Factorising Quadratic Expressions

Factorising expressions of the form x² + bx + c and ax² + bx + c into two brackets using the grid method.

In the exam for Edexcel IGCSE Maths 4MA1, Cambridge O Level Maths 4024, Cambridge IGCSE Maths 0580, Edexcel Further Pure Maths 4PM1 · 40 real past-paper questions below

اقرأ هذا الدرس مشروحاً بالعربي ←

What Factorising Means

You already know how to expand: $(x + 3) (x + 2)$ becomes $x^{2} + 5 x + 6$ .

Factorising is the same journey backwards. You start with $x^{2} + 5 x + 6$ and find the two brackets it came from.

Example

Think of 12. Writing it as $3 \times 4$ is factorising a number. Factorising a quadratic is the same idea: write it as one bracket times another.

Rule

Why bother? In the exam you factorise to solve an equation and to simplify a fraction. It is a tool you will use again and again.

Key idea

Factorising undoes expanding. You can always check your brackets by expanding them back.

Key idea

About 3,800 years ago, a scribe in Babylon (in today's Iraq) set this problem on a clay tablet: the area of a square plus its side makes three quarters. That is $x^{2} + x = \frac{3}{4}$ , a quadratic equation, solved step by step without a single letter.

How: Sum and Product

Look at how two brackets expand. The two numbers inside do two jobs at once.

Formula

(x + p) (x + q) = x^{2} + (p + q) x + pq

$p$ and $q$ are the numbers in the brackets. They add to make the middle number, $p + q$ , and multiply to make the last number, $pq$ . For example, $(x + 3) (x + 2) = x^{2} + 5 x + 6$ : 3 + 2 = 5 and 3 × 2 = 6.

Steps

1
Find two numbers that multiply to the last number.
2
Of those, pick the pair that adds to the middle number.
3
Put one number in each bracket.
4
Check by expanding.

Worked example

1
$x^{2} + 2 x - 35$ — The question. We need two numbers that multiply to −35 and add to +2.
2
$1 \times (- 35), (- 1) \times 35, 5 \times (- 7), (- 5) \times 7$ — List every pair that multiplies to −35. One number must be negative, because only + × − gives a minus answer.
3
$- 5 + 7 = 2$ — Add each pair. Only −5 and 7 make +2, so they are the pair.
4
$(x - 5) (x + 7)$ — Put one number in each bracket with its sign: −5 goes in as x − 5, and 7 as x + 7.
5
$x^{2} + 7 x - 5 x - 35$ — Check by expanding: x × x, then x × 7, then −5 × x, then −5 × 7.
6
$= x^{2} + 2 x - 35$ — 7x − 5x = 2x. It matches the question, so the brackets are right.

Last number positive

Both signs are the same as the middle sign.

y^{2} - 9 y + 14

= (y - 2) (y - 7)

Last number negative

One + and one −. The bigger number takes the middle sign.

x^{2} + 2 x - 35

= (x - 5) (x + 7)

Common mistake

Picking a pair that adds correctly but has the wrong signs. Always expand to check — it takes ten seconds.

Both Numbers Negative

Many exam questions have a minus in the middle and a plus at the end, like $x^{2} - 10 x + 16$ . Then both numbers in the brackets are negative.

Worked example

1
$x^{2} - 10 x + 16$ — The question. We need two numbers that multiply to +16 and add to −10.
2
$(- 1) \times (- 16), (- 2) \times (- 8), (- 4) \times (- 4)$ — The product is +, so both signs are the same. The sum is −, so both must be negative. List the negative pairs.
3
$(- 2) + (- 8) = - 10$ — Add each pair: −17, −10, −8. Only −2 and −8 make −10.
4
$(x - 2) (x - 8)$ — One number in each bracket. Both brackets get a minus.
5
$x^{2} - 8 x - 2 x + 16$ — Check by expanding. The last term is $(- 2) \times (- 8) = + 16$ , because minus times minus is plus.
6
$= x^{2} - 10 x + 16$ — −8x − 2x = −10x. It matches the question.

Common mistake

Writing $(x + 2) (x + 8)$ . It multiplies to +16 but adds to +10, not −10. A minus in the middle with a plus at the end means two minus signs.

Example

Two letters work the same way. For $p^{2} - 8 pq + 15 q^{2}$ , find two numbers that multiply to 15 and add to −8: −3 and −5. Each one carries a q, so $p^{2} - 8 pq + 15 q^{2} = (p - 3 q) (p - 5 q)$ .

When There Is a Number in Front of x²

Some questions start with $3 x^{2}$ or $6 y^{2}$ instead of just $x^{2}$ . Then one extra step helps: split the middle term.

Steps

1
Multiply the first number by the last number.
2
Find two numbers that multiply to that and add to the middle number.
3
Use them to split the middle term into two.
4
Take out a common factor from each half. The brackets will match.

Worked example

1
$3 x^{2} - 10 x - 8$ — The question. The number in front of x² is 3 and the last number is −8.
2
$3 \times (- 8) = - 24$ — Multiply first × last. Now we need two numbers that multiply to −24 and add to the middle number, −10.
3
$(1, - 24), (2, - 12), (3, - 8), (4, - 6)$ — List the pairs in order, starting from 1. The sum −10 is negative, so the bigger number gets the minus. Their sums are −23, −10, −5, −2.
4
$2 + (- 12) = - 10$ — The pair is 2 and −12. Listing in order always finds it, even when a × c is large.
5
$3 x^{2} - 12 x + 2 x - 8$ — Split −10x into −12x + 2x. The value has not changed, because −12x + 2x = −10x.
6
$3 x (x - 4) + 2 x - 8$ — Take 3x out of the first two terms: 3x² ÷ 3x = x and −12x ÷ 3x = −4.
7
$3 x (x - 4) + 2 (x - 4)$ — Take 2 out of the last two terms: 2x ÷ 2 = x and −8 ÷ 2 = −4. Both brackets are (x − 4), so you are on track.
8
$(3 x + 2) (x - 4)$ — (x − 4) is in both parts, so take it out. What is left, 3x + 2, is the other bracket.
9
$3 x^{2} - 12 x + 2 x - 8 = 3 x^{2} - 10 x - 8$ — Check by expanding. It matches the question.

Key idea

If the two brackets after splitting do not match, you made a slip. Go back and check the pair of numbers.

Perfect Squares

Sometimes the first and last terms are both squares, like $9 x^{2}$ and 25. Then the answer may be one bracket squared.

Formula

a^{2} + 2 ab + b^{2} = (a + b)^{2} a^{2} - 2 ab + b^{2} = (a - b)^{2}

$a$ and $b$ are the things being squared. The middle term must be 2 × a × b — that is how you test if the pattern fits.

Worked example

1
$9 x^{2} - 30 x + 25$ — The question. The first and last terms are both squares, so test the pattern.
2
$9 x^{2} = (3 x)^{2}, 25 = 5^{2}$ — Find what is squared: a = 3x and b = 5.
3
$2 \times 3 x \times 5 = 30 x$ — Test the middle: 2ab = 30x. The question has −30x: the same size, so the pattern fits. The minus decides the sign next.
4
$(3 x - 5)^{2}$ — The middle term has a minus, so use the $(a - b)^{2}$ form.
5
$(3 x - 5) (3 x - 5) = 9 x^{2} - 15 x - 15 x + 25$ — Check by expanding: −15x − 15x = −30x. It matches the question.

Common mistake

Writing $(3 x - 5) (3 x + 5)$ . That expands to $9 x^{2} - 25$ with no middle term. A perfect square has the same sign in both brackets.

Key idea

Not sure it is a perfect square? The split method still works: 9 × 25 = 225, and (−15) × (−15) = 225 and (−15) + (−15) = −30.

Take Out a Common Factor First

When the question says factorise completely, look for a number that divides every term. Take it out first — the brackets become much easier.

Worked example

1
$3 x^{2} - 6 x - 45$ — The question. 3, 6 and 45 are all in the 3 times table.
2
$3 (x^{2} - 2 x - 15)$ — Take out the 3: 3x² ÷ 3 = x², −6x ÷ 3 = −2x and −45 ÷ 3 = −15.
3
$(- 5) \times 3 = - 15, - 5 + 3 = - 2$ — Factorise the bracket: we need two numbers that multiply to −15 and add to −2. They are −5 and 3.
4
$3 (x - 5) (x + 3)$ — Put them in brackets and keep the 3 in front. It is part of the answer.
5
$3 (x^{2} - 2 x - 15) = 3 x^{2} - 6 x - 45$ — Check: (x − 5)(x + 3) = x² − 2x − 15, then × 3 gives the question back.

Common mistake

Stopping at $(3 x - 15) (x + 3)$ . It expands correctly, but 3x − 15 still has a common 3, so it is not factorised completely and loses a mark.

Key idea

If x² has a minus, take out −1 the same way. It flips every sign inside: $- x^{2} + 2 x + 15 = - (x^{2} - 2 x - 15) = - (x - 5) (x + 3)$ .

When the Exam Says “Hence Solve”

Very often the exam asks you to factorise, then hence solve the equation = 0. “Hence” means: use the brackets you just found.

Rule

If two things multiply to make 0, one of them must be 0. So each bracket can be set equal to 0.

Worked example

1
$x^{2} + 2 x - 35 = 0$ — The equation. It must equal 0 before the brackets can help, and it already does.
2
$(x - 5) (x + 7) = 0$ — Swap in the brackets we found earlier. “Hence” means use them — do not start again.
3
$x - 5 = 0 or x + 7 = 0$ — Two brackets multiply to 0, so one of them must be 0. Set each one equal to 0.
4
$x = 5$ — Solve the first: add 5 to both sides.
5
$x = - 7$ — Solve the second: take 7 from both sides.
6
$5^{2} + 2 \times 5 - 35 = 0$ — Check one answer in the equation: 25 + 10 − 35 = 0. Correct.

Common mistake

Giving $x = - 5$ and $x = 7$ . The signs flip when you solve: $x - 5 = 0$ gives $x = 5$ .

Key idea

With a number in front it works the same way: $3 x + 2 = 0$ gives $3 x = - 2$ , so $x = - \frac{2}{3}$ .

Which Method? Read the Question

Before you start, look at the number in front of $x^{2}$ and at the words of the question.

You see	You do
Every term shares a number	Take it out first, then factorise the bracket
Just $x^{2}$	Two numbers: multiply to the last, add to the middle
A number in front, like $3 x^{2}$	Multiply first × last, then split the middle
First and last terms are squares	Try $(a + b)^{2}$ or $(a - b)^{2}$
“Hence solve … = 0”	Set each bracket equal to 0

Key idea

Factorising is usually 2 marks. Brackets with the right product or the right sum already earn 1 of them, so always write your brackets down. “Hence solve” is 1 more mark and uses your brackets — expand to check them first.

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