N.B.Nota Bene

Home  /  Extended Maths  /  MYP 4

Advanced Factoring

Once single brackets feel easy, these are the three factorising moves that unlock almost everything else in algebra: grouping four terms, splitting a trinomial (even when the leading number is not 1), and spotting a difference of two squares.

MYP 4Extended MathsAlgebraCriteria A · C · D~10 min read

Written by Fayek · MYP student

Why factorise at all

To factorise an expression is to rewrite it as a product: things multiplied together, rather than added. It is the reverse of expanding brackets. The reason it matters so much is the null factor law: if a product equals zero, then one of the factors must be zero. So the moment an expression is in factor form, solving equations, simplifying fractions and finding roots all become easy.

Factor
A quantity that divides exactly into an expression. In \(6x^2\) the factors include \(2\), \(3\), \(x\) and \(6x\). In \((x+1)(x-4)\) the two brackets are the factors.

The very first thing to check on any expression is a common factor across every term. Always take that out before trying anything cleverer.

\[ 6x^2 + 9x = 3x(2x + 3) \]

Where this is assessed

Factorising sits mostly in Criterion A (choosing and applying the right technique) and Criterion C (setting your working out with correct notation). When a question wraps it in a context or asks you to solve and interpret, it reaches into Criterion D.

Factorising by grouping

When an expression has four terms and no factor common to all of them, try grouping: split the four terms into two pairs, factorise each pair, and hope that both pairs leave the same bracket behind. That shared bracket is then itself a common factor.

Worked example

Factorise \(2x^3 + 6x^2 + x + 3\).

1
Split into two pairs: \((2x^3 + 6x^2) + (x + 3)\).
2
Take the common factor out of each pair. From the first, \(2x^2\): \(2x^2(x + 3)\). The second pair \((x+3)\) is already a bracket, so factor out \(1\): \(1(x + 3)\).
3
Now both pieces carry the same bracket \((x+3)\): \(2x^2(x + 3) + 1(x + 3)\).
4
Factor out that shared bracket: \((x + 3)(2x^2 + 1)\).
\((x + 3)(2x^2 + 1)\)

If the brackets do not match

Try reordering the terms before grouping, and watch the signs. If a pair starts with a minus, factor out a negative so the inside bracket flips to match the other pair. Expanding your answer back out is the fastest way to check.

Quadratic trinomials

A trinomial is a three-term quadratic, \(ax^2 + bx + c\). How you factorise it depends on whether the leading coefficient \(a\) is \(1\) or something else.

When \(a = 1\)

Look for two numbers that multiply to \(c\) and add to \(b\). Those two numbers go straight into the brackets.

\[ x^2 + 7x + 12 = (x + 3)(x + 4) \]

Here \(3 \times 4 = 12\) and \(3 + 4 = 7\), so the split is immediate.

When \(a \ne 1\): the AC or cross method

When the leading number is not \(1\), the two-number trick needs one extra move. Multiply \(a\) by \(c\), find two numbers that multiply to that product \(ac\) and still add to \(b\), then use them to split the middle term into two. Now you have four terms, so you finish by grouping.

Worked example

Factorise \(6x^2 + 11x + 3\).

1
Identify \(a = 6\), \(b = 11\), \(c = 3\). Compute \(ac = 6 \times 3 = 18\).
2
Find two numbers multiplying to \(18\) and adding to \(11\): those are \(9\) and \(2\).
3
Split the middle term \(11x\) into \(9x + 2x\): \(6x^2 + 9x + 2x + 3\).
4
Group in pairs and factor each: \(3x(2x + 3) + 1(2x + 3)\).
5
Take out the shared bracket: \((2x + 3)(3x + 1)\).
\((2x + 3)(3x + 1)\)

Mind the signs

When \(c\) is negative, the two numbers have opposite signs; when \(c\) is positive but \(b\) is negative, both numbers are negative. Get the signs right at step 2 and the rest falls into place. Always expand to confirm: \((2x+3)(3x+1) = 6x^2 + 2x + 9x + 3 = 6x^2 + 11x + 3\). Correct.

Difference of two squares

This is the pattern worth memorising, because it is instant once you recognise it. Any expression that is one perfect square minus another factorises into conjugate brackets:

\[ a^2 - b^2 = (a - b)(a + b) \]

The two brackets are identical except for the sign in the middle. Note there is no middle term in the original: it is only two terms, connected by a minus. A sum of two squares, \(a^2 + b^2\), does not factorise over the real numbers.

Worked example

Factorise \(9x^2 - 16\).

1
Recognise each term as a perfect square: \(9x^2 = (3x)^2\) and \(16 = 4^2\).
2
Match the pattern with \(a = 3x\) and \(b = 4\).
3
Write the conjugate brackets: \((3x - 4)(3x + 4)\).
\((3x - 4)(3x + 4)\)

Common factor first, always

Sometimes the difference of squares is hidden. In \(50 - 2x^2\) it does not look like a difference of squares until you pull out the \(2\): \(2(25 - x^2) = 2(5 - x)(5 + x)\). Take common factors out before you go hunting for a pattern.

Check yourself

Cover the answers and try each one. The working matters as much as the final brackets.

1. Factorise \(x^3 - 4x^2 + 3x - 12\) by grouping. +

Split into pairs: \((x^3 - 4x^2) + (3x - 12)\). Factor each: \(x^2(x - 4) + 3(x - 4)\). Both share \((x - 4)\), so the answer is \((x - 4)(x^2 + 3)\).

2. Factorise \(2x^2 + 7x + 3\) using the AC method. +

Here \(ac = 2 \times 3 = 6\), and the pair multiplying to \(6\) and adding to \(7\) is \(6\) and \(1\). Split the middle: \(2x^2 + 6x + x + 3\). Group: \(2x(x + 3) + 1(x + 3)\). The answer is \((x + 3)(2x + 1)\).

3. Factorise \(25x^2 - 49\). +

This is a difference of two squares: \(25x^2 = (5x)^2\) and \(49 = 7^2\). So it factorises to \((5x - 7)(5x + 7)\).

Part of the Extended Maths library. Spotted an error or want a topic added? That feedback makes the notes better.