How to factorise: common factors and quadratics

Find the HCF of the terms (numbers and letters) and put it outside:

• $6x + 9 = 3(2x + 3)$
• $4x^2 + 10x = 2x(2x + 5)$
• $12a^2b - 8ab^2 = 4ab(3a - 2b)$

"Factorise fully" means take out the highest common factor. $4x^2 + 10x = 2(2x^2 + 5x)$ isn't fully factorised. For the number part, see how to find the HCF.

Find two numbers that multiply to $c$ and add to $b$:

• $x^2 + 7x + 12$: $3 \times 4 = 12$ and $3 + 4 = 7$, so $(x + 3)(x + 4)$.
• $x^2 - x - 12$: $-4 \times 3 = -12$ and $-4 + 3 = -1$, so $(x - 4)(x + 3)$.
• $x^2 - 10x + 21$: $-3 \times -7 = 21$ and $-3 + -7 = -10$, so $(x - 3)(x - 7)$.

Sign shortcut: if $c$ is positive, both numbers have the same sign as $b$. If $c$ is negative, the signs are different.

$a^2 - b^2 = (a + b)(a - b)$

• $x^2 - 49 = (x + 7)(x - 7)$
• $4x^2 - 25 = (2x + 5)(2x - 5)$
• $2x^2 - 18 = 2(x^2 - 9) = 2(x + 3)(x - 3)$ (common factor first)

A sum of two squares, like $x^2 + 9$, doesn't factorise this way.

$2x^2 + 7x + 3$ (GCSE Higher)

1. Multiply $a \times c$: $2 \times 3 = 6$.
2. Find two numbers that multiply to 6 and add to 7: 6 and 1.
3. Split the middle term: $2x^2 + 6x + x + 3$.
4. Factorise in pairs: $2x(x + 3) + 1(x + 3)$.
5. Take out the shared bracket: $(2x + 1)(x + 3)$.

If a quadratic won't factorise, use the formula. Our quadratic equation solver shows the factorised form when there is one.

If $x^2 + 7x + 12 = 0$, then $(x + 3)(x + 4) = 0$, so one bracket must be zero: $x = -3$ or $x = -4$. Notice the signs flip. More in the quadratic equations topic guide.

1. $5x - 20$
2. $x^2 + 2x - 15$
3. $9x^2 - 1$
4. $3x^2 - 10x + 8$

Answers: $5(x - 4)$, $(x + 5)(x - 3)$, $(3x + 1)(3x - 1)$, $(3x - 4)(x - 2)$. Check each by expanding.