# The Factor Theorem – A Level Maths Revision

The factor theorem is a quick way to find linear factors of a polynomial. It can be used alongside your preferred method of polynomial division to fully factorise polynomials (we also have a full blog on how to factorise cubics).

The factor theorem states that:

For a polynomial f(𝑥):

If f(𝑎) = 0 then (𝑥 – 𝑎) is a factor.
If f($$\frac{b}{c}$$) = 0 then (𝑐𝑥 – 𝑏) is a factor.

This allows you to establish whether a linear expression is a factor of a polynomial by substituting a given value of 𝑥 into the polynomial.

Example Question 1

Show that (𝑥 – 2) is a factor of 2𝑥3 + 3𝑥2 – 11𝑥 – 6.

Start by saying that f(𝑥) = 2𝑥3 + 3𝑥2 – 11𝑥 – 6.

According to the factor theorem, if (𝑥 – 2) is a factor then f(2) = 0.

\begin{aligned} f(2) &= 2 \times 2^3 + 3 \times 2^2 -11 \times 2 -6 \\ &= 16+12-22-6\\&= 0\end{aligned}

Therefore, (𝑥 – 2) is a factor of 2𝑥3 + 3𝑥2 – 11𝑥 – 6.

You can also use the factor theorem to factorise a polynomial, by considering the constant term.

Example Question 2

Fully factorise 2𝑥3 + 13𝑥2 + 17𝑥 – 12.

As before, start by saying that f(𝑥) = 2𝑥3 + 13𝑥2 + 17𝑥 – 12.

In this example, the constant term is -12. As the function is a cubic, it will fully factorise to three linear expressions. The constant terms of those three linear expressions must multiply to -12, so the constant terms in the factors must be factors of -12.

Start by checking if (𝑥 – 2) is a factor:

\begin{aligned}f(2) &= 2 \times 2^3 + 13 \times 2^2 + 17 \times 2 -12 \\&= 90 \end{aligned}

As f(2) ≠ 0, (𝑥 – 2) is not a factor.

Next, try (𝑥 + 3):

\begin{aligned}f(-3) &= 2 \times(-3)^3 + 13 \times (-3)^2 + 17 \times (-3) -12 \\&= 0 \end{aligned}

As f(-3) = 0, then (𝑥 + 3) is a factor. This means you can divide f(𝑥) by (𝑥 + 3) and there will not be a remainder. Here we’ll use algebraic long division, but you can use your preferred method of algebraic division.

2𝑥3 + 13𝑥2 + 17𝑥 – 12 = (𝑥 + 3)(2𝑥2 + 7𝑥 – 4)

To complete the factorisation, factorise 2𝑥2 + 7𝑥 – 4 as (2𝑥 – 1)(𝑥 + 4).

2𝑥3 + 13𝑥2 + 17𝑥 – 12 = (𝑥 + 3)(2𝑥 – 1)(𝑥 + 4)

According to the factor theorem, f(-4) = 0 and f($$\frac{1}{2}$$) = 0; the process of finding an initial factor is trial and error but would always lead to the same final result.

Practice Questions

1. Determine whether (𝑥 – 2) is a factor of 𝑥3 + 9𝑥2 – 3x + 16.

2. Determine whether (𝑥 + 3) is a factor of 2𝑥3 + 7𝑥2 + 9.

3. Determine whether (2𝑥 – 1) is a factor of 3𝑥3 – 5𝑥2 – 14𝑥 + 8.

4. Show that (𝑥 + 2) is a factor of 𝑥3 + 4𝑥2 + 𝑥 – 6 and hence factorise the expression fully.

5. Show that (3𝑥 – 2) is a factor of 3𝑥3 – 20𝑥2 + 27𝑥 – 10 and hence factorise the expression fully.

6. Solve the equation 𝑥3 + 3𝑥2 – 6𝑥 – 8 = 0

7. Solve the equation 2𝑥3 + 3𝑥2 – 39𝑥 – 20 = 0

8. f(𝑥) = 𝑥3 + 5𝑥2 – 𝑎𝑥 – 56
Given that (𝑥 – 4) is a factor of f(𝑥), calculate the value of 𝑎.

9. f(𝑥) = 2𝑥3 + 𝑥2 – 18𝑥 – 𝑏
Given that (𝑥 + 3) is a factor of f(𝑥), calculate the value of 𝑏.

10. f(𝑥) = 2𝑥3 + 𝑎𝑥2 + 𝑏𝑥 + 30
Given that (𝑥 – 2) and (𝑥 – 5) are factors of f(𝑥), find the value of 𝑎 and 𝑏 and hence factorise f(𝑥) fully.

1. Determine whether (𝑥 – 2) is a factor of 𝑥3 + 9𝑥2 – 3x + 16.

Let f(𝑥) = 𝑥3 + 9𝑥2 – 3x + 16.
f(2) = 54
No, it is not a factor.

2. Determine whether (𝑥 + 3) is a factor of 2𝑥3 + 7𝑥2 + 9.

Let f(𝑥) = 2𝑥3 + 7𝑥2 + 9.
f(-3) = 18
No, it is not a factor.

3. Determine whether (2𝑥 – 1) is a factor of 3𝑥3 – 5𝑥2 – 14𝑥 + 8.

Let f(𝑥) = 3𝑥3 – 5𝑥2 – 14𝑥 + 8.
f($$\boldsymbol{\frac{1}{2}}$$) = $$\boldsymbol{\frac{1}{8}}$$
No, it is not a factor.

4. Show that (𝑥 + 2) is a factor of 𝑥3 + 4𝑥2 + 𝑥 – 6 and hence factorise the expression fully.

Let f(𝑥) = 𝑥3 + 4𝑥2 + 𝑥 – 6.
(-2)3 + 4(-2)2 + -2 – 6 = 0
f(-2) = 0, so (𝑥 + 2) is a factor.
(𝑥3 + 4𝑥2 + x – 6) ÷ (𝑥 + 2) = 𝑥2 + 2𝑥 – 3
𝑥2 + 2𝑥 – 3 = (𝑥 + 3)(𝑥 – 1)
𝑥3 + 4𝑥2 + 𝑥 – 6 = (𝑥 + 3)(𝑥 – 1)(𝑥 + 2)

5. Show that (3𝑥 – 2) is a factor of 3𝑥3 – 20𝑥2 + 27𝑥 – 10 and hence factorise the expression fully.

Let f(𝑥) = 3𝑥3 – 20𝑥2 + 27𝑥 – 10.
3($$\boldsymbol{\frac{2}{3}}$$)3 – 20($$\boldsymbol{\frac{2}{3}}$$)2 + 27($$\boldsymbol{\frac{2}{3}}$$) – 10 = 0
f($$\boldsymbol{\frac{2}{3}}$$) = 0, so (3𝑥 – 2) is a factor.
(3𝑥3 – 20𝑥2 + 27𝑥 – 10) ÷ (3𝑥 – 2) = 𝑥2 – 6𝑥 + 5
𝑥2 – 6𝑥 + 5 = (𝑥 – 1)(𝑥 – 5)
3𝑥3 – 20𝑥2 + 27𝑥 – 10 = (3𝑥 – 2)(𝑥 – 1)(𝑥 – 5)

6. Solve the equation 𝑥3 + 3𝑥2 – 6𝑥 – 8 = 0.

𝑥3 + 3𝑥2 – 6𝑥 – 8 = (𝑥 – 2)(𝑥 + 4)(𝑥 + 1)
The solutions are 𝑥 = 2, 𝑥 = -4, 𝑥 = -1
.

7. Solve the equation 2𝑥3 + 3𝑥2 – 39𝑥 – 20 = 0.

2𝑥3 + 3𝑥2 – 39𝑥 – 20 = (𝑥 – 4)(𝑥 + 5)(2𝑥 + 1)
The solutions are 𝑥 = 4, 𝑥 = -5, 𝑥 = -$$\boldsymbol{\frac{1}{2}}$$
.

8. f(𝑥) = 𝑥3 + 5𝑥2 – 𝑎𝑥 – 56
Given that (𝑥 – 4) is a factor of f(𝑥), calculate the value of 𝑎.

If (𝑥 – 4) is a factor of f(𝑥), then f(4) = 0.
f(4) = 88 – 4𝑎
88 – 4𝑎 = 0
𝑎 = 22

9. f(𝑥) = 2𝑥3 + 𝑥2 – 18𝑥 – 𝑏
Given that (𝑥 + 3) is a factor of f(𝑥), calculate the value of 𝑏.

If (𝑥 + 3) is a factor of f(𝑥), then f(-3) = 0.
f(-3) = 9 – 𝑏
9 – 𝑏 = 0
𝑏 = 9

10. f(𝑥) = 2𝑥3 + 𝑎𝑥2 + 𝑏𝑥 + 30
Given that (𝑥 – 2) and (𝑥 – 5) are factors of f(𝑥), find the value of 𝑎 and 𝑏 and hence factorise f(𝑥) fully.

f(2) = 4𝑎 + 2𝑏 + 46
f(5) = 25𝑎 + 5𝑏 + 280
4𝑎 + 2𝑏 + 46 = 0
25𝑎 + 5𝑏 + 280 = 0
Solving simultaneously gives 𝑎 = -11, 𝑏 = -1.
(2𝑥3 – 11𝑥2 – 𝑥 + 30) ÷ (𝑥 – 2) = 2𝑥2 – 7𝑥 – 15
(2𝑥3 – 11𝑥2 – 𝑥 + 30) = (2𝑥 + 3)(𝑥 – 2)(𝑥 – 5)

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