POP

G7.5 Factorising Harder Quadratics: $ax^2+bx+c$

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01 — Explanation

Why is factorising harder once a>1?

When $a=1$ (e.g. $x^2-4x-21$ ), you just need two numbers that multiply to $c$ and add to $b$ : here $(3,-7)$ gives $(x+3)(x-7)$ . Once $a>1$ , the brackets could split the $x$ -term many different ways — e.g. $6x^2+11x+3$ could come from $(6x\ldots)(x\ldots)$ or $(3x\ldots)(2x\ldots)$ — so guessing becomes unreliable and you need a systematic method.

02 — Explanation

What should I always check before factorising?

Take out any greatest common factor (GCF) first — it shrinks the numbers you're working with.

12x^2+18x+6 = 6(2x^2+3x+1)

Exam Tip: Don't confuse an expression with an equation. You can divide both sides of $4x^2+8x+4=0$ by 4, but you cannot "divide" the expression $4x^2+8x+4$ by 4 and drop it — the GCF must stay in the final factorised answer: $4(x+1)^2$ , not $(x+1)^2$ .

03 — Explanation

How does the 'AC method' factorise ax^2+bx+c?

Find the key product $P=a\times c$ .
Find two numbers $m,n$ that multiply to $P$ and add to $b$ .
Split the middle term using $m$ and $n$ , then factorise by grouping.

Worked example: Factorise $6x^2+17x+5$ .

$P=6\times5=30$ ; need factors of 30 adding to 17 → $2$ and $15$ .

6x^2+2x+15x+5 = 2x(3x+1)+5(3x+1) = (3x+1)(2x+5)

With negatives: Factorise $3x^2-11x-4$ . $P=-12$ ; factors adding to $-11$ → $1$ and $-12$ .

3x^2+x-12x-4 = x(3x+1)-4(3x+1) = (3x+1)(x-4)

Exam Tip: When the third of your four terms is negative, you'll almost always factor a negative number out of the second pair — check the result still leaves matching brackets.

04 — Explanation

How does the grid (box) method do the same thing visually?

Draw a 2×2 grid: put $ax^2$ top-left, $c$ bottom-right, and the two split terms ( $mx$ , $nx$ ) in the other two cells. Factorise each row and column — the sign of each factor must match the cell nearest it. Read the factors from the top and side of the grid.

For $2p^2-p-10$ ( $P=-20$ , factors $-5,4$ ):

\begin{array}{c|c|c} \\ & p & +2\\\hline \\ 2p & 2p^2 & +4p\\\hline \\ -5 & -5p & -10 \\ \end{array}

Factors: $(p+2)(2p-5)$ . This grid makes the sign of each GCF visual, which avoids the "negative grouping" slip that trips people up in the AC method.

05 — Explanation

How do I spot a special-case shortcut?

Difference of two squares ( $A^2-B^2$ , i.e. $b=0$ ): $25x^2-16=(5x+4)(5x-4)$ .

Perfect square trinomial ( $A^2\pm2AB+B^2$ ): check whether $ax^2$ and $c$ are both perfect squares and whether the middle term equals $2AB$ . For $4x^2-12x+9$ : $A=2x$ , $B=3$ , and $2(2x)(3)=12x$ matches, so it's $(2x-3)^2$ .

06 — Worked example

Worked exam-style question

Factorise $6y^2-y-5$ .

GCF: none.
$P=6\times(-5)=-30$ ; need factors adding to $-1$ → $-6$ and $5$ .
Split: $6y^2-6y+5y-5 = 6y(y-1)+5(y-1) = (6y+5)(y-1)$ .
Check by expanding: $(6y+5)(y-1)=6y^2-6y+5y-5=6y^2-y-5$ ✓.

Exam Tip: Even a partial attempt at the $ac$ product and splitting earns a method mark, so always write out your factor pairs rather than only writing a final (possibly wrong) answer.

07 — Common mistakes

Common mistakes to avoid

Dividing an expression by a common factor instead of factoring it out.
Finding factors of $ac$ that add to $b$ but transcribing them with the wrong sign.
Forgetting to check for GCF first, leaving unnecessarily large numbers to split.
Not expanding your final answer to double-check it matches the original expression.