A-Level — mathsStandard2 min read

Polynomials and the Factor Theorem

Divide polynomials, apply the factor theorem and remainder theorem at A-Level. Factorise cubics and quartics.

Tutor AI Team2026-02-25T01:53:03.156Z

Polynomial division, the factor theorem, and the remainder theorem are essential A-Level tools for factorising higher-degree polynomials.

Core Concepts

Factor Theorem

If $f(a) = 0$ , then $(x - a)$ is a factor of $f(x)$ .

Remainder Theorem

When $f(x)$ is divided by $(x - a)$ , the remainder is $f(a)$ .

Polynomial Division

Long division or synthetic division to divide $f(x)$ by $(x - a)$ .

Factorising Cubics

Step 1: Try $f(1), f(-1), f(2), f(-2), ...$ to find a root. Step 2: Use the factor theorem: if $f(a) = 0$ , then $(x-a)$ is a factor. Step 3: Divide by $(x-a)$ to get a quadratic. Step 4: Factorise the quadratic.

Worked Example: Example 1

Problem

$f(x) = x^3 - 6x^2 + 11x - 6$ .

$f(1) = 1 - 6 + 11 - 6 = 0$ ✓ → $(x-1)$ is a factor.

Divide: $x^3 - 6x^2 + 11x - 6 = (x-1)(x^2 - 5x + 6) = (x-1)(x-2)(x-3)$ .

Solution

Worked Example: Remainder

Problem

Find the remainder when $x^3 + 2x - 5$ is divided by $(x - 2)$ .

$f(2) = 8 + 4 - 5 = 7$ . Remainder = 7.

Solution

Practice Problems

1. Show $(x + 2)$ is a factor of $x^3 + 3x^2 - 4$ .
1. Fully factorise $2x^3 - x^2 - 5x - 2$ .
1. Find the remainder when $x^4 - 3x^2 + 2$ is divided by $(x - 1)$ .

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Key Takeaways

✓
Factor theorem: $f(a) = 0$ → $(x-a)$ is a factor.
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Remainder theorem: remainder = $f(a)$ .
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To factorise cubics: find a root, divide, factorise the quotient.

Tags:

#a-level#maths#pure#algebra#polynomials#factor-theorem

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