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 $x^2+4x+3$.

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 $(x+2)(x-4)$.

Use the AC method (splitting the middle term):

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

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

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

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 $\sqrt9=3$.

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: $2x^2-18$.

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

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.