Solving Cubic Equations Revision Notes for VCE SSCE Mathematical Methods

Solving Cubic Equations

Introduction

To solve cubic equations, the primary strategy is to factorise the expression first. Once we have factorised the cubic into linear factors, we can use the null factor theorem to find the solutions. The null factor theorem states that if a product of factors equals zero, then at least one of those factors must equal zero.

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A cubic equation can have up to three solutions (also called roots or zeros), depending on how it factorises. This is different from quadratic equations, which can have at most two solutions.

Solving cubics already in factorised form

When a cubic equation is already presented in factorised form, we can apply the null factor theorem directly to find all solutions. This is the simplest case, as the factorisation work has already been done for us.

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Worked Example: Solving a Factorised Cubic

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

Solution:

By applying the null factor theorem, if the product equals zero, then at least one factor must equal zero.

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

This means:

x - 2 = 0 \text{ or } x + 1 = 0 \text{ or } x + 3 = 0

Solving each equation:

x = 2 or x = -1 or x = -3

Therefore, the solutions are $x = 2, -1$ and $-3$ .

Taking out a common factor

When solving cubic equations, always look for common factors first. If all terms contain a common factor (often $x$ ), we can factor it out to simplify the problem. This is often the quickest path to a solution.

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Common before complex: Always check for common factors before trying other factorisation methods. This can save you significant time and effort!

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Worked Example: Common Factor

Solve $2x^3 - x^2 - x = 0$ .

Solution:

First, notice that every term contains $x$ as a factor. We can factor this out:

x(2x^2 - x - 1) = 0

Now we need to factorise the quadratic expression $2x^2 - x - 1$ . This factorises to:

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

Using the null factor theorem:

x = 0 \text{ or } 2x + 1 = 0 \text{ or } x - 1 = 0

Therefore:

x = 0 or $x = -\frac{1}{2}$ or x = 1

Sometimes after taking out the common factor, the remaining quadratic doesn't factorise with integers. In these cases, we can use other techniques like completing the square or the quadratic formula.

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Worked Example: Common Factor with Completing the Square

Solve $x^3 + 2x^2 - 10x = 0$ .

Solution:

Factor out the common factor $x$ :

x(x^2 + 2x - 10) = 0

The quadratic $x^2 + 2x - 10$ doesn't factorise easily with integers. We can complete the square:

x(x^2 + 2x + 1 - 11) = 0

x[(x + 1)^2 - 11] = 0

This is a difference of two squares:

x(x + 1 - \sqrt{11})(x + 1 + \sqrt{11}) = 0

Therefore:

x = 0 or $x = -1 + \sqrt{11}$ or $x = -1 - \sqrt{11}$

Using grouping to factorise

Grouping is a powerful technique where we rearrange terms to identify common factors. This method works when the cubic can be written as pairs of terms that share factors. The key is to look for patterns where terms can be grouped to reveal a common factor.

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Worked Example: Grouping Method

Solve $x^3 - 4x^2 - 11x + 44 = 0$ .

Solution:

Group the terms in pairs:

x^2(x - 4) - 11(x - 4) = 0

Notice that both groups contain the factor $(x - 4)$ :

(x - 4)(x^2 - 11) = 0

The factor $x^2 - 11$ is a difference of squares:

(x - 4)(x - \sqrt{11})(x + \sqrt{11}) = 0

Therefore:

x = 4 or $x = \sqrt{11}$ or $x = -\sqrt{11}$

Which can be written as: $x = 4$ or $x = \pm\sqrt{11}$

The grouping method also works when the equation contains parameters or constants.

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Worked Example: Grouping with a Parameter

Solve $x^3 - ax^2 - 11x + 11a = 0$ where $a$ is a constant.

Solution:

Group the terms:

x^2(x - a) - 11(x - a) = 0

Factor out $(x - a)$ :

(x - a)(x^2 - 11) = 0

Therefore:

$x = a$ or $x = \pm\sqrt{11}$

Using the factor theorem

The factor theorem states that if $P(a) = 0$ for a polynomial $P(x)$ , then $(x - a)$ is a factor of the polynomial. This is particularly useful when the cubic doesn't have obvious common factors or grouping patterns.

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Test the easy ones first: When using the factor theorem, test small integer values in this order: $\pm 1, \pm 2, \pm 3$ . These are the most common factors in exam questions.

The strategy is:

Test small integer values until you find one that gives $P(a) = 0$
Use polynomial division to find the remaining quadratic factor
Factorise the quadratic or use the quadratic formula

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Worked Example: Factor Theorem

Solve $x^3 - 4x^2 - 11x + 30 = 0$ .

Solution:

Let $P(x) = x^3 - 4x^2 - 11x + 30$

Test some values:

P(1) = 1 - 4 - 11 + 30 = 16 \neq 0

P(-1) = -1 - 4 + 11 + 30 = 36 \neq 0

P(2) = 8 - 16 - 22 + 30 = 0

Since $P(2) = 0$ , we know that $(x - 2)$ is a factor.

Now we can find the remaining factors by division or inspection:

x^3 - 4x^2 - 11x + 30 = (x - 2)(x^2 - 2x - 15)

The quadratic $x^2 - 2x - 15$ factorises to $(x - 5)(x + 3)$

Therefore:

(x - 2)(x - 5)(x + 3) = 0

The solutions are:

x = 2, 5 or x = -3

Not all cubic equations have three solutions. Sometimes the quadratic factor cannot be factorised further, particularly when its discriminant is negative.

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Worked Example: Cubic with Only One Solution

Solve $2x^3 - 5x^2 + 5x - 2 = 0$ .

Solution:

Let $P(x) = 2x^3 - 5x^2 + 5x - 2$

Test values:

P(1) = 2 - 5 + 5 - 2 = 0

Therefore $(x - 1)$ is a factor.

By division:

2x^3 - 5x^2 + 5x - 2 = (x - 1)(2x^2 - 3x + 2)

So we have:

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

For the quadratic $2x^2 - 3x + 2$ , let's check the discriminant:

\Delta = b^2 - 4ac = (-3)^2 - 4(2)(2) = 9 - 16 = -7

Since the discriminant is negative, the quadratic has no real solutions and cannot be factorised further.

Therefore, the only solution is:

x = 1

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Exam Tip: You can use a CAS calculator to solve cubic equations. On a TI-Nspire, use the solve() function from the Algebra menu. On a Casio ClassPad, use the solve icon. However, you should understand the manual factorisation methods as questions may require you to show working.

Summary of factorisation techniques

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Key Factorisation Methods:

Cubic polynomial equations can be solved by using appropriate factorisation techniques. The main methods include:

Taking out a common factor - Always check first if all terms share a common factor
Using grouping of terms - Group terms in pairs to identify common factors
Using the factor theorem - Test integer values to find linear factors
Polynomial division or equating coefficients - Used after finding one factor to determine remaining factors
Sum or difference of two cubes - Special factorisation formulas for specific forms
Using the quadratic formula - To complete the factorisation when the remaining quadratic doesn't factorise with integers

Remember!

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Key Points to Remember:

The first step in solving cubic equations is always to factorise. Look for common factors before trying other methods.
The null factor theorem tells us that if a product equals zero, at least one factor must be zero. This is how we find solutions from factorised form.
When using the factor theorem, test small integers first ( $\pm 1, \pm 2, \pm 3$ ). These are the most likely factors in exam questions.
A cubic equation can have one, two, or three real solutions depending on its factorisation. Check the discriminant of any remaining quadratic to determine if further solutions exist.
After finding one linear factor using the factor theorem, always use division to find the remaining quadratic factor, then factorise or use the quadratic formula to complete the solution.

Solving Cubic Equations (VCE SSCE Mathematical Methods): Revision Notes