Solving Cubic Equations (Cambridge International IGCSE Further Maths)

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

  • $ax^{3}+bx^{2}+cx+d=0$

• 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 cubic

STEP 1 Use factor theorem. Find a value $p$ such that $\mathrm{f} ( p ) = 0$.

STEP 2 Use polynomial division. Divide $\mathrm{f} ( x )$ by $( x - p )$. (It is possible to do this step 'by inspection', see the worked example below)

STEP 3 Use the result of your division to write $\mathrm{f} ( x ) = ( x - p ) ( a x^{2} + b x + c )$.

STEP 4 If the quadratic $( a x^{2} + b x + c )$ can be factorised, do so. $\mathrm{f} ( x )$ 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.

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, $( x - p )$

  • $x-p=0$

  • so $x=p$ is a solution

  • This is the factor theorem!

  • For a quadratic factor, $( a x^{2} + b x + c )$

    • $ax^{2}+bx+c=0$

    • 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,

  • $x^{3} + 4 x^{2} - 11 x - 30 = ( x + 2 ) ( x + 5 ) ( x - 3 )$

  • so the solutions to the cubic equation $x^{3}+4x^{2}-11x-30=0$ are

    • $x=-2,x=-5$ and $x=3$

• Cubic equations can have equal (repeated) solutions

  • e.g. $( x - 2 )^{2} ( x + 1 )$ has two (equal and real) roots, $x=2$ (repeated) and $x=-1$
  • e.g. $( 2 x - 1 )^{3}$ has three (equal and real) roots, $x=\frac{1}{2}$

Exam Hint

• When $d=0$ (i.e. there is no constant term) then $x$ is a factor of the cubic function, and so $x=0$ is a solution

  • This is a special case of factor theorem, where $\mathrm{f} ( 0 ) = 0$

    • 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 $x^{3}-10x^{2}+12x+8=0$.

STEP 1 - use factor theorem with $\mathrm{f} ( x ) = x^{3} - 10 x^{2} + 12 x + 8$

$\mathrm{f} ( 1 ) = ( 1 )^{3} - 10 ( 1 )^{2} + 12 ( 1 ) + 8 = 11 \text{eq} 0$

$\mathrm{f} ( - 1 ) = ( - 1 )^{3} - 10 ( - 1 )^{2} + 12 ( - 1 ) + 8 = - 15 \text{eq} 0$

$\mathrm{f} ( 2 ) = ( 2 )^{3} - 10 ( 2 )^{2} + 12 ( 2 ) + 8 = 0$

$\therefore ( x - 2 )$ is a factor of $\mathrm{f} ( x )$

STEP 2 - polynomial division ($f ( x ) \div ( x - 2 )$) or 'by inspection' By inspection ...

$\mathrm{f} ( x ) = ( x - 2 ) ( a x^{2} + b x + c )$

('cubic' ÷ 'linear' = 'quadratic')

$a=1$

(because the $x^{3}$* is generated only from $x\times ax^{2}$)*

$c=-4$

(because the constant term is generated only from $-2\times c$)

Equate coefficients of $x$ (or $x^{2}$) terms to find $b$,

$12=c-2b$

$b=\frac{-4-12}{2}=-8$

STEP 3

$\mathrm{f} ( x ) = ( x - 2 ) ( x^{2} - 8 x - 4 )$

STEP 4 - the quadratic does not factorise

STEP 5 - Use the factors to find the solutions

$\begin{gathered} x - 2 & = 0 , x = 2 \end{gathered}$

$x^{2}-8x-4=0,x=\frac{8\pm\sqrt{64+16}}{2},x=4\pm2\sqrt{5}$

(using the quadratic formula)

Final answer: The solutions to$x^{3}-10x^{2}+12x+8=0$are$x=2,x=4+2\sqrt{5}$and$x=4-2\sqrt{5}$.