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Introduction

In grades 10 and 11, you learnt how to solve different types of equations. Most of the solutions, relied on being able to factorise some expression and the factorisation of quadratics was studied in detail. This chapter focusses on the factorisation of cubic polynomials, that is expressions with the highest power equal to 3.

The factor theorem

The factor theorem describes the relationship between the root of a polynomial and a factor of the polynomial.

Factor Theorem

For any polynomial, f ( x ) , for all values of a which satisfy f ( a ) = 0 , ( x - a ) is a factor of f ( x ) . Or, more concisely:

f ( x ) = ( x - a ) q ( x )

is a polynomial.

In other words: If the remainder when dividing f ( x ) by ( x - a ) is zero, then ( x - a ) is a factor of f ( x ) .

So if f ( - b a ) = 0 , then ( a x + b ) is a factor of f ( x ) .

Use the Factor Theorem to determine whether y - 1 is a factor of f ( y ) = 2 y 4 + 3 y 2 - 5 y + 7 .

  1. In order for y - 1 to be a factor, f ( 1 ) must be 0.

  2. f ( y ) = 2 y 4 + 3 y 2 - 5 y + 7 f ( 1 ) = 2 ( 1 ) 4 + 3 ( 1 ) 2 - 5 ( 1 ) + 7 = 2 + 3 - 5 + 7 = 7
  3. Since f ( 1 ) 0 , y - 1 is not a factor of f ( y ) = 2 y 4 + 3 y 2 - 5 y + 7 .

Using the Factor Theorem, verify that y + 4 is a factor of g ( y ) = 5 y 4 + 16 y 3 - 15 y 2 + 8 y + 16 .

  1. In order for y + 4 to be a factor, g ( - 4 ) must be 0.

  2. g ( y ) = 5 y 4 + 16 y 3 - 15 y 2 + 8 y + 16 g ( - 4 ) = 5 ( - 4 ) 4 + 16 ( - 4 ) 3 - 15 ( - 4 ) 2 + 8 ( - 4 ) + 16 = 5 ( 256 ) + 16 ( - 64 ) - 15 ( 16 ) + 8 ( - 4 ) + 16 = 1280 - 1024 - 240 - 32 + 16 = 0
  3. Since g ( - 4 ) = 0 , y + 4 is a factor of g ( y ) = 5 y 4 + 16 y 3 - 15 y 2 + 8 y + 16 .

Factorisation of cubic polynomials

A cubic polynomial is a polynomial of the form

a x 3 + b x 2 + c x + d

where a is nonzero. We have seen in Grade 10 that the sum and difference of cubes is factorised as follows.:

( x + y ) ( x 2 - x y + y 2 ) = x 3 + y 3

and

( x - y ) ( x 2 + x y + y 2 ) = x 3 - y 3

We also saw that the quadratic term does not have rational roots.

There are many methods of factorising a cubic polynomial. The general method is similar to that used to factorise quadratic equations. If you have a cubic polynomial of the form:

f ( x ) = a x 3 + b x 2 + c x + d

then in an ideal world you would get factors of the form:

( A x + B ) ( C x + D ) ( E x + F ) .

But sometimes you will get factors of the form:

( A x + B ) ( C x 2 + E x + D )

We will deal with simplest case first. When a = 1 , then A = C = E = 1 , and you only have to determine B , D and F . For example, find the factors of:

x 3 - 2 x 2 - 5 x + 6 .

In this case we have

a = 1 b = - 2 c = - 5 d = 6

The factors will have the general form shown in [link] , with A = C = E = 1 . We can then use values for a , b , c and d to determine values for B , D and F . We can re-write [link] with A = C = E = 1 as:

( x + B ) ( x + D ) ( x + F ) .

If we multiply this out we get:

( x + B ) ( x + D ) ( x + F ) = ( x + B ) ( x 2 + D x + F x + D F ) = x 3 + D x 2 + F x 2 + B x 2 + D F x + B D x + B F x + B D F = x 3 + ( D + F + B ) x 2 + ( D F + B D + B F ) x + B D F

We can therefore write:

b = - 2 = D + F + B c = - 5 = D F + B D + B F d = 6 = B D F .

This is a set of three equations in three unknowns. However, we know that B , D and F are factors of 6 because B D F = 6 . Therefore we can use a trial and error method to find B , D and F .

This can become a very tedious method, therefore the Factor Theorem can be used to find the factors of cubic polynomials.

Factorise f ( x ) = x 3 + x 2 - 9 x - 9 into three linear factors.

  1. Try

    f ( 1 ) = ( 1 ) 3 + ( 1 ) 2 - 9 ( 1 ) - 9 = 1 + 1 - 9 - 9 = - 16

    Therefore ( x - 1 ) is not a factor

    Try

    f ( - 1 ) = ( - 1 ) 3 + ( - 1 ) 2 – 9 ( - 1 ) – 9 = – 1 + 1 + 9 – 9 = 0

    Thus ( x + 1 ) is a factor, because f ( - 1 ) = 0 .

    Now divide f ( x ) by ( x + 1 ) using division by inspection:

    Write x 3 + x 2 - 9 x - 9 = ( x + 1 ) ( )

    The first term in the second bracket must be x 2 to give x 3 if one works backwards.

    The last term in the second bracket must be - 9 because + 1 × - 9 = - 9 .

    So we have x 3 + x 2 - 9 x - 9 = ( x + 1 ) ( x 2 + ? x - 9 ) .

    Now, we must find the coefficient of the middle term ( x ).

    ( + 1 ) ( x 2 ) gives the x 2 in the original polynomial. So, the coefficient of the x -term must be 0.

    So f ( x ) = ( x + 1 ) ( x 2 - 9 ) .

  2. x 2 - 9 can be further factorised to ( x - 3 ) ( x + 3 ) ,

    and we are now left with f ( x ) = ( x + 1 ) ( x - 3 ) ( x + 3 )

In general, to factorise a cubic polynomial, you find one factor by trial and error. Use the factor theorem to confirm that the guess is a root. Then divide the cubic polynomial by the factor to obtain a quadratic. Once you have the quadratic, you can apply the standard methods to factorise the quadratic.

For example the factors of x 3 - 2 x 2 - 5 x + 6 can be found as follows: There are three factors which we can write as

( x - a ) ( x - b ) ( x - c ) .

Use the Factor Theorem to factorise

x 3 - 2 x 2 - 5 x + 6 .
  1. Try

    f ( 1 ) = ( 1 ) 3 - 2 ( 1 ) 2 - 5 ( 1 ) + 6 = 1 - 2 - 5 + 6 = 0

    Therefore ( x - 1 ) is a factor.

  2. x 3 - 2 x 2 - 5 x + 6 = ( x - 1 ) ( )

    The first term in the second bracket must be x 2 to give x 3 if one works backwards.

    The last term in the second bracket must be - 6 because - 1 × - 6 = + 6 .

    So we have x 3 - 2 x 2 - 5 x + 6 = ( x - 1 ) ( x 2 + ? x - 6 ) .

    Now, we must find the coefficient of the middle term ( x ).

    ( - 1 ) ( x 2 ) gives - x 2 . So, the coefficient of the x -term must be - 1 .

    So f ( x ) = ( x - 1 ) ( x 2 - x - 6 ) .

  3. x 2 - x - 6 can be further factorised to ( x - 3 ) ( x + 2 ) ,

    and we are now left with x 3 - 2 x 2 - 5 x + 6 = ( x - 1 ) ( x - 3 ) ( x + 2 )

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Source:  OpenStax, Siyavula textbooks: grade 12 maths. OpenStax CNX. Aug 03, 2011 Download for free at http://cnx.org/content/col11242/1.2
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