 x 2 x²² – 5 5 x – 6 6 You need to find a value of x for which f( x x)) = 0.
f(1) = 1 + 2 – 5 5 – 6 6 = -8 f(-1) = -1 + 2 + 5 – 6 6 = 0
f(-1) = 0 so by the factor theorem x + x + 1 is a factor of f( x f( x). ). The next step is to factorise f( x x)) into the linear factor
x + x + 1 and a quadratic factor.
x³ x³ + 2 x² x² – 5 5 x – 6 6 = ( x + x + 1) quadratic factor. Let the quadratic factor be ax² ax² + bx + bx + c. x³ x³ + 2 x 2 x²² – 5 x 5 x – 6 6 = ( x ( x + + 1)(ax 1)(ax²² + bx + bx + c) = ax³ ax³ + bx² bx² + cx + cx + ax² ax² + bx + bx + c = ax³ ax³ + (a (a + b) x² x² + (b (b + c) x + x + c
Multiply out the brackets
Equating coefficients of x of x³³ a = 1 Equating constant term c = -6 Equating coefficients of x of x²² a + b = 2 b = 1 Check coefficient of x of x:: b + c = 1 – 6 6 = -5 x³ x³ + 2 x 2 x²² – 5 x 5 x – 6 6 = ( x ( x + + 1)( x 1)( x²² + x + x – 6) 6) = ( x ( x + + 1)( x 1)( x – 2)( 2)( x + x + 3)
Factorise the quadratic factor
The solutions of the equation are x = x = -1, x -1, x = = 2 and x and x = = -3. (ii) Part (i) shows shows that the graph of of y = y = x x³³ + 2 x² x² – 5 5 x – 6 6 crosses the x the x-axis -axis at (-3, 0), (-1, 0) and (2, 0). By putting x = x = 0 you can see that it crosses the y the y-axis -axis at (0, -6). This information allows you to sketch the graph.
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AQA C1 Polynomials 2 Notes and Examples
In the example above, the quadratic factor fa ctor was found by equating coefficients. There are a number of other methods of finding this quadratic factor. You can do it by polynomial division, or by inspection (which means doing it in your head). Which one you use is really down to t o personal preference. You can see some different approaches using the Flash resources r esources P o l y n o m i a l d i v i s i on o n b y i n s p e c ti ti o n , P o l y n o m i a l d i v i s i o n – b o x m e t h o d ( n o r e m a i n d e r ) and ac t o r i s i n g a c u b i c and F ac , and the PowerPoint presentation F ac ac t o r i s i n g p o l y n o m i a l s . The Geogebra resource P o l y n o m i a l d i v i s i o n shows the box method. You can also look at the S o l v i n g c u b i c s v i d e o . et c h i n g f a c t o r i s ed e d c u b i c s and the Mathcentre video The Flash resource S k et Polynomial functions may also be useful. may
For some additional practice, try the in teractive questions S k e t c h i n g polyn om ial curves and F i n d i n g a p o l y n o m i a l f r o m i t s r o o t s . and
Example 2 f( x f( x)) = 2 x 2 x³³ + px + px²² + 5 x 5 x – 6 6 has a factor x – 2. 2. Find the value of p and p and hence factorise f( x f( x)) as far as possible. Solution x – 2 2 is a factor of f( x f( x)) f(2) = 0 f(2) = 16 + 4 p 4 p + + 10 – 6 6 = 20 + 4 p 4 p
20 + 4 p 4 p = = 0 p = p = -5 f( x f( x)) = 2 x 2 x³³ - 5 x² x² + 5 x 5 x – 6 6 2 x³ x³ - 5 x² x² + 5 x – 6 6 = ( x ( x – 2)(ax 2)(ax²² + bx + bx + c) = ax³ ax³ + bx² bx² + cx – 2ax 2ax²² - 2bx 2bx – 2c 2c
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AQA C1 Polynomials 2 Notes and Examples = ax³ ax³ + (b (b – 2a 2a) x² x² + (c (c – 2b 2b) x – 2c 2c Equating coefficients of x of x³³ a = 2 Equating constant terms -2c -2c = -6 c = 3 Equating coefficients of x of x²² b – 2a 2a = -5 b – 4 4 = -5 b = -1 Check coefficient of x of x:: c – 2b 2b = 3 + 2 = 5 2 x³ x³ - 5 x² x² + 5 x 5 x – 6 6 = ( x ( x – 2)(2 2)(2 x² x² – x + x + 3) The discriminant of the quadratic factor is ( -1)² - 4 2 3 = 1 – 24 24 = -23 As this is negative, the quadratic factor cannot be factorised further.
Dividing polynomials When you divide one polynomial by another, it may divide exactly, or there may be a remainder, just as in arithmetic. For example:
26 is called the dividend dividend,, 6 is called the divisor , 4 is the quotient and 2 is the remainder .
26 6 = 4 remainder 2
You could rewrite this statement as: 26 = 6 4 + 2
This rearrangement helps to give one method of dividing polynomials. When working with polynomials, remember the following points: If you you are dividing by a linear expression, expression, the quotient is of order order one less than the dividend (e.g. for a quartic, the quotient is cubic) and the remainder, if any, is a constant term. If you are dividing by a quadratic term, the quotient quotient is of of order two less than the dividend (e.g. for a quartic, the quotient is quadratic) and the remainder, if any, could be linear or a constant term. This idea idea can can be be extended extended to a polynomial polynomial divisor of of any order. Most, if not all, of the examples you meet will only involve dividing by a linear expression. When you divide a cubic expression by a linear expression, as in this example, examp le, the quotient is a quadratic expression and the remainder, if any, is a constant term. term.
Example 3 Divide 2 x 2 x³³ + 3 x 3 x²² - x - x + + 1 by x by x + + 2
Solution Let the quotient be ax² ax² + bx + bx + c and the remainder be d .
2 x³ x³ + 3 x² x² – x x + 1
= ( x + x + 2)(ax 2)(ax²² + bx + bx + c) + d = x( x(ax² ax² + bx + bx + c) + 2(ax 2(ax²² + bx + bx + c) + d = ax³ ax³ + bx² bx² + cx + cx + 2ax 2ax²² + 2bx 2bx + + 2c 2c + d = ax³ ax³ + (b (b + 2a 2a) x² x² + (c (c + 2b 2b) x + x + 2c 2c + d
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AQA C1 Polynomials 2 Notes and Examples Equating coefficients of x of x³³ Equating coefficients of x of x²² Equating coefficients of x of x Equating constant terms
a = 2 2a = 3 b + 4 = 3 b = -1 b + 2a 2b = -1 c – 2 2 = -1 c = 1 c + 2b 2c + d = = 1 2 + d = = 1 d = = -1 2c
2 x³ x³ + 3 x² x² – x + x + 1 = ( x ( x + + 2)(2 x² x² – x + x + 1) – 1 1 The quotient is 2 x 2 x²² – x + x + 1 and the remainder is -1.
Don’t be put off by a negative remainder, as in this example – this – this is quite acceptable in polynomial division! There are other approaches to polynomial division. The PowerPoint in g p o l y n o m i a l s presentation D i v i d in shows (1) algebraic long division and (2) shows ( 2) polynomial division by inspection inspection (i.e. in your head). You can also look at more examples using the Flash resources P o l y n o m i a l d i v i s i o n ( l o n g d i v i s i o n ) and e m a i n d e r ) . and P o l y n o m i a l d i v i s i o n – b o x m e t h o d ( r em There is also a P o l y n o m i a l d i v i s i o n v i d e o – this – this deals with the long l ong division method only. For some extra practice in examples like the one above, try the interactive in g p o l y n o m i a l s resource D i v i d in .
The remainder theorem The factor theorem is a special case of the remainder theorem. f( x x)) is divided by a linear expression ( x – a), then if ( x – a) When a polynomial f( x), ), then there will be a remainder. is not a factor of f( x f( x x)) = ( x ( x – a) quotient + remainder x = = a into this expression gives: Substituting x
i.e.
f(a) = 0 quotient + remainder f(a f(a f( a) = remainder
The remainder theorem says: For a polynomial f( x f( x), ), f(a f(a) is the remainder when f( x f( x)) is divided by
x a
.
f ( x)remainder f ( a) then . x) x a g( is x – a is a factor of f( f( x x). ). If the zero, x), ), then f( f(a a) = 0. So if x – a is a factor of f( x This is the factor theorem.
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AQA C1 Polynomials 2 Notes and Examples Example 4 4 Find the remainder when f( x f( x)) = x = x – 3 3 x³ x³ + x + x²² – 4 4 is divided by (i) x – 2 2 (ii) x + x + 1 Solution (i) By the remainder theorem the remainder is f(2) f(2) = 24 – 3 32³ + 2² - 4 = 16 – 24 24 + 4 – 4 4 = -8 Remainder = -8
(ii) By the remainder theorem the remainder is f(-1) 4 f(-1) = (-1) – 3(-1)³ 3(-1)³ + (-1)² – 4 4 = 1 + 3 + 1 – 4 4 = 1 Remainder = 1
remain der You can look at similar examples using the t he Flash resource The remain theorem .
For extra practice in examples like the one above, try the interactive questions F i n d i n g a r e m a i n d e r .
Example 5 f( x f( x)) = 2 x 2 x³³ + ax² ax² + bx + bx + 1 When f( x f( x)) is divided by x – 1 the remainder is 7 When f( x f( x)) is divided by x + x + 3 the remainder is -5 Find the values of a and b. Solution f(1) = 7
f(-3) = -5
Adding:
2 + a + b + 1 = 7 a + b = 4 (-3)²a – 3b 3b + 1 = -5 2(-3)³ + (-3)²a -54 + 9a 9a – 3b 3b + 1 = -5 9a – 3b 3b = 48 9a 3a – b = 16 3a
a + b = 4 3a – b = 16 4a 4a = 20 a = 5, b = -1
For extra practice in examples like the t he one above, try the interactive questions F i n d i n g a p o l y n o m i a l , g i v e n t h e r e m a i n d e r .
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