Solving by factorisation

The idea: a product equal to zero

If two numbers multiply to give $0$, at least one of them must be $0$. This simple fact solves quadratic equations.

Suppose $(x - 3)(2x + 5) = 0$. Then

• either $x - 3 = 0$, which gives $x = 3$,
• or $2x + 5 = 0$, which gives $x = -\dfrac{5}{2}$.

So the roots are $3$ and $-\dfrac{5}{2}$. The whole job is to write the quadratic as a product of two linear factors. You met factorising in the chapter on polynomials. Here we use it to solve.

Splitting the middle term

To factorise $ax^2 + bx + c$, find two numbers $p$ and $q$ with

Then write $bx$ as $px + qx$ and group the terms in pairs.

A worked example. Solve $6x^2 - x - 12 = 0$.

1. The target. $a \times c = 6 \times (-12) = -72$, and $b = -1$.
2. Find the pair. Two numbers with product $-72$ and sum $-1$: $-9$ and $8$.
3. Split. $6x^2 - 9x + 8x - 12 = 0$.
4. Group. $3x(2x - 3) + 4(2x - 3) = 0$, so $(3x + 4)(2x - 3) = 0$.
5. Solve. $3x + 4 = 0$ gives $x = -\dfrac{4}{3}$, and $2x - 3 = 0$ gives $x = \dfrac{3}{2}$.
6. Check. For $x = \dfrac{3}{2}$: $6 \times \dfrac{9}{4} - \dfrac{3}{2} - 12 = \dfrac{27}{2} - \dfrac{3}{2} - 12 = 0$.

Signs help you search. If $a \times c$ is positive, $p$ and $q$ have the same sign as $b$. If $a \times c$ is negative, one is positive and one is negative.

Special cases

• No constant term. $5x^2 - 15x = 0$ gives $5x(x - 3) = 0$, so $x = 0$ or $x = 3$.
• A difference of squares. $x^2 - 49 = 0$ gives $(x - 7)(x + 7) = 0$, so $x = 7$ or $x = -7$.
• A perfect square. $x^2 - 10x + 25 = 0$ gives $(x - 5)^2 = 0$. Both factors give $x = 5$: the equation has one repeated root.

Do not divide by $x$. From $x^2 = 4x$, dividing by $x$ gives only $x = 4$. But $x = 0$ is also a root! Instead, write $x^2 - 4x = 0$, so $x(x - 4) = 0$, and both roots appear.

Practise with new numbers

Try it in Python

Finding the pair by hand can take a while when $a \times c$ is large. This program tries every whole number $p$ that divides $a \times c$, and prints the pair whose sum is $b$.

Common mistakes

• Using $c$ instead of $a \times c$. For $6x^2 - x - 12$, the pair must multiply to $-72$, not $-12$.
• Wrong signs in the roots. The factor $(x + 4)$ gives $x = -4$, not $x = 4$.
• Dividing by $x$. It loses the root $x = 0$.
• Not checking. A quick check in the original equation catches most slips.

Remember these

• If a product is $0$, one of its factors is $0$.
• Split the middle term: $p + q = b$ and $p \times q = a \times c$.
• Each linear factor gives one root. A perfect square gives a repeated root.
• Bring every term to one side. Never divide both sides by $x$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Solve $x^2 + 2x - 15 = 0$.

2. Solve $3x^2 - 5x + 2 = 0$.

3. Solve $x^2 - 16 = 0$ by factorising.

4. Ravi divides both sides of $x^2 = 5x$ by $x$ and writes $x = 5$. Which root has he lost?

5. Solve $9x^2 - 6x + 1 = 0$.