Solving quadratics by factorising

Foundation and Higher

The idea

If two things multiply to make zero, one of them must be zero. So once $x^2 - x - 12 = 0$ is factorised as $(x - 4)(x + 3) = 0$ , the solutions are $x = 4$ and $x = -3$ .

Rearrange so one side is $0$ .
Factorise the quadratic.
Set each bracket equal to zero and solve both.

Watch out $(x - 4) = 0$ gives $x = 4$ , not $x = -4$ .

See it worked

Worked exampleNon-calculator3 marks

Solve each equation.

(a) Solve $x^2 + 5x + 6 = 0$

(b) Solve $x^2 - 16 = 0$

(a) Factorise. $(x + 2)(x + 3) = 0$
One of the brackets must be zero: $x + 2 = 0$ or $x + 3 = 0$ , so $x = -2$ or $x = -3$ .
(b) $(x + 4)(x - 4) = 0$ , so $x = -4$ or $x = 4$ .

Answer (a) $x = -2$ or $x = -3$ ; (b) $x = 4$ or $x = -4$

Your turn

Question 1Non-calculator4 marks

Solve each equation.

(a) Solve $x^2 - 7x = 0$

(b) Solve $x^2 - 2x - 15 = 0$

Question 2Calculator allowed1 mark

The equation $x^{2}$ + kx + 18 = 0 has two positive integer roots. The roots differ by 3. What is k?

Question 3Calculator allowed3 marks

The two roots of $x^{2}$ $-$ 6x + c = 0 are positive. One root is three times the other. Find c.

In past papers

57 official questions in 55 papers since 2017. The latest:

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