Solving Quadratics by Factorising

Solving quadratics by factorising: finding the pair, making it equal zero first, and reading the answers with the right signs.

In the exam for Cambridge O Level Maths 4024, Edexcel IGCSE Maths 4MA1, Cambridge IGCSE Maths 0580, Edexcel Further Pure Maths 4PM1, IGCSE Additional Maths 0606 · 68 real past-paper questions below

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Why Factorising Solves It

If two numbers multiply to make 0, one of them must be 0. That is the whole idea.

Rule

If $(x + 6) (x - 4) = 0$ then $x + 6 = 0$ or $x - 4 = 0$ . So each bracket gives one answer.

Key idea

Most questions come in two parts: factorise (2 marks), then “hence solve” (1 mark). Always write the brackets line. Most of the marks are for it, so answers with no brackets, found by guessing or on a calculator, can score 0.

Factorise, Then Solve

Find two numbers that multiply to the last number and add to the middle number.

Worked example

1
$x^{2} + 2 x - 24 = 0$ — the equation
2
$6 \times - 4 = - 24, 6 + (- 4) = 2$ — multiply to −24, add to 2
3
$(x + 6) (x - 4) = 0$ — the two brackets
4
$x = - 6 or x = 4$ — each bracket equal to 0

Common mistake

Giving x = 6 and x = −4. The answers have the opposite sign to the numbers in the brackets.

Finding the Pair

If the pair is not obvious, list the factor pairs of the last number and check which adds to the middle.

Worked example

1
$x^{2} - 5 x - 36 = 0$ — multiply to −36, add to −5
2
$(3, - 12), (4, - 9), (6, - 6)$ — try pairs: only 4 and −9 add to −5
3
$(x + 4) (x - 9) = 0$ — the brackets
4
$x = - 4 or x = 9$ — opposite signs

Key idea

If the last number is negative, the pair has one plus and one minus. If the last number is positive and the middle is negative, both numbers are negative: $x^{2} - 18 x + 45 = (x - 3) (x - 15)$ .

Make It Equal Zero First

If the equation is not “= 0”, move everything to one side before you factorise.

Worked example

1
$x^{2} + 3 x = 18$ — not equal to 0 yet
2
$x^{2} + 3 x - 18 = 0$ — take 18 from both sides
3
$(x + 6) (x - 3) = 0$ — 6 × −3 = −18, 6 − 3 = 3
4
$x = - 6 or x = 3$ — two answers

Exams also put x² and x terms on the right. Move every term, not just the number.

Worked example

1
$2 x^{2} + x = x^{2} + 5 x + 21$ — Now $x^{2}$ and x terms are on both sides. Bring them all to the left, one move at a time.
2
$x^{2} + x = 5 x + 21$ — Take x² from both sides.
3
$x^{2} - 4 x = 21$ — Take 5x from both sides: x − 5x = −4x.
4
$x^{2} - 4 x - 21 = 0$ — Take 21 from both sides. Now it equals 0.
5
$(x - 7) (x + 3) = 0$ — −7 × 3 = −21 and −7 + 3 = −4.
6
$x = 7 or x = - 3$ — Each bracket equal to 0.

Common mistake

Factorising x(x + 3) = 18 and saying x = 18. That only works when the right side is 0.

Common mistake

Dividing $x^{2} = 3 x$ by x to get x = 3. You lose x = 0. Instead make it $x^{2} - 3 x = 0$ , take out x: $x (x - 3) = 0$ , so x = 0 or x = 3.

Reject the Impossible Answer

Many exam questions are about a shape, a price or a time. Solving gives two answers, but only one can be real.

A rectangle is x cm wide and (x + 3) cm long. Its area is 40 cm². Find x.

Worked example

1
$x (x + 3) = 40$ — Area = width × length.
2
$x^{2} + 3 x = 40$ — Expand the bracket: x × x and x × 3.
3
$x^{2} + 3 x - 40 = 0$ — Take 40 from both sides so it equals 0.
4
$(x + 8) (x - 5) = 0$ — 8 × (−5) = −40 and 8 + (−5) = 3.
5
$x = - 8 or x = 5$ — Each bracket equal to 0.
6
$x = 5$ — Reject −8: a width cannot be negative.
7
$5 \times 8 = 40$ — Check: width 5, length 8, area 40. It works.

Common mistake

Leaving both answers. Write the one that makes sense and say why the other is rejected. The next part of the question often uses it.

Key idea

Story questions and equations with fractions that turn into quadratics are in the lesson “Equations That Become Quadratics”.

A Number in Front of x²

Harder questions put a number in front of x². The brackets then look like (2x − 3), and one answer is a fraction.

Worked example

1
$2 x^{2} + 7 x - 15 = 0$ — The question. A 2 in front of x², so one bracket starts with 2x.
2
$(2 x - 3) (x + 5) = 0$ — Factorise as in “Factorising Quadratic Expressions”. Check: 2x × 5 − 3 × x = 10x − 3x = 7x.
3
$2 x - 3 = 0 or x + 5 = 0$ — Each bracket equal to 0.
4
$2 x = 3$ — Add 3 to both sides.
5
$x = \frac{3}{2}$ — Divide by 2. The answer is a fraction, and that is fine.
6
$x = \frac{3}{2} or x = - 5$ — The second bracket gives −5 as usual.

Common mistake

Giving x = 3 from (2x − 3). You must also divide by the 2, so x = 3/2.

Which Method? Read the Question

Check the right side is 0, then look at the numbers.

You see	You do
“Factorise … hence solve”	Brackets, then each = 0
Not equal to 0	Move terms to make = 0
Hard-to-find pair	List factor pairs
It will not factorise	See Quadratic Formula
A length, price or time	Reject the impossible answer
A number in front of x²	Brackets like (2x − 3), then divide

Key idea

Check an answer: x = 4 in x² + 2x − 24 gives 16 + 8 − 24 = 0.

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