Solving Cubic Equations (CIE IGCSE Additional Maths)

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Solving Cubic Equations

What is a cubic equation?

  • A cubic function is an polynomial of degree 3
    • i.e.  the highest power of x is 3 
  • A cubic equation can be written in the form
    • a x cubed plus b x squared plus c x plus d equals 0
  • Solving a cubic equation involves factorising the cubic function first.

How do I factorise a cubic function?

  • Factorising a cubic (function) combines the factor theorem with the method of polynomial division
  • The example below shows the steps for factorising a cubicfactorising a cubic - question for worked example

STEP 1
Use factor theorem.
Find a value p such that straight f open parentheses p close parentheses equals 0.

factorising a cubic - step 1

STEP 2
Use polynomial division.
Divide straight f open parentheses x close parentheses by open parentheses x minus p close parentheses.
(It is possible to do this step 'by inspection', see the worked example below)

factorising a cubic - step 2

STEP 3
Use the result of your division to write straight f open parentheses x close parentheses equals open parentheses x minus p close parentheses open parentheses a x squared plus b x plus c close parentheses.factorising a cubic - step 3

STEP 4
If the quadratic open parentheses a x squared plus b x plus c close parentheses can be factorised, do so.
straight f open parentheses x close parentheses can then be written as the product of three linear factors.
If the quadratic cannot be factorised, then the result from STEP 3 is the final factorisation.

factorising a cubic - final answer

How do I solve a cubic equation?

  • A cubic equation will have either 1, 2 or 3 (real) solutions
    • (The cubic function will have either 1, 2 or 3 (real) roots)
  • Once the cubic function is factorised using the four steps above, there is one more step to carry out

STEP 5
Find the solutions to the cubic equation by making each factor equal to zero

    • For each linear factor, open parentheses x minus p close parentheses
      • x minus p equals 0 
      • so x equals p is a solution
      • This is the factor theorem!
    • For a quadratic factor, open parentheses a x squared plus b x plus c close parentheses
      • a x squared plus b x plus c equals 0 
      • use either the quadratic formula or completing the square (as it won't factorise)
        • this will give two of the solutions to the cubic equation
      • if there are no solutions to the quadratic equation there are no solutions other than that from the linear factor

  • From the example above,
    • x cubed plus 4 x squared minus 11 x minus 30 equals open parentheses x plus 2 close parentheses open parentheses x plus 5 close parentheses open parentheses x minus 3 close parentheses
    • so the solutions to the cubic equation x cubed plus 4 x squared minus 11 x minus 30 equals 0 are
      • x equals negative 2 comma space x equals negative 5 and x equals 3
  • Cubic equations can have equal (repeated) solutions
    • e.g.   open parentheses x minus 2 close parentheses squared open parentheses x plus 1 close parentheses has two (equal and real) roots, x equals 2 (repeated) and x equals negative 1 
    • e.g.   open parentheses 2 x minus 1 close parentheses cubed has three (equal and real) roots, x equals 1 half

Examiner Tip

  • When d equals 0 (i.e. there is no constant term) then x is a factor of the cubic function, and so x equals 0 is a solution
    • This is a special case of factor theorem, where straight f open parentheses 0 close parentheses equals 0
      • spotting the factor of x means there is no need to test values
    • Take out a factor of x and a quadratic function will remain
    • Deal with the quadratic in any of the usual ways

Worked example

a)
Solve the cubic equation x cubed minus 10 x squared plus 12 x plus 8 equals 0.

STEP 1 - use factor theorem with straight f open parentheses x close parentheses equals x cubed minus 10 x squared plus 12 x plus 8

straight f open parentheses 1 close parentheses equals open parentheses 1 close parentheses cubed minus 10 open parentheses 1 close parentheses squared plus 12 open parentheses 1 close parentheses plus 8 equals 11 space space bold not equal to bold 0

straight f open parentheses negative 1 close parentheses equals open parentheses negative 1 close parentheses cubed minus 10 open parentheses negative 1 close parentheses squared plus 12 open parentheses negative 1 close parentheses plus 8 equals negative 15 space space bold not equal to bold 0

straight f open parentheses 2 close parentheses equals open parentheses 2 close parentheses cubed minus 10 open parentheses 2 close parentheses squared plus 12 open parentheses 2 close parentheses plus 8 equals 0

therefore space space open parentheses x minus 2 close parentheses is a factor of straight f open parentheses x close parentheses

STEP 2 - polynomial division (f open parentheses x close parentheses space divided by space open parentheses x minus 2 close parentheses) or 'by inspection'
By inspection ...

straight f open parentheses x close parentheses equals open parentheses x minus 2 close parentheses open parentheses a x squared plus b x plus c close parentheses

('cubic' ÷ 'linear' = 'quadratic')

a equals 1

(because the x cubed is generated only from x space cross times space a x squared)

c equals negative 4

(because the constant term is generated only from negative 2 space cross times space c)

Equate coefficients of x (or x squared) terms to find b,

12 equals c minus 2 b

b equals fraction numerator negative 4 minus 12 over denominator 2 end fraction equals negative 8

STEP 3

straight f open parentheses x close parentheses equals open parentheses x minus 2 close parentheses open parentheses x squared minus 8 x minus 4 close parentheses

STEP 4 - the quadratic does not factorise

STEP 5 - Use the factors to find the solutions

table row cell x minus 2 end cell equals cell 0 comma space space space space space x equals 2 end cell end table

x squared minus 8 x minus 4 equals 0 comma space space space space space x equals fraction numerator 8 plus-or-minus square root of 64 plus 16 end root over denominator 2 end fraction comma space space space space space x equals 4 plus-or-minus 2 square root of 5

(using the quadratic formula)

The solutions to bold italic x to the power of bold 3 bold minus bold 10 bold italic x to the power of bold 2 bold plus bold 12 bold italic x bold plus bold 8 bold equals bold 0 are bold italic x bold equals bold 2 bold comma bold space bold italic x bold equals bold 4 bold plus bold 2 square root of bold 5 and bold italic x bold equals bold 4 bold minus bold 2 square root of bold 5.

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Paul

Author: Paul

Expertise: Maths

Paul has taught mathematics for 20 years and has been an examiner for Edexcel for over a decade. GCSE, A level, pure, mechanics, statistics, discrete – if it’s in a Maths exam, Paul will know about it. Paul is a passionate fan of clear and colourful notes with fascinating diagrams – one of the many reasons he is excited to be a member of the SME team.