Solving Cubic Equations

This lesson teaches you the techniques for solving cubic equations with complex roots.

📚 Complex Numbers 🎯 Difficulty: ⭐⭐⭐ ⏱️ Learn time: 60 minutes 📋 Exam Board: Edexcel & AQA

📚 Introduction

Cubic equations are polynomial equations of degree 3. They may seem daunting at first but you'll soon see that they are pretty manageable. In my experience, they come up almost 100% of the time in past papers, so I suggest learning them to a high degree.

🎯 Learning Objectives

By the end of this lesson, you will be able to:

Solve cubic equations using the factor theorem
Find all roots of a cubic equation when given one root
Apply polynomial division to reduce cubics to quadratics

🔑 Key Concepts

Basic Rules

When solving cubic equations, keep the following rules in mind:
If a polynomial has one complex root, then another root is the root's complex conjugate.
E.g: if $2 + 3i$ is a root, then $2 - 3i$ is also a root.

📖 Rules of Cubics:

Cubics can have either:

3 real roots
1 real root and 2 complex conjugate roots

A cubic is given as: (where $a, b, c, d$ are constants)

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

Problems with Cubics

Even though cubics are not too different to quadratics, they are much harder to solve when given no roots (solutions). In your exam, you will always be given at least one root, so you can use the factor theorem to find the other roots.

💡 The Factor Theorem

The Factor Theorem says:
If $a$ is a solution to $f(a) = 0$, then $(x - a)$ is a factor of $f(x)$.

Let me explain in English:

Let's say $f(x) = x^3 - 6x^2 + 11x - 6$.
If we set $x$ to any number, and the polynomial equals 0, then that number is a root of the polynomial, and its factor is written as: $(x - (\text{that number}))$.

Quick tip:

If a factor is $(x + a)$ instead of $(x - a)$, then the root is negative because: $(x + a) = (x - (-a))$.

📝 Worked Examples

📋 Example 1: Cubic Equation

Question: $f(x) = x^3 - 3x^2 - 4x + 12. \quad f(2) = 0$. Find all possible solutions.

Solution:

Step 1:

Understand the problem. The question says that $f(2) = 0$, which means it's a root, and we can write:

f(x) = x^3 - 3x^2 - 4x + 12$$ $$f(2) = (2)^3 - 3(2)^2 - 4(2) + 12 = 0$$ $$\Rightarrow f(2)\text{ is a solution } \Rightarrow (x - 2)\text{ is a factor}

Step 2:

Now we know a factor, we can reduce it to a quadratic by dividing.
You can use any method to divide; I'll use long division.

\text{Divide } f(x) \text{ by the known factor, } (x - 2):$$ $$\begin{array}{r} \boxed{x^2 - x - 6\phantom{)}} \\[4pt] x - 2\;{\overline{\smash{\big)}\,x^3 - 3x^2 - 4x + 12}} \\[4pt] \underline{\phantom{+}x^3 - 2x^2}\quad\downarrow\quad\quad\downarrow \\ \phantom{+}0 - x^2 - 4x \quad\quad\downarrow \\ \phantom{+} \underline{- x^2 + 2x + 12} \\ \phantom{+}0 - 6x + 12 \\ \underline{\phantom{+} -6x + 12} \\ \phantom{+}0 \end{array}$$ $$\uparrow \text{You know you did it right when the remainder is 0.}\uparrow

(It was really hard to format that long division in LaTeX, so apologies if it's messy)

Final Answer:

Now that we have a quadratic from that cubic, we can solve it using the quadratic formula:

x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ $$\text{So:}$$ $$x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-6)}}{2(1)}$$ $$x = \frac{1 \pm \sqrt{25}}{2}$$ $$x = \frac{1 \pm 5}{2} \rightarrow x = 3 \text{ or } x = -2$$ $$\text{So the final solutions are: } \boxed{x = 2, 3, -2}$$ $$\text{The factors are: } \boxed{(x - 2)(x - 3)(x + 2)}

📋 Example 2: Cubic Equation knowledge

Question: Give the general form of a cubic equation. (1 mark)
A root of a cubic is $4 + 6i$. What's another root? (1 mark)
Given that $(x + 5)$ is a factor of $f(x)$, write this in the form $f(a) = 0$ where $a$ is a solution. (2 marks)

Solution:

Answer:

Part 1: General form of a cubic equation:

f(x) = ax^3 + bx^2 + cx + d$$ $$\text{where } a, b, c, d \text{ are constants and } a \neq 0

Part 2: Since complex roots come in conjugate pairs, if $4 + 6i$ is a root, then another root is:

$$4 - 6i$$

Part 3: If $(x + 5)$ is a factor, then the root is $-5$, so we can write:

f(-5) = 0$$ $$\text{It's negative because } (x + 5) = (x - (-5))

⚠️ Common Mistakes

🚫 Mistake 1: Not verifying the given root

What students do wrong: Assuming the given root is correct without substituting it back into the equation.

How to avoid it: Always verify $f(a) = 0$ before using $(x - a)$ as a factor. This saves marks in Edexcel papers.

🚫 Mistake 2: Polynomial division errors

What students do wrong: Rushing through long division and making sign errors or misaligning terms.

How to avoid it: Work systematically and check your division by expanding $(x - a) \times \text{quotient}$ equals the original cubic.

💪 Practice Problems

Try these problems to test your understanding:

🎯 Practice Question 1

(A) Solve $f(x) = x^3 + 2x^2 - x - 2$ given that $f(-1) = 0$.

(B) Hence, find all the roots of the equation.

🔍 Show Solution

Part A: Since $f(-1) = 0$, we know $(x + 1)$ is a factor. Using polynomial division: $f(x) = (x + 1)(x^2 + x - 6) = \boxed{(x + 1)(x + 3)(x - 2)}$.
Part B: The roots of the equation are $\boxed{x = -1, -3, 2}$.

🎯 Practice Question 2

$f(g) = g^3 + 4g^2 + 4g + 16 = 0$, where $f(-2) = 0$
(A) Give another root without working it out.
(B) Find all other possible roots.

🔍 Show Solution

Part A: Since $f(-2) = 0$, we know that $(g + 2)$ is a factor.
Part B: Using polynomial division: $f(g) = (g + 2)(g^2 + 2g + 8)$. Solving the quadratic using the quadratic formula gives the other roots as $\boxed{-1 + \sqrt{7}i}$ and $\boxed{-1 - \sqrt{7}i}$.

📋 Summary

🎯 Key Takeaways

Cubic equations can be solved using the factor theorem when given one root
Polynomial long division reduces a cubic to a quadratic that can be solved using the quadratic formula
Complex roots always come in conjugate pairs for polynomials with real coefficients
Always verify that a given value is actually a root before proceeding with factorization

📚 What's Next?

Now that you understand cubic equations, you can move on to quartic equations, which use similar techniques but require additional factorization steps. You should also practice applying these skills to complex number problems in exam contexts.

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