Solving Cubic Equations

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 , and can be written as a product of three linear factors
- - Any two of the factors could be multiplied together to give a quadratic factor
A cubic with one real root can be written as the product of a linear and a quadratic factor
- - The quadratic factor will not have any real roots
  - So its discriminant $q^{2} - 4 p r$ will be negative

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 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 by 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
- If that equation has no real solutions, then $x = α$ is the only real solution of the cubic

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

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

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}

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 the 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}

Solving Cubic Equations (Edexcel IGCSE Further Maths)