# 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(π₯):

If f(π) = 0 then (π₯ – π) is a factor.
If f($$\frac{b}{c}$$) = 0 then (ππ₯ – π) is a factor.

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

Example Question 1

Show that (π₯ β 2) is a factor of 2π₯3 + 3π₯2 β 11π₯ β 6.

Start by saying that f(π₯) = 2π₯3 + 3π₯2 β 11π₯ β 6.

According to the factor theorem, if (π₯ – 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, (π₯ – 2) is a factor of 2π₯3 + 3π₯2 β 11π₯ β 6.

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

Example Question 2

Fully factorise 2π₯3 + 13π₯2 + 17π₯ β 12.

As before, start by saying that f(π₯) = 2π₯3 + 13π₯2 + 17π₯ β 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 (π₯ – 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) β  0, (π₯ – 2) is not a factor.

Next, try (π₯ + 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 (π₯ + 3) is a factor. This means you can divide f(π₯) by (π₯ + 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.

2π₯3 + 13π₯2 + 17π₯ β 12 = (π₯ + 3)(2π₯2 + 7π₯ β 4)

To complete the factorisation, factorise 2π₯2 + 7π₯ β 4 as (2π₯ β 1)(π₯ + 4).

2π₯3 + 13π₯2 + 17π₯ β 12 = (π₯ + 3)(2π₯ β 1)(π₯ + 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 (π₯ β 2) is a factor of π₯3 + 9π₯2 β 3x + 16.

2. Determine whether (π₯ + 3) is a factor of 2π₯3 + 7π₯2 + 9.

3. Determine whether (2π₯ β 1) is a factor of 3π₯3 β 5π₯2 β 14π₯ + 8.

4. Show that (π₯ + 2) is a factor of π₯3 + 4π₯2 + π₯ β 6 and hence factorise the expression fully.

5. Show that (3π₯ β 2) is a factor of 3π₯3 β 20π₯2 + 27π₯ β 10 and hence factorise the expression fully.

6. Solve the equation π₯3 + 3π₯2 β 6π₯ β 8 = 0

7. Solve the equation 2π₯3 + 3π₯2 β 39π₯ β 20 = 0

8. f(π₯) = π₯3 + 5π₯2 β ππ₯ β 56
Given that (π₯ β 4) is a factor of f(π₯), calculate the value of π.

9. f(π₯) = 2π₯3 + π₯2 β 18π₯ β π
Given that (π₯ + 3) is a factor of f(π₯), calculate the value of π.

10. f(π₯) = 2π₯3 + ππ₯2 + ππ₯ + 30
Given that (π₯ β 2) and (π₯ β 5) are factors of f(π₯), find the value of π and π and hence factorise f(π₯) fully.

1. Determine whether (π₯ β 2) is a factor of π₯3 + 9π₯2 β 3x + 16.

Let f(π₯) = π₯3 + 9π₯2 β 3x + 16.
f(2) = 54
No, it is not a factor.

2. Determine whether (π₯ + 3) is a factor of 2π₯3 + 7π₯2 + 9.

Let f(π₯) = 2π₯3 + 7π₯2 + 9.
f(-3) = 18
No, it is not a factor.

3. Determine whether (2π₯ β 1) is a factor of 3π₯3 β 5π₯2 β 14π₯ + 8.

Let f(π₯) = 3π₯3 β 5π₯2 β 14π₯ + 8.
f($$\boldsymbol{\frac{1}{2}}$$) = $$\boldsymbol{\frac{1}{8}}$$
No, it is not a factor.

4. Show that (π₯ + 2) is a factor of π₯3 + 4π₯2 + π₯ β 6 and hence factorise the expression fully.

Let f(π₯) = π₯3 + 4π₯2 + π₯ β 6.
(-2)3 + 4(-2)2 + -2 β 6 = 0
f(-2) = 0, so (π₯ + 2) is a factor.
(π₯3 + 4π₯2 + x β 6) Γ· (π₯ + 2) = π₯2 + 2π₯ β 3
π₯2 + 2π₯ β 3 = (π₯ + 3)(π₯ β 1)
π₯3 + 4π₯2 + π₯ β 6 = (π₯ + 3)(π₯ β 1)(π₯ + 2)

5. Show that (3π₯ β 2) is a factor of 3π₯3 β 20π₯2 + 27π₯ β 10 and hence factorise the expression fully.

Let f(π₯) = 3π₯3 β 20π₯2 + 27π₯ β 10.
3($$\boldsymbol{\frac{2}{3}}$$)3 β 20($$\boldsymbol{\frac{2}{3}}$$)2 + 27($$\boldsymbol{\frac{2}{3}}$$) β 10 = 0
f($$\boldsymbol{\frac{2}{3}}$$) = 0, so (3π₯ β 2) is a factor.
(3π₯3 β 20π₯2 + 27π₯ β 10) Γ· (3π₯ β 2) = π₯2 β 6π₯ + 5
π₯2 β 6π₯ + 5 = (π₯ β 1)(π₯ β 5)
3π₯3 β 20π₯2 + 27π₯ β 10 = (3π₯ β 2)(π₯ β 1)(π₯ β 5)

6. Solve the equation π₯3 + 3π₯2 β 6π₯ β 8 = 0.

π₯3 + 3π₯2 β 6π₯ β 8 = (π₯ β 2)(π₯ + 4)(π₯ + 1)
The solutions are π₯ = 2, π₯ = -4, π₯ = -1
.

7. Solve the equation 2π₯3 + 3π₯2 β 39π₯ β 20 = 0.

2π₯3 + 3π₯2 β 39π₯ β 20 = (π₯ β 4)(π₯ + 5)(2π₯ + 1)
The solutions are π₯ = 4, π₯ = -5, π₯ = -$$\boldsymbol{\frac{1}{2}}$$
.

8. f(π₯) = π₯3 + 5π₯2 β ππ₯ β 56
Given that (π₯ β 4) is a factor of f(π₯), calculate the value of π.

If (π₯ β 4) is a factor of f(π₯), then f(4) = 0.
f(4) = 88 β 4π
88 β 4π = 0
π = 22

9. f(π₯) = 2π₯3 + π₯2 β 18π₯ β π
Given that (π₯ + 3) is a factor of f(π₯), calculate the value of π.

If (π₯ + 3) is a factor of f(π₯), then f(-3) = 0.
f(-3) = 9 β π
9 β π = 0
π = 9

10. f(π₯) = 2π₯3 + ππ₯2 + ππ₯ + 30
Given that (π₯ β 2) and (π₯ β 5) are factors of f(π₯), find the value of π and π and hence factorise f(π₯) fully.

f(2) = 4π + 2π + 46
f(5) = 25π + 5π + 280
4π + 2π + 46 = 0
25π + 5π + 280 = 0
Solving simultaneously gives π = -11, π = -1.
(2π₯3 β 11π₯2 β π₯ + 30) Γ· (π₯ β 2) = 2π₯2 β 7π₯ β 15
(2π₯3 β 11π₯2 β π₯ + 30) = (2π₯ + 3)(π₯ β 2)(π₯ β 5)

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