POP

G5.7 Expanding and Factorising Quadratics

Image · 914 × 270

01 — Explanation

How do I expand two brackets together?

Multiply every term in the first bracket by every term in the second. The grid method is the most reliable way to track signs: draw a $2\times2$ grid, label rows/columns with each bracket's terms (keeping their signs), multiply to fill each cell, then collect like terms.

Example: $(x+1)(x+3)$ → cells give $x^2, 3x, x, 3$ → total $x^2+4x+3$ .

Exam Tip

$(x+k)^2 \ne x^2+k^2$ — this is one of the most heavily tested algebra errors at GCSE. Always rewrite $(x+3)^2$ as $(x+3)(x+3)$ first; the middle term $2kx$ comes from adding the Outer and Inner products and is easy to drop if you skip this step.

02 — Explanation

How do I factorise a quadratic when the x^2 coefficient is 1?

For $x^2+bx+c$ , find two numbers $m,n$ with $m\times n=c$ and $m+n=b$ . The sign of $c$ tells you whether $m,n$ have the same sign (if $c>0$ ) or opposite signs (if $c<0$ ); the sign of $b$ then tells you which.

Example: Factorise $x^2-2x-8$ . Factors of $-8$ summing to $-2$ : $+2$ and $-4$ . So $(x+2)(x-4)$ .

Exam Tip

Before factorising into two brackets, always check for a common factor across all terms first — $3x^2+6x-9$ must become $3(x^2+2x-3)$ before the quadratic inside is factorised, or the answer isn't "fully factorised" and loses marks.

03 — Explanation

How do I factorise when the x^2 coefficient isn't 1?

Use the AC method (splitting the middle term):

Multiply $a\times c$ .
Find two factors of $ac$ that sum to $b$ .
Rewrite the middle term $bx$ using those two factors.
Factorise the resulting four-term expression by grouping in pairs.

Example: Factorise $2x^2+5x+3$ .

$ac=2\times3=6$ .
Factors of 6 summing to 5: $2$ and $3$ .
Rewrite: $2x^2+2x+3x+3$ .
Group: $2x(x+1)+3(x+1) = (x+1)(2x+3)$ .

Exam Tip

After grouping, the two brackets must be identical — if they're not, you've made a sign error earlier and should re-check your factor pair.

04 — Explanation

What's special about the "difference of two squares"?

When a quadratic has no middle term and both remaining terms are perfect squares separated by a minus sign, $x^2-k^2$ factorises instantly as $(x-k)(x+k)$ — because the Inner and Outer products cancel out. This extends to non-monic cases too: $4x^2-9 = (2x-3)(2x+3)$ , since $\sqrt{4x^2}=2x$ and $\sqrt9=3$ .

05 — Explanation

How do I check my factorisation is correct?

Always expand your answer back out using FOIL or the grid method — if it doesn't match the original expression exactly, an error has crept in. This "Factorise → Expand → Verify" loop should be automatic every time.

Worked Example

Factorise fully: $2x^2-18$ .

Extract the common factor first: $2(x^2-9)$ .
Recognise the difference of two squares: $x^2-9=(x-3)(x+3)$ .
Final answer: $2(x-3)(x+3)$ .

Exam Tip

Forgetting the initial factor of 2 is one of the most common ways marks are lost here — "fully factorised" means nothing further can be taken out of any part of the answer.

06 — Explanation

Why does factorising matter beyond this topic on its own?

Factorising a quadratic is rarely the final goal — it's usually a tool used inside a bigger problem, most commonly simplifying an algebraic fraction. A question might ask you to simplify $\dfrac{x^2-5x+6}{x-3}$ : factorising the numerator gives $\dfrac{(x-2)(x-3)}{x-3}$ , and the common $(x-3)$ factor cancels, leaving $x-2$ . Recognising when a numerator or denominator can be factorised — rather than treating expansion and factorisation as isolated exercises — is what connects this topic to the harder algebraic-fraction and equation-solving questions later in the course.