Cubic equations and factorising cubics

Q: How do you find the first linear factor of a cubic expression?

Use the Factor Theorem : trial small integers (like \(1, -1, 2, -2\)) until you find a value \(k\) such that \(f(k) = 0\), meaning \((x - k)\) is a factor.

Q: Once you have found one linear factor, what is the next step in factorising the cubic?

Perform algebraic division to divide the cubic expression by the known linear factor to find the remaining quadratic quotient.

Q: According to the syllabus, what kind of roots will exam-style cubic equations always contain?

They will contain at least one rational root , ensuring that trial and error with integers or simple fractions will always yield the first factor.

Further Pure Mathematics · Pearson Edexcel IGCSE · 4 min read · Updated: 7 Sept 2026

Introduction to Cubic Equations

In your previous studies, you mastered quadratic equations (where the highest power is \(x^2\)). Now, we are stepping up to cubic equations. A cubic equation is any equation where the highest power of the variable is 3, usually written in the form \(ax^3 + bx^2 + cx + d = 0\).

Why do we care about cubics? They appear everywhere in physics and engineering, especially when calculating volumes or modeling complex curves. In this chapter, we will learn how to break these large expressions down into manageable pieces using factorisation and then solve them.

The Goal: Breaking it Down

Factorising a cubic means writing it as a product of its factors, typically looking like this:
\((x - p)(x - q)(x - r) = 0\)
Don't worry if this seems tricky at first! Unlike quadratics, where we have the "quadratic formula" to do the heavy lifting, cubics require a bit more detective work. We use a three-step process: Trial, Division, and Quadratic Factorisation.

Step 1: Finding the First Factor

To start factorising a cubic like \(f(x) = x^3 + 3x^2 - 4\), we need to find one value of \(x\) that makes the whole equation equal zero. This is where the Factor Theorem comes in.

Quick Review: The Factor Theorem states that if \(f(k) = 0\), then \((x - k)\) is a factor of the expression. (Note: For more detail on this, see the "Factor and Remainder Theorems" chapter).

How to do it: Use trial and error! Start with small integers like \(1, -1, 2, -2\).
For \(f(x) = x^3 + 3x^2 - 4\):
Try \(x = 1\): \(f(1) = (1)^3 + 3(1)^2 - 4 = 1 + 3 - 4 = 0\).
Since the result is \(0\), we know for sure that \((x - 1)\) is a factor!

Step 2: Algebraic Division

Once we have one factor, we need to "divide" the cubic by that factor to find what's left. This is called Algebraic Division. You are essentially turning a cubic into \((a \text{ factor}) \times (a \text{ quadratic})\).

If we divide \(x^3 + 3x^2 - 4\) by \((x - 1)\), we are looking for a quadratic \(ax^2 + bx + c\) such that:
\(x^3 + 3x^2 - 4 = (x - 1)(ax^2 + bx + c)\)

Common Mistake to Avoid: When dividing, make sure you account for "missing" terms. If there is no \(x\) term (like in our example), write it as \(0x\) so your columns stay lined up: \(x^3 + 3x^2 + 0x - 4\).

After performing the division (covered in the "Simple Algebraic Division" chapter), we find:
\(x^3 + 3x^2 - 4 = (x - 1)(x^2 + 4x + 4)\)

Step 3: Factorising the Quadratic

Now we are back on familiar ground! We just need to factorise the quadratic part: \(x^2 + 4x + 4\).
We look for two numbers that multiply to \(4\) and add to \(4\). These numbers are \(2\) and \(2\).
So, \(x^2 + 4x + 4 = (x + 2)(x + 2)\), which we can write as \((x + 2)^2\).

Final Factorised Form:
\(x^3 + 3x^2 - 4 = (x - 1)(x + 2)^2\)

Solving the Equation

If the question asks you to solve the equation \(x^3 + 3x^2 - 4 = 0\), you simply take your factors and set each one to zero:
1. \(x - 1 = 0 \implies x = 1\)
2. \(x + 2 = 0 \implies x = -2\)
So the solutions (roots) are \(x = 1\) and \(x = -2\).

Exam Tip: The syllabus states that cubic equations set in exams will contain at least one rational root. This is great news! It means that your "Trial and Error" step will always eventually work if you try simple fractions or integers.

Working with \(ax \pm b\) Factors

Sometimes the first factor isn't as simple as \((x - 1)\). It might be something like \((2x - 3)\).
If you are asked to determine the remainder when \(f(x)\) is divided by \((ax + b)\), or if you are given a factor in that form, remember the Remainder Theorem trick:
To find the remainder when dividing by \((2x - 3)\), calculate \(f(\frac{3}{2})\).
If \(f(\frac{3}{2}) = 0\), then \((2x - 3)\) is a factor.

Summary Checklist

1. Trial and Error: Find a value \(k\) such that \(f(k) = 0\).
2. Identify the Factor: If \(f(k) = 0\), then \((x - k)\) is a factor. (Watch the signs! If \(f(-2) = 0\), the factor is \((x + 2)\)).
3. Divide: Use algebraic division to find the quadratic quotient.
4. Factorise Fully: Factorise the quadratic if possible.
5. Solve: Set each factor to zero to find the values of \(x\).

Key Takeaway

Think of factorising a cubic like peeling an onion. You use the Factor Theorem to peel off the first layer (the first linear factor), leaving you with a quadratic core that you already know how to handle from your previous math classes!

Quick check

Can you answer these now?

Open each question to check the key ideas from this chapter.

What is the general form of a cubic equation?

A cubic equation is written in the form \(ax^3 + bx^2 + cx + d = 0\), where the highest power of the variable is 3.

What does it mean to factorise a cubic expression fully?

Factorising a cubic means expressing it as a product of its linear factors, typically in the form \((x - p)(x - q)(x - r)\).

How do you find the first linear factor of a cubic expression?

Use the Factor Theorem: trial small integers (like \(1, -1, 2, -2\)) until you find a value \(k\) such that \(f(k) = 0\), meaning \((x - k)\) is a factor.

Once you have found one linear factor, what is the next step in factorising the cubic?

Perform algebraic division to divide the cubic expression by the known linear factor to find the remaining quadratic quotient.

What should you do if a cubic expression is missing a specific power of \(x\) during algebraic division?

Use a placeholder of \(0\) for the missing term (e.g., write \(x^3 - 4\) as \(x^3 + 0x^2 + 0x - 4\)) to keep the columns aligned.

How do you find the final solutions (roots) of the cubic equation after factorising?

Set each individual factor equal to zero and solve for \(x\).

According to the syllabus, what kind of roots will exam-style cubic equations always contain?

They will contain at least one rational root, ensuring that trial and error with integers or simple fractions will always yield the first factor.

How do you test if \((ax - b)\) is a factor of a function \(f(x)\)?

Check if \(f(\frac{b}{a}) = 0\).

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