GCSEFurther MathsAQARevision Notes2. Algebra & FunctionsPolynomials & Factor TheoremApplications of Factor Theorem

Applications of Factor Theorem (AQA GCSE Further Maths): Revision Note

Exam code: 8365

Written by: Mark Curtis

Reviewed by: Dan Finlay

Updated on 27 June 2025

Applications of Factor Theorem

What is factorising by inspection?

Factorising by inspection means expanding brackets in your head to find only one possibility
If (2x + 1) is a factor of 2x² + 7x + 3, then...
- $2 x^{2} + 7 x + 3$ ≡ $(2 x + 1) (. . . + . . .)$
- by inspection, the only possibility is $(2 x + 1) (x + 3)$
  - this gives 2x² and +3
The answer can be written down with no working

How do I factorise a cubic when one linear factor is given?

If you know a linear factor of a cubic expression, you can use that...
- ... "cubic expression" ≡ "linear factor" × "quadratic factor"
If (x - 2) is a factor of 2x³ + 7x² - 17x - 10,
- 2x³ + 7x² - 17x - 10 ≡ (x - 2)(ax² + bx + c)
- By inspection, a = 2 and c = 5 (this give 2x³and -10)
- Find b by equating coefficients of x²
  - 7x² on the left
  - $- 2 \times 2 x^{2}$ and $x \times b x$ on the right (search for x² terms made up of something from the first bracket and something from the second bracket)
  - this gives $- 4 x^{2} + b x^{2}$ on the right
  - b must be 11
- 2x³ + 7x² - 17x - 10 ≡ (x - 2)(2x² + 11x + 5)
- The final step is to factorise the quadratic factor
  - 2x² + 11x - 5 ≡ (2x + 1)(x + 5)
  - so the cubic 2x³ + 7x² - 17x - 10 factorises to (x - 2)(2x + 1)(x + 5)

How do I factorise a cubic without knowing any linear factors?

Find a linear factor using the Factor Theorem (then use the method above)
If (x - a) is a factor, then f(a) = 0, so test different values of a to find a factor
To find a linear factor of f(x) = x³ - 6x² - x + 30, work out f(1), f(-1), f(2), f(-2), f(3), f(-3)... etc until you get f(a) = 0
- $f (1) = 1^{3} - 6 \times 1^{2} - 1 + 30 = 24 f (- 1) = {(- 1)}^{3} - 6 \times {(- 1)}^{2} - (- 1) + 30 = 24 f (2) = 2^{3} - 6 \times 2^{2} - 2 + 30 = 12 f (- 2) = {(- 2)}^{3} - 6 \times {(- 2)}^{2} - (- 2) + 30 = 0$
- the first zero came from f(-2) = 0 so, by the Factor Theorem, (x + 2) is a factor
- You only need to test ± whole numbers that divide "30" (the number at the end of the cubic)
Cubics with a number in front of x³, such as f(x) = 3x³ + 4x² - 5x - 2, are harder
- try factors of (3x + 1), (3x - 1), (3x + 2), (3x - 2), ... as well as (x + 1), (x - 1), (x + 2), (x - 2)...

How do I solve a cubic equation?

Factorise the cubic using the method above, then set each bracket equal to zero and solve for x
To solve 3x³ + 4x² - 5x - 2 = 0
- f(1) = 0 so (x - 1) is a factor, factorising to (x - 1)(x + 2)(3x + 1) = 0
- Solve each bracket equal to zero
  - x - 1 = 0 gives x = 1
  - x + 2 = 0 gives x = -2
  - 3x + 1 = 0 gives $x = - \frac{1}{3}$
- the solutions are $x = - 2, - \frac{1}{3} and 1$

Examiner Tips and Tricks

Beware of trying to find all three linear factors by just testing numbers
- you could find f(1) = 0, f(-1) = 0 and f(2) = 0 from $f (x) = 2 x^{3} - 4 x^{2} - 2 x + 4$ and think it factorises to (x - 1)(x + 1)(x - 2) but it doesn't (expand and check)
- some cubics only have one factor (so you'd be testing an infinite number of other integers trying to find non-existent factors!)

Worked Example

Solve $x^{3} - 5 x^{2} - 16 x + 80 = 0$

Set the polynomial equal to $f (x)$ and find the first linear factor by testing positive and negative factors of 80, starting with the smallest values.

Test f(1) and f(-1).

$\begin{array}{rcl} f (1) & = & {(1)}^{3} - 5 {(1)}^{2} - 16 (1) + 80 = 1 - 5 - 16 + 80 = 60 \\ f (- 1) & = & {(- 1)}^{3} - 5 {(- 1)}^{2} - 16 (- 1) + 80 = - 1 - 5 + 16 + 80 = 90 \end{array}$

Test f(2) and f(-2).

$\begin{array}{rcl} f (2) & = & {(2)}^{3} - 5 {(2)}^{2} - 16 (2) + 80 = 8 - 20 - 32 + 80 = 36 \\ f (- 2) & = & {(- 2)}^{3} - 5 {(- 2)}^{2} - 16 (- 2) + 80 = - 8 - 20 + 32 + 80 = 84 \end{array}$

Test f(4) and f(-4), there is no need to test f(3) and f(-3) as 3 is not a factor of 80.

$\begin{array}{rcl} f (4) & = & {(4)}^{3} - 5 {(4)}^{2} - 16 (4) + 80 = 64 - 80 - 64 + 80 = 0 \end{array}$

The linear factor is found so there is no need to test anymore.

$f (4) = 0$ so $x - 4$ is a factor.

$\begin{array}{rcl} f (x) & = & (x - 4) (a x^{2} + b x + c) \end{array}$

Set the two expressions equal to each other and equate coefficients.

$\begin{array}{rcl} x^{3} - 5 x^{2} - 16 x + 80 & = & (x - 4) (a x^{2} + b x + c) \end{array}$

Equating the first terms, $a x^{3} = x^{3}$ therefore $a = 1$ .

$\begin{array}{rcl} x^{3} - 5 x^{2} - 16 x + 80 & = & (x - 4) (x^{2} + b x + c) \end{array}$

Equating the last (constant) terms, $- 4 c = 80$ therefore $c = \frac{80}{- 4} = - 20$ .

$\begin{array}{rcl} f (x) & = & (x - 4) (x^{2} + b x - 20) \end{array}$

Consider the $x^{2}$ terms by multiplying out these parts.

$\begin{array}{rcl} - 5 x^{2} & = & - 4 a x^{2} + b x^{2} \end{array}$

Substitute $a = 1$ and solve for $b$ .

$\begin{array}{rcl} - 5 & = & - 4 + b \\ b & = & - 1 \end{array}$

Substitute in and factorise the quadratic.

$\begin{array}{rcl} f (x) & = & (x - 4) (x^{2} - x - 20) \\ = & (x - 4) (x + 4) (x - 5) \end{array}$

The solutions are the opposite signs of each factor.

$x = - 4, x = 4, x = 5$

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