Solving Quadratics by Factorising

How do I solve a quadratic equation using factorisation?

Rearrange it into the form ax² + bx + c = 0
- zero must be on one side
  - it is easier to use the side where a is positive
Factorise the quadratic and solve each bracket equal to zero
- If (x + 4)(x - 1) = 0, then either x + 4 = 0 or x - 1 = 0
  - Because if A × B = 0, then either A = 0 or B = 0
To solve $(x - 3) (x + 7) = 0$
- …solve “first bracket = 0”:
  - x – 3 = 0
  - add 3 to both sides: x = 3
- …and solve “second bracket = 0”
  - x + 7 = 0
  - subtract 7 from both sides: x = -7
- The two solutions are x = 3 or x = -7
  - The solutions have the opposite signs to the numbers in the brackets
To solve $(2 x - 3) (3 x + 5) = 0$
- …solve “first bracket = 0”
  - 2x – 3 = 0
  - add 3 to both sides: 2x = 3
  - divide both sides by 2: x = $\frac{3}{2}$
- …solve “second bracket = 0”
  - 3x + 5 = 0
  - subtract 5 from both sides: 3x = -5
  - divide both sides by 3: x = $- \frac{5}{3}$
- The two solutions are x = $\frac{3}{2}$ or x = $- \frac{5}{3}$
To solve $x (x - 4) = 0$
- it may help to think of x as (x – 0) or (x)
- …solve “first bracket = 0”
  - (x) = 0, so x = 0
- …solve “second bracket = 0”
  - x – 4 = 0
  - add 4 to both sides: x = 4
- The two solutions are x = 0 or x = 4
  - It is a common mistake to divide both sides by x at the beginning - you will lose a solution (the x = 0 solution)

Examiner Tip

Use a calculator to check your final solutions!
- Calculators also help you to factorise (if you're struggling with that step)
- A calculator gives solutions to $6 x^{2} + x - 2 = 0$ as x = $- \frac{2}{3}$ and x = $\frac{1}{2}$
  - "Reverse" the method above to factorise!
  - $6 x^{2} + x - 2 = (3 x + 2) (2 x - 1)$
- Warning: a calculator gives solutions to 12x² + 2x – 4 = 0 as x = $- \frac{2}{3}$ and x = $\frac{1}{2}$
  - But 12x² + 2x – 4 ≠ $(3 x + 2) (2 x - 1)$ as these brackets expand to 6x² + ... not 12x² + ...
  - Multiply by 2 to correct this
  - 12x² + 2x – 4 = $2 (3 x + 2) (2 x - 1)$

Worked example

(a)

Solve $(x - 2) (x + 5) = 0$

Set the first bracket equal to zero

x – 2 = 0

Add 2 to both sides

x = 2

Set the second bracket equal to zero

x + 5 = 0

Subtract 5 from both sides

x = -5

Write both solutions together using “or”

x = 2 or x = -5

(b)

Solve

(8 x + 7) (2 x - 3) = 0

Set the first bracket equal to zero

8x + 7 = 0

Subtract 7 from both sides

8x = -7

Divide both sides by 8

x = $- \frac{7}{8}$

Set the second bracket equal to zero

2x - 3 = 0

Add 3 to both sides

2x = 3

Divide both sides by 2

x = $\frac{3}{2}$

Write both solutions together using “or”

x = $- \frac{7}{8}$ or x = $\frac{3}{2}$

(c)

Solve $x (5 x - 1) = 0$

Do not divide both sides by x (this will lose a solution at the end)
Set the first “bracket” equal to zero

(x) = 0

Solve this equation to find x

x = 0

Set the second bracket equal to zero

5x - 1 = 0

Add 1 to both sides

5x = 1

Divide both sides by 5

x = $\frac{1}{5}$

Write both solutions together using “or”

x = 0 or x = $\frac{1}{5}$

Solving Quadratics by Factorising (Cambridge O Level Maths)

Revision Note