Solving Cubic Equations (Edexcel IGCSE Further Maths)

Spec point — 3B

Solving Cubic Equations

How many real solutions can a cubic equation have?

• A cubic equation$ax^{3}+bx^{2}+cx+d=0$ 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

  • $a x^{3} + b x^{2} + c x + d = ( x - \\alpha ) ( x - \\beta ) ( x - \\gamma )$

    • 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

  • $a x^{3} + b x^{2} + c x + d = ( x - \\alpha ) ( p x^{2} + q x + r )$

    • The quadratic factor will not have any real roots
    • So its discriminant$q^{2}-4pr$ will be negative

How do I solve cubic equations?

• Suppose you have a cubic equation $ax^{3}+bx^{2}+cx+d=0$

• 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 $( x - \\alpha )$ is a factor

• You can then divide $ax^{3}+bx^{2}+cx+d$ by $( x - \\alpha )$ to find a quadratic factor

  • $a x^{3} + b x^{2} + c x + d = ( x - \\alpha ) ( p x^{2} + q x + r )$
  • Use algebraic division, or factorise by inspection or by comparing coefficients

• Then you can find any other roots by solving $px^{2}+qx+r=0$

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

Exam Hint

• 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 $2x^{3}-x^{2}-8x+4=0$.

Answer:

Substitute $x=2$ into the equation

Final answer: $\begin{gathered} 2 ( 2 )^{3} - ( 2 )^{2} - 8 ( 2 ) + 4 & = 16 - 4 - 16 + 4 = 0 \end{gathered}$

Therefore$x=2$is a solution

(b) Find the other solutions to the equation.

Answer:

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

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

You can find the values of $p$, $q$ and $r$ by comparing coefficients

• You might also be able to do this by inspection

  • or else you could use algebraic division

• Start by expanding the brackets

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

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

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

• 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 you can write the cubic in factorised form

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

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

• Any quadratic solving method would work
• But it is easiest here if you can 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 $( 2 x - 1 ) = 2 ( x - 1 \frac{1}{2} )$

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