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Factorising Quadratics

A quadratic like \(x^2 + bx + c\) can often be rewritten as two brackets multiplied together. That form is the key to solving it, because a product equal to zero hands you the answers directly.

MYP 4Standard MathsAlgebraCriteria A · C · D~11 min read

Written by Fayek Jawad Haider · MYP student

What we are doing

Expanding \((x + 3)(x + 4)\) gives \(x^2 + 7x + 12\). Factorising a quadratic runs that backwards: starting from \(x^2 + 7x + 12\), you find the two brackets it came from.

Quadratic expression
An expression whose highest power is \(x^2\), here of the form \(x^2 + bx + c\).

For a quadratic that starts with a single \(x^2\), the two brackets both begin with \(x\): \((x + p)(x + q)\). The whole job is finding the right numbers \(p\) and \(q\).

Finding the two numbers

Expanding \((x + p)(x + q)\) gives \(x^2 + (p + q)x + pq\). So you need two numbers that multiply to \(c\) and add to \(b\). Find them and the factorising is done.

Worked example

Factorise \(x^2 + 7x + 12\).

1
You need two numbers that multiply to \(12\) and add to \(7\).
2
List factor pairs of \(12\): \(1\times12\), \(2\times6\), \(3\times4\). The pair \(3\) and \(4\) adds to \(7\).
3
Place them in the brackets: \((x + 3)(x + 4)\).
\((x + 3)(x + 4)\)

Product first, then check the sum

Always list the pairs that multiply to \(c\) first; there are only a few. Then pick the pair whose sum matches \(b\). Expanding your brackets back is a quick way to confirm.

Handling the signs

The signs of the two numbers follow from the signs of \(b\) and \(c\). Reading them correctly is what separates a right answer from a near miss.

Sign of \(c\)Sign of \(b\)The two numbers
positivepositiveboth positive
positivenegativeboth negative
negativeeitherone positive, one negative
Worked example

Factorise \(x^2 - x - 6\).

1
Here \(c = -6\) is negative, so the two numbers have opposite signs, and they add to \(b = -1\).
2
Pairs multiplying to \(-6\): \((-3, 2)\) adds to \(-1\). That is the pair.
3
Write the brackets: \((x - 3)(x + 2)\).
\((x - 3)(x + 2)\)

Solving by the null factor law

The null factor law says that if two things multiply to give zero, at least one of them must be zero. So once a quadratic equation is factorised and equal to zero, set each bracket to zero in turn.

\[ (x + p)(x + q) = 0 \quad\Rightarrow\quad x + p = 0 \ \text{ or } \ x + q = 0 \]
Worked example

Solve \(x^2 + 5x + 6 = 0\).

1
Two numbers multiply to \(6\) and add to \(5\): that is \(2\) and \(3\). Factorise: \((x + 2)(x + 3) = 0\).
2
By the null factor law, \(x + 2 = 0\) or \(x + 3 = 0\).
3
Solving each gives \(x = -2\) or \(x = -3\).
\(x = -2 \ \text{ or } \ x = -3\)

Check yourself

1. Factorise \(x^2 + 8x + 15\). +

Two numbers multiply to \(15\) and add to \(8\): that is \(3\) and \(5\). Answer: \((x + 3)(x + 5)\).

2. Factorise \(x^2 - 2x - 8\). +

Since \(c = -8\) is negative, the numbers have opposite signs and add to \(-2\): that is \(-4\) and \(2\). Answer: \((x - 4)(x + 2)\).

3. Solve \(x^2 + x - 12 = 0\). +

Numbers multiply to \(-12\) and add to \(1\): that is \(4\) and \(-3\), so \((x + 4)(x - 3) = 0\). By the null factor law, Answer: \(x = -4\) or \(x = 3\).

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