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# Factorising Cubic Polynomials - Grade 12

## 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.

Definition 1: Factor Theorem

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

f ( x ) = ( x - a ) q ( x ) f ( x ) = ( x - a ) q ( x )
(1)

is a polynomial.

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

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

### Exercise 1: Factor Theorem

Use the Factor Theorem to determine whether y-1y-1 is a factor of f(y)=2y4+3y2-5y+7f(y)=2y4+3y2-5y+7.

#### Solution

1. Step 1. Determine how to approach the problem :

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

2. Step 2. Calculate f(1)f(1) :
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 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
(2)
3. Step 3. Conclusion :

Since f(1)0f(1)0, y-1y-1 is not a factor of f(y)=2y4+3y2-5y+7f(y)=2y4+3y2-5y+7.

### Exercise 2: Factor Theorem

Using the Factor Theorem, verify that y+4y+4 is a factor of g(y)=5y4+16y3-15y2+8y+16g(y)=5y4+16y3-15y2+8y+16.

#### Solution

1. Step 1. Determine how to approach the problem :

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

2. Step 2. Calculate f(1)f(1) :
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 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)
3. Step 3. Conclusion :

Since g(-4)=0g(-4)=0, y+4y+4 is a factor of g(y)=5y4+16y3-15y2+8y+16g(y)=5y4+16y3-15y2+8y+16.

## Factorisation of Cubic Polynomials

A cubic polynomial is a polynomial of the form

a x 3 + b x 2 + c x + d a x 3 + b x 2 + c x + d
(4)

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 ( x + y ) ( x 2 - x y + y 2 ) = x 3 + y 3
(5)

and

( x - y ) ( x 2 + x y + y 2 ) = x 3 - y 3 ( x - y ) ( x 2 + x y + y 2 ) = x 3 - y 3
(6)

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 f ( x ) = a x 3 + b x 2 + c x + d
(7)

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

( A x + B ) ( C x + D ) ( E x + F ) . ( A x + B ) ( C x + D ) ( E x + F ) .
(8)

But sometimes you will get factors of the form:

( A x + B ) ( C x 2 + E x + D ) ( A x + B ) ( C x 2 + E x + D )
(9)

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

x 3 - 2 x 2 - 5 x + 6 . x 3 - 2 x 2 - 5 x + 6 .
(10)

In this case we have

a = 1 b = - 2 c = - 5 d = 6 a = 1 b = - 2 c = - 5 d = 6
(11)

The factors will have the general form shown in Equation 8, with A=C=E=1A=C=E=1. We can then use values for aa, bb, cc and dd to determine values for BB, DD and FF. We can re-write Equation 8 with A=C=E=1A=C=E=1 as:

( x + B ) ( x + D ) ( x + F ) . ( x + B ) ( x + D ) ( x + F ) .
(12)

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 ( 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
(13)

We can therefore write:

b = - 2 = D + F + B c = - 5 = D F + B D + B F d = 6 = B D F . b = - 2 = D + F + B c = - 5 = D F + B D + B F d = 6 = B D F .
(14)

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

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

### Exercise 3: Factorisation of Cubic Polynomials

Factorise f(x)=x3+x2-9x-9f(x)=x3+x2-9x-9 into three linear factors.

#### Solution

1. Step 1. By trial and error using the factor theorem to find a factor :

Try

f ( 1 ) = ( 1 ) 3 + ( 1 ) 2 - 9 ( 1 ) - 9 = 1 + 1 - 9 - 9 = - 16 f ( 1 ) = ( 1 ) 3 + ( 1 ) 2 - 9 ( 1 ) - 9 = 1 + 1 - 9 - 9 = - 16
(15)

Therefore (x-1)(x-1) is not a factor

Try

f ( - 1 ) = ( - 1 ) 3 + ( - 1 ) 2  9 ( - 1 )  9 =  1 + 1 + 9  9 = 0 f ( - 1 ) = ( - 1 ) 3 + ( - 1 ) 2  9 ( - 1 )  9 =  1 + 1 + 9  9 = 0
(16)

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

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

Write x3+x2-9x-9=(x+1)()x3+x2-9x-9=(x+1)()

The first term in the second bracket must be x2x2 to give x3x3 if one works backwards.

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

So we have x3+x2-9x-9=(x+1)(x2+?x-9)x3+x2-9x-9=(x+1)(x2+?x-9).

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

(+1)(x2)(+1)(x2) gives the x2x2 in the original polynomial. So, the coefficient of the xx-term must be 0.

So f(x)=(x+1)(x2-9)f(x)=(x+1)(x2-9).

2. Step 2. Factorise fully :

x2-9x2-9 can be further factorised to (x-3)(x+3)(x-3)(x+3),

and we are now left with f(x)=(x+1)(x-3)(x+3)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 x3-2x2-5x+6x3-2x2-5x+6 can be found as follows: There are three factors which we can write as

( x - a ) ( x - b ) ( x - c ) . ( x - a ) ( x - b ) ( x - c ) .
(17)

### Exercise 4: Factorisation of Cubic Polynomials

Use the Factor Theorem to factorise

x 3 - 2 x 2 - 5 x + 6 . x 3 - 2 x 2 - 5 x + 6 .
(18)

#### Solution

1. Step 1. Find one factor using the Factor Theorem :

Try

f ( 1 ) = ( 1 ) 3 - 2 ( 1 ) 2 - 5 ( 1 ) + 6 = 1 - 2 - 5 + 6 = 0 f ( 1 ) = ( 1 ) 3 - 2 ( 1 ) 2 - 5 ( 1 ) + 6 = 1 - 2 - 5 + 6 = 0
(19)

Therefore (x-1)(x-1) is a factor.

2. Step 2. Division by inspection :

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

The first term in the second bracket must be x2x2 to give x3x3 if one works backwards.

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

So we have x3-2x2-5x+6=(x-1)(x2+?x-6)x3-2x2-5x+6=(x-1)(x2+?x-6).

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

(-1)(x2)(-1)(x2) gives -x2-x2. So, the coefficient of the xx-term must be -1-1.

So f(x)=(x-1)(x2-x-6)f(x)=(x-1)(x2-x-6).

3. Step 3. Factorise fully :

x2-x-6x2-x-6 can be further factorised to (x-3)(x+2)(x-3)(x+2),

and we are now left with x3-2x2-5x+6=(x-1)(x-3)(x+2)x3-2x2-5x+6=(x-1)(x-3)(x+2)

## Exercises - Using Factor Theorem

1. Find the remainder when 4x3-4x2+x-54x3-4x2+x-5 is divided by (x+1)(x+1).
2. Use the factor theorem to factorise x3-3x2+4x3-3x2+4 completely.
3. f ( x ) = 2 x 3 + x 2 - 5 x + 2 f ( x ) = 2 x 3 + x 2 - 5 x + 2
1. Find f(1)f(1).
2. Factorise f(x)f(x) completely
4. Use the Factor Theorem to determine all the factors of the following expression:
x3+x2-17x+15x3+x2-17x+15
(20)
5. Complete: If f(x)f(x) is a polynomial and pp is a number such that f(p)=0f(p)=0, then (x-p)(x-p) is .....

## Solving Cubic Equations

Once you know how to factorise cubic polynomials, it is also easy to solve cubic equations of the kind

a x 3 + b x 2 + c x + d = 0 a x 3 + b x 2 + c x + d = 0
(21)

### Exercise 5: Solution of Cubic Equations

Solve

6 x 3 - 5 x 2 - 17 x + 6 = 0 . 6 x 3 - 5 x 2 - 17 x + 6 = 0 .
(22)

#### Solution

1. Step 1. Find one factor using the Factor Theorem :

Try

f ( 1 ) = 6 ( 1 ) 3 - 5 ( 1 ) 2 - 17 ( 1 ) + 6 = 6 - 5 - 17 + 6 = - 10 f ( 1 ) = 6 ( 1 ) 3 - 5 ( 1 ) 2 - 17 ( 1 ) + 6 = 6 - 5 - 17 + 6 = - 10
(23)

Therefore (x-1)(x-1) is NOT a factor.

Try

f ( 2 ) = 6 ( 2 ) 3 - 5 ( 2 ) 2 - 17 ( 2 ) + 6 = 48 - 20 - 34 + 6 = 0 f ( 2 ) = 6 ( 2 ) 3 - 5 ( 2 ) 2 - 17 ( 2 ) + 6 = 48 - 20 - 34 + 6 = 0
(24)

Therefore (x-2)(x-2) IS a factor.

2. Step 2. Division by inspection :

6 x 3 - 5 x 2 - 17 x + 6 = ( x - 2 ) ( ) 6 x 3 - 5 x 2 - 17 x + 6 = ( x - 2 ) ( )

The first term in the second bracket must be 6x26x2 to give 6x36x3 if one works backwards.

The last term in the second bracket must be -3-3 because -2×-3=+6-2×-3=+6.

So we have 6x3-5x2-17x+6=(x-2)(6x2+?x-3)6x3-5x2-17x+6=(x-2)(6x2+?x-3).

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

(-2)(6x2)(-2)(6x2) gives -12x2-12x2. So, the coefficient of the xx-term must be 7.

So, 6x3-5x2-17x+6=(x-2)(6x2+7x-3)6x3-5x2-17x+6=(x-2)(6x2+7x-3).

3. Step 3. Factorise fully :

6x2+7x-36x2+7x-3 can be further factorised to (2x+3)(3x-1)(2x+3)(3x-1),

and we are now left with 6x3-5x2-17x+6=(x-2)(2x+3)(3x-1)6x3-5x2-17x+6=(x-2)(2x+3)(3x-1)

4. Step 4. Solve the equation :
6 x 3 - 5 x 2 - 17 x + 6 = 0 ( x - 2 ) ( 2 x + 3 ) ( 3 x - 1 ) = 0 x = 2 ; 1 3 ; - 3 2 6 x 3 - 5 x 2 - 17 x + 6 = 0 ( x - 2 ) ( 2 x + 3 ) ( 3 x - 1 ) = 0 x = 2 ; 1 3 ; - 3 2
(25)

Sometimes it is not possible to factorise the trinomial ("second bracket"). This is when the quadratic formula

x = - b ± b 2 - 4 a c 2 a x = - b ± b 2 - 4 a c 2 a
(26)

can be used to solve the cubic equation fully.

For example:

### Exercise 6: Solution of Cubic Equations

Solve for xx: x3-2x2-6x+4=0.x3-2x2-6x+4=0.

#### Solution

1. Step 1. Find one factor using the Factor Theorem :

Try

f ( 1 ) = ( 1 ) 3 - 2 ( 1 ) 2 - 6 ( 1 ) + 4 = 1 - 2 - 6 + 4 = - 1 f ( 1 ) = ( 1 ) 3 - 2 ( 1 ) 2 - 6 ( 1 ) + 4 = 1 - 2 - 6 + 4 = - 1
(27)

Therefore (x-1)(x-1) is NOT a factor.

Try

f ( 2 ) = ( 2 ) 3 - 2 ( 2 ) 2 - 6 ( 2 ) + 4 = 8 - 8 - 12 + 4 = - 8 f ( 2 ) = ( 2 ) 3 - 2 ( 2 ) 2 - 6 ( 2 ) + 4 = 8 - 8 - 12 + 4 = - 8
(28)

Therefore (x-2)(x-2) is NOT a factor.

f ( - 2 ) = ( - 2 ) 3 - 2 ( - 2 ) 2 - 6 ( - 2 ) + 4 = - 8 - 8 + 12 + 4 = 0 f ( - 2 ) = ( - 2 ) 3 - 2 ( - 2 ) 2 - 6 ( - 2 ) + 4 = - 8 - 8 + 12 + 4 = 0
(29)

Therefore (x+2)(x+2) IS a factor.

2. Step 2. Division by inspection :

x 3 - 2 x 2 - 6 x + 4 = ( x + 2 ) ( ) x 3 - 2 x 2 - 6 x + 4 = ( x + 2 ) ( )

The first term in the second bracket must be x2x2 to give x3x3.

The last term in the second bracket must be 2 because 2×2=+42×2=+4.

So we have x3-2x2-6x+4=(x+2)(x2+?x+2)x3-2x2-6x+4=(x+2)(x2+?x+2).

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

(2)(x2)(2)(x2) gives 2x22x2. So, the coefficient of the xx-term must be -4-4. (2x2-4x2=-2x22x2-4x2=-2x2)

So x3-2x2-6x+4=(x+2)(x2-4x+2)x3-2x2-6x+4=(x+2)(x2-4x+2).

x2-4x+2x2-4x+2 cannot be factorised any futher and we are now left with

( x + 2 ) ( x 2 - 4 x + 2 ) = 0 ( x + 2 ) ( x 2 - 4 x + 2 ) = 0

3. Step 3. Solve the equation :
( x + 2 ) ( x 2 - 4 x + 2 ) = 0 ( x + 2 ) = 0 o r ( x 2 - 4 x + 2 ) = 0 ( x + 2 ) ( x 2 - 4 x + 2 ) = 0 ( x + 2 ) = 0 o r ( x 2 - 4 x + 2 ) = 0
(30)
4. Step 4. Apply the quadratic formula for the second bracket :

Always write down the formula first and then substitute the values of a,ba,b and cc.

x = - b ± b 2 - 4 a c 2 a = - ( - 4 ) ± ( - 4 ) 2 - 4 ( 1 ) ( 2 ) 2 ( 1 ) = 4 ± 8 2 = 2 ± 2 x = - b ± b 2 - 4 a c 2 a = - ( - 4 ) ± ( - 4 ) 2 - 4 ( 1 ) ( 2 ) 2 ( 1 ) = 4 ± 8 2 = 2 ± 2
(31)
5. Step 5. Final solutions :

x=-2orx=2±2x=-2orx=2±2

### Exercises - Solving of Cubic Equations

1. Solve for xx: x3+x2-5x+3=0x3+x2-5x+3=0
2. Solve for yy: y3-3y2-16y-12=0y3-3y2-16y-12=0
3. Solve for mm: m3-m2-4m-4=0m3-m2-4m-4=0
4. Solve for xx: x3-x2=3(3x+2)x3-x2=3(3x+2)

#### Tip:

:
Remove brackets and write as an equation equal to zero.
5. Solve for xx if 2x3-3x2-8x=32x3-3x2-8x=3

## End of Chapter Exercises

1. Solve for xx: 16(x+1)=x2(x+1)16(x+1)=x2(x+1)
1. Show that x-2x-2 is a factor of 3x3-11x2+12x-43x3-11x2+12x-4
2. Hence, by factorising completely, solve the equation
3x3-11x2+12x-4=03x3-11x2+12x-4=0
(32)
2. 2x3-x2-2x+2=Q(x).(2x-1)+R2x3-x2-2x+2=Q(x).(2x-1)+R for all values of xx. What is the value of RR?
1. Use the factor theorem to solve the following equation for mm:
8m3+7m2-17m+2=08m3+7m2-17m+2=0
(33)
2. Hence, or otherwise, solve for xx:
23x+3+7·22x+2=17·2x23x+3+7·22x+2=17·2x
(34)
3. A challenge: Determine the values of pp for which the function
f(x)=3p3-(3p-7)x2+5x-3f(x)=3p3-(3p-7)x2+5x-3
(35)
leaves a remainder of 9 when it is divided by (x-p)(x-p).

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