Solving Cubic Equations (Edexcel IGCSE Further Pure Maths)

Revision Note

Roger B

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 cubed plus b x squared plus c x plus d equals 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 alpha, beta and gamma can be written as a product of three linear factors

    • a x cubed plus b x squared plus c x plus d equals open parentheses x minus alpha close parentheses open parentheses x minus beta close parentheses open parentheses x minus gamma close parentheses

      • Any two of the factors could be multiplied together to give a quadratic factor

  • A cubic with one real root alpha can be written as the product of a linear and a quadratic factor

    • a x cubed plus b x squared plus c x plus d equals open parentheses x minus alpha close parentheses open parentheses p x squared plus q x plus r close parentheses

      • The quadratic factor will not have any real roots

      • So its discriminant q squared minus 4 p r will be negative

How do I solve cubic equations?

  • Suppose you have a cubic equation a x cubed plus b x squared plus c x plus d equals 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 equals alpha by substituting values into the equation until it equals 0

      • Try small integer values (1 comma space minus 1 comma space 2 comma space minus 2 comma...)

  • If you know a root then you know a factor

    • This is because of the factor theorem

      • If x equals alpha is a root, then open parentheses x minus alpha close parentheses is a factor

  • You can then divide a x cubed plus b x squared plus c x plus d by open parentheses x minus alpha close parentheses to find a quadratic factor

    • a x cubed plus b x squared plus c x plus d equals open parentheses x minus alpha close parentheses open parentheses p x squared plus q x plus r close parentheses

    • Use algebraic division, or factorise by inspection or by comparing coefficients

  • Then you can find any other roots by solving  p x squared plus q x plus r equals 0

    • If that equation has no real solutions, then x equals alpha 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 equals 2 is a solution to the cubic equation 2 x cubed minus x squared minus 8 x plus 4 equals 0.

Substitute x equals 2 into the equation

table row cell 2 open parentheses 2 close parentheses cubed minus open parentheses 2 close parentheses squared minus 8 open parentheses 2 close parentheses plus 4 end cell equals cell 16 minus 4 minus 16 plus 4 equals 0 end cell end table

Therefore bold italic x bold equals bold 2 is a solution

b) Find the other solutions to the equation.

By the factor theorem, we know that open parentheses x minus 2 close parentheses is a factor of 2 x cubed minus x squared minus 8 x plus 4 

table row cell 2 x cubed minus x squared minus 8 x plus 4 end cell equals cell open parentheses x minus 2 close parentheses open parentheses p x squared plus q x plus r close parentheses end cell end table

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

table row cell 2 x cubed minus x squared minus 8 x plus 4 end cell equals cell p x cubed plus open parentheses q minus 2 p close parentheses x squared plus open parentheses r minus 2 q close parentheses x minus 2 r end cell end table

The x cubed coefficient and constant term give us p and r right away

2 x cubed equals p x cubed space space space rightwards double arrow space space space p equals 2

4 equals negative 2 r space space rightwards double arrow space space r equals negative 2

Use either the x squared or x coefficients to find the value of q

negative x squared equals open parentheses q minus 2 p close parentheses x squared
 table attributes columnalign right center left columnspacing 0px end attributes row cell q minus 2 p end cell equals cell negative 1 end cell row cell q minus 2 open parentheses 2 close parentheses end cell equals cell negative 1 end cell row cell q minus 4 end cell equals cell negative 1 end cell row q equals 3 end table

Now we can right the cubic in factorised form

open parentheses x minus 2 close parentheses open parentheses 2 x squared plus 3 x minus 2 close parentheses equals 0

To find other solutions we need to solve  2 x squared plus 3 x minus 2 equals 0

Any quadratic solving method would work

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

open parentheses x minus 2 close parentheses open parentheses x plus 2 close parentheses open parentheses 2 x minus 1 close parentheses equals 0

So the other solutions are 

x equals negative 2,  from the factor open parentheses x plus 2 close parentheses

x equals 1 half,  from the factor open parentheses 2 x minus 1 close parentheses equals 2 open parentheses x minus 1 half close parentheses

bold italic x bold equals bold minus bold 2 bold space bold space bold and bold space bold space bold italic x bold equals bold 1 over bold 2

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Roger B

Author: Roger B

Expertise: Maths

Roger's teaching experience stretches all the way back to 1992, and in that time he has taught students at all levels between Year 7 and university undergraduate. Having conducted and published postgraduate research into the mathematical theory behind quantum computing, he is more than confident in dealing with mathematics at any level the exam boards might throw at you.

Dan Finlay

Author: Dan Finlay

Expertise: Maths Lead

Dan graduated from the University of Oxford with a First class degree in mathematics. As well as teaching maths for over 8 years, Dan has marked a range of exams for Edexcel, tutored students and taught A Level Accounting. Dan has a keen interest in statistics and probability and their real-life applications.