Solving Cubic Equations (Edexcel IGCSE Further Pure Maths)

Revision Note

Written by: Roger B

Reviewed by: Dan Finlay

Solving Cubic Equations

How many real solutions can a cubic equation have?

A cubic equation $a x^{3} + b x^{2} + c x + d = 0$ will always have either one or three real roots (or solutions)
- Some of these roots may be repeated
- So it is possible to have one, two, or three unique solutions
A cubic with three real roots $α$ , $β$ and $γ$ can be written as a product of three linear factors
- $a x^{3} + b x^{2} + c x + d = (x - α) (x - β) (x - γ)$
  - Any two of the factors could be multiplied together to give a quadratic factor
A cubic with one real root $α$ can be written as the product of a linear and a quadratic factor
- $a x^{3} + b x^{2} + c x + d = (x - α) (p x^{2} + q x + r)$
  - The quadratic factor will not have any real roots
  - So its discriminant $q^{2} - 4 p r$ will be negative

How do I solve cubic equations?

Suppose you have a cubic equation $a x^{3} + b x^{2} + c x + d = 0$
An exam question will often give you one root
- You may be asked to show that the root is a solution to the equation
- Or you might have to find a root $x = α$ by substituting values into the equation until it equals 0
  - Try small integer values ( $1, - 1, 2, - 2, . . .$ )
If you know a root then you know a factor
- This is because of the factor theorem
  - If $x = α$ is a root, then $(x - α)$ is a factor
You can then divide $a x^{3} + b x^{2} + c x + d$ by $(x - α)$ to find a quadratic factor
- $a x^{3} + b x^{2} + c x + d = (x - α) (p x^{2} + q x + r)$
- Use algebraic division, or factorise by inspection or by comparing coefficients
Then you can find any other roots by solving $p x^{2} + q x + r = 0$
- If that equation has no real solutions, then $x = α$ is the only real solution of the cubic

Examiner Tips and Tricks

Solving cubic questions may also include a graph of the cubic function
- Remember that roots correspond to x-intercepts on the graph

Worked Example

a) Show that $x = 2$ is a solution to the cubic equation $2 x^{3} - x^{2} - 8 x + 4 = 0$ .

Substitute $x = 2$ into the equation

$\begin{array}{rcl} 2 {(2)}^{3} - {(2)}^{2} - 8 (2) + 4 & = & 16 - 4 - 16 + 4 = 0 \end{array}$

Therefore $x = 2$ is a solution

b) Find the other solutions to the equation.

By the factor theorem, we know that $(x - 2)$ is a factor of $2 x^{3} - x^{2} - 8 x + 4$

$\begin{array}{rcl} 2 x^{3} - x^{2} - 8 x + 4 & = & (x - 2) (p x^{2} + q x + r) \end{array}$

We'll do this here by comparing coefficients

(You might also be able to do it by inspection, or else by algebraic division)
Start by expanding the brackets

$\begin{array}{rcl} 2 x^{3} - x^{2} - 8 x + 4 & = & p x^{3} + (q - 2 p) x^{2} + (r - 2 q) x - 2 r \end{array}$

The $x^{3}$ coefficient and constant term give us $p$ and $r$ right away

$2 x^{3} = p x^{3} \Rightarrow p = 2 4 = - 2 r \Rightarrow r = - 2$

Use either the $x^{2}$ or $x$ coefficients to find the value of $q$

$- x^{2} = (q - 2 p) x^{2}$
$\begin{array}{rcl} q - 2 p & = & - 1 \\ q - 2 (2) & = & - 1 \\ q - 4 & = & - 1 \\ q & = & 3 \end{array}$

Now we can right the cubic in factorised form

$(x - 2) (2 x^{2} + 3 x - 2) = 0$

To find other solutions we need to solve $2 x^{2} + 3 x - 2 = 0$

Any quadratic solving method would work

But here it would be easiest if you could spot the factorisation

$(x - 2) (x + 2) (2 x - 1) = 0$

So the other solutions are

$x = - 2$ , from the factor $(x + 2)$

$x = \frac{1}{2}$ , from the factor $(2 x - 1) = 2 (x - \frac{1}{2})$

$x = - 2 and x = \frac{1}{2}$

Last updated: 19 October 2024

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