Algebra

Factorising Quadratic Equations

Introduction

A quadratic equation in one unknown has the form \(ax^2 + bx + c = 0\) where \(a\), \(b\) and \(c\) are real numbers and \(a \neq 0\).

The simplest method of solution, but one that is not always applicable, is by factorising. To use this method we need to use the zero-factor theorem that states:

If \(a\) and \(b\) are both complex numbers, and if \(ab = 0\), then either \(a = 0\) or \(b = 0\), or both.

Examples

Example

Factorise \((3y - 2)(y + 5) = 0\)

Question

Factorise \((3y - 2)(y + 5) = 0\)

Solution

1. Step 1 — Using the zero-factor theorem, equate each factor to 0
   \[ (3y - 2)(y + 5) = 0 \]

   Then either

   \[ 3y - 2 = 0 \Rightarrow y = \frac{2}{3} \]

   or

   \[ y + 5 = 0 \Rightarrow y = -5 \]
2. Step 2 — State the solutions

   The solutions are \(y = \frac{2}{3}\) or \(y = -5\).

Example

Factorise \(4x^2 + 9x + 2 = 0\)

Question

Factorise \(4x^2 + 9x + 2 = 0\)

Solution

1. Step 1 — Factorise
   \[ 4x^2 + 9x + 2 = 0 \]
   \[ (x + 2)(4x + 1) = 0 \]
2. Step 2 — Using the zero-factor theorem, equate each factor to 0

   Then either

   \[ x + 2 = 0 \Rightarrow x = -2 \]

   or

   \[ 4x + 1 = 0 \Rightarrow x = \frac{-1}{4} \]
3. Step 3 — State the solutions

   The solutions are \(x = -2\) or \(x = \frac{-1}{4}\).

Example

Factorise \(5y = 36 - y^2\)

Question

Factorise \(5y = 36 - y^2\)

Solution

1. Step 1 — Rearrange in the form \(ax^2 + bx + c = 0\)
   \[ 5y = 36 - y^2 \]
   \[ y^2 + 5y - 36 = 0 \]
2. Step 2 — Factorise
   \[ (y - 4)(y + 9) = 0 \]
3. Step 3 — Using the zero-factor theorem, equate each factor to 0

   Then either

   \[ y - 4 = 0 \Rightarrow y = 4 \]

   or

   \[ y + 9 = 0 \Rightarrow y = -9 \]
4. Step 4 — State the solutions

   The solutions are \(y = 4\) or \(y = -9\).

Example

Factorise \(4x^2 - 2x - 20 = x^2 + 9x\)

Question

Factorise \(4x^2 - 2x - 20 = x^2 + 9x\)

Solution

1. Step 1 — Rearrange in the form \(ax^2 + bx + c = 0\) and combine similar terms
   \[ 4x^2 - 2x - 20 = x^2 + 9x \]
   \[ 3x^2 - 11x - 20 = 0 \]
2. Step 2 — Factorise
   \[ (3x + 4)(x - 5) = 0 \]
3. Step 3 — Using the zero-factor theorem, equate each factor to 0

   Then either

   \[ 3x + 4 = 0 \Rightarrow x = \frac{-4}{3} \]

   or

   \[ x - 5 = 0 \Rightarrow x = 5 \]
4. Step 4 — State the solutions

   The solutions are \(x = \frac{-4}{3}\) or \(x = 5\).

Example

Factorise \(x + 12 = \frac{12}{x+1}\)

Question

Factorise \(x + 12 = \dfrac{12}{x+1}\)

Solution

1. Step 1 — Multiply both sides of the equation by \(x + 1\)
   \[ x + 12 = \frac{12}{x+1} \]
   \[ (x + 1)(x + 12) = 12 \]
2. Step 2 — Expand the brackets
   \[ x^2 + 13x + 12 = 12 \]
3. Step 3 — Subtract 12 from both sides
   \[ x^2 + 13x = 0 \]
4. Step 4 — Factorise
   \[ x(x + 13) = 0 \]
5. Step 5 — Using the zero-factor theorem, equate each factor to 0

   Then either

   \[ x = 0 \]

   or

   \[ x + 13 = 0 \Rightarrow x = -13 \]
6. Step 6 — State the solutions

   The solutions are \(x = 0\) or \(x = -13\).