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).

If you’d like to download this blog in PDF or PowerPoint formats, click here. To practice some of the prior knowledge needed to use the factor theorem, try this multiple-choice quiz.

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 (ππ₯ – 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 ππ₯ 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} + ππ₯^{2} + ππ₯ + 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.

Answers

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