Solving Cubic Equations Revision Notes for Grade 12 NSC Matric Mathematics

Solving Cubic Equations

What is a cubic equation?

A cubic equation is a polynomial equation of degree 3, which has the general form:

$ax^3 + bx^2 + cx + d = 0$

Where $a$ , $b$ , $c$ , and $d$ are constants, and $a \neq 0$ .

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The coefficient $a$ must not equal zero, otherwise the equation would not be cubic but would reduce to a quadratic or lower degree polynomial.

Method for solving cubic equations

To solve cubic equations, we use a systematic approach that combines factorisation techniques with the factor theorem. The process involves finding one linear factor, then factoring the remaining quadratic expression.

Step-by-step process

Step 1: Use the factor theorem to find one factor

Test simple values like $\pm 1$ , $\pm 2$ , $\pm 3$ , etc. by substituting them into the equation
If $f(a) = 0$ , then $(x - a)$ is a factor of the polynomial

Step 2: Factorise by inspection

Once you have found one factor, divide the cubic polynomial by this factor
This will give you a linear factor multiplied by a quadratic factor

Step 3: Factorise the quadratic (if possible)

Try to factorise the quadratic expression further
If factorisation is not possible, prepare to use the quadratic formula

Step 4: Solve the equation

Set each factor equal to zero
Solve the linear equations directly
Use the quadratic formula for quadratic factors that cannot be factorised

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The key to solving cubic equations is finding that first factor using the factor theorem. Always start by testing simple integer values like $\pm 1$ , $\pm 2$ , $\pm 3$ .

Worked example 1

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Worked Example: Solving $6x^3 - 5x^2 - 17x + 6 = 0$

Step 1: Find one factor using the factor theorem

Let $f(x) = 6x^3 - 5x^2 - 17x + 6$

Try $f(1)$ : $f(1) = 6(1)^3 - 5(1)^2 - 17(1) + 6 = 6 - 5 - 17 + 6 = -10 \neq 0$

Therefore $(x - 1)$ is not a factor.

Try $f(2)$ : $f(2) = 6(2)^3 - 5(2)^2 - 17(2) + 6 = 48 - 20 - 34 + 6 = 0$

Therefore (x - 2) is a factor.

Step 2: Factorise by inspection

$6x^3 - 5x^2 - 17x + 6 = (x - 2)(6x^2 + 7x - 3)$

Step 3: Factorise the quadratic fully

$6x^2 + 7x - 3 = (2x + 3)(3x - 1)$

Therefore: $6x^3 - 5x^2 - 17x + 6 = (x - 2)(2x + 3)(3x - 1)$

Step 4: Solve the equation

$(x - 2)(2x + 3)(3x - 1) = 0$

This gives us: x = 2 or x = -3/2 or x = 1/3

Worked example 2

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Worked Example: Solving $x^3 - 2x^2 - 6x + 4 = 0$

Step 1: Use the factor theorem to find a factor

Let $f(x) = x^3 - 2x^2 - 6x + 4$

Try $f(-2)$ : $f(-2) = (-2)^3 - 2(-2)^2 - 6(-2) + 4 = -8 - 8 + 12 + 4 = 0$

Therefore (x + 2) is a factor.

Step 2: Factorise by inspection

$x^3 - 2x^2 - 6x + 4 = (x + 2)(x^2 - 4x + 2)$

Step 3: Check if the quadratic can be factorised

$x^2 - 4x + 2$ cannot be factorised further using simple factors.

Step 4: Apply the quadratic formula

For $x^2 - 4x + 2 = 0$ , we have $a = 1$ , $b = -4$ , $c = 2$

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$x = \frac{4 \pm \sqrt{16 - 8}}{2} = \frac{4 \pm \sqrt{8}}{2} = \frac{4 \pm 2\sqrt{2}}{2} = 2 \pm \sqrt{2}$

Final solutions: x = -2 or x = 2 + √2 or x = 2 - √2

When factorisation isn't possible

Sometimes the quadratic expression obtained after finding one factor cannot be factorised using inspection. In these cases, you need to use alternative methods.

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When the quadratic factor cannot be factorised by inspection:

Use the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
Always write down the formula first, then substitute the values
Simplify your answer completely
Check that $b^2 - 4ac \geq 0$ for real solutions

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If $b^2 - 4ac < 0$ , the quadratic has no real solutions, which means the cubic equation has only one real solution (the one found using the factor theorem).

Exam tips

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Essential Exam Strategies:

Always test simple integer values first when using the factor theorem
Show all your working - marks are awarded for method
Check your factors by substituting back into the original equation
Simplify surds completely in your final answer
If you get stuck finding a factor, try negative values or fractions like $\pm \frac{1}{2}$ , $\pm \frac{1}{3}$
Double-check your arithmetic at each step, especially when calculating $f(a)$

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Key Points to Remember:

Cubic equations have the form $ax^3 + bx^2 + cx + d = 0$
Use the factor theorem to find one factor by testing simple values
Once you have one factor, factorise by inspection to get (linear factor) × (quadratic factor)
Solve linear factors directly and use the quadratic formula for quadratics that won't factorise
A cubic equation can have up to 3 real solutions
Always check your solutions by substituting back into the original equation

Solving Cubic Equations (Grade 12 NSC Matric Mathematics): Revision Notes