The Logic of Breaking Down a Cubic#
A cubic equation like $x^3 - 6x^2 + 11x - 6 = 0$ has up to 3 solutions (roots). But you can’t use the quadratic formula on it directly — it has an $x^3$ term!
The strategy is always the same: reduce it to a quadratic by finding one factor first.
1. The Complete Strategy#
Step 1: Find the First Factor (Trial and Error)#
Use the Factor Theorem: if $f(c) = 0$, then $(x - c)$ is a factor.
Which values to try? Look at the constant term (the number without an $x$). The factors of the constant term are your best guesses.
For $f(x) = x^3 - 6x^2 + 11x - 6$:
- Constant term = $-6$
- Try: $\pm 1, \pm 2, \pm 3, \pm 6$
Test $x = 1$:
$$ f(1) = 1 - 6 + 11 - 6 = 0 \checkmark $$So $(x - 1)$ is a factor!
Step 2: Divide to Get the Quadratic#
Use either long division or synthetic division (inspection) to divide the cubic by your factor:
$$ x^3 - 6x^2 + 11x - 6 = (x - 1)(x^2 - 5x + 6) $$Step 3: Factor the Quadratic#
$$ x^2 - 5x + 6 = (x - 2)(x - 3) $$Step 4: Write the Full Factorisation#
$$ x^3 - 6x^2 + 11x - 6 = (x - 1)(x - 2)(x - 3) = 0 $$Roots: $x = 1$, $x = 2$, $x = 3$.
2. Synthetic Division (Inspection Method)#
This is faster than long division for dividing by $(x - c)$.
How it works#
To divide $2x^3 + 3x^2 - 11x - 6$ by $(x - 2)$:
Write the coefficients: $2, 3, -11, -6$
| Step | Bring down | Multiply by 2 | Add to next |
|---|---|---|---|
| 1 | 2 | ||
| 2 | $2 \times 2 = 4$ | $3 + 4 = 7$ | |
| 3 | $7 \times 2 = 14$ | $-11 + 14 = 3$ | |
| 4 | $3 \times 2 = 6$ | $-6 + 6 = 0$ ← remainder |
The quotient coefficients are: $2, 7, 3$ → $2x^2 + 7x + 3$
So: $2x^3 + 3x^2 - 11x - 6 = (x - 2)(2x^2 + 7x + 3)$
Factor the quadratic:
$$ 2x^2 + 7x + 3 = (2x + 1)(x + 3) $$Full factorisation:
$$ 2x^3 + 3x^2 - 11x - 6 = (x - 2)(2x + 1)(x + 3) $$Roots: $x = 2$, $x = -\frac{1}{2}$, $x = -3$.
3. Long Division Method#
For those who prefer the formal method:
Divide $x^3 + 2x^2 - 5x - 6$ by $(x + 1)$:
$$ \begin{array}{r} x^2 + x - 6 \\\\ x + 1 \overline{) x^3 + 2x^2 - 5x - 6} \\\\ \underline{x^3 + x^2} \\\\ x^2 - 5x \\\\ \underline{x^2 + x} \\\\ -6x - 6 \\\\ \underline{-6x - 6} \\\\ 0 \end{array} $$Result: $x^3 + 2x^2 - 5x - 6 = (x + 1)(x^2 + x - 6) = (x + 1)(x + 3)(x - 2)$
4. When the Leading Coefficient is Not 1#
If $f(x) = 2x^3 - x^2 - 13x - 6$:
Step 1: Try values. The factors of the constant ($-6$) divided by factors of the leading coefficient ($2$) give you the rational root candidates: $\pm 1, \pm 2, \pm 3, \pm 6, \pm\frac{1}{2}, \pm\frac{3}{2}$.
Test $x = 3$: $f(3) = 54 - 9 - 39 - 6 = 0$ ✓
Step 2: Divide by $(x - 3)$ using synthetic division:
Coefficients: $2, -1, -13, -6$
| Bring down | Multiply by 3 | Add |
|---|---|---|
| 2 | ||
| $2 \times 3 = 6$ | $-1 + 6 = 5$ | |
| $5 \times 3 = 15$ | $-13 + 15 = 2$ | |
| $2 \times 3 = 6$ | $-6 + 6 = 0$ |
Quotient: $2x^2 + 5x + 2 = (2x + 1)(x + 2)$
Full: $2x^3 - x^2 - 13x - 6 = (x - 3)(2x + 1)(x + 2)$
Roots: $x = 3$, $x = -\frac{1}{2}$, $x = -2$.
5. Nature of Cubic Roots#
A cubic equation always has at least one real root. It can have:
| Scenario | Roots |
|---|---|
| 3 distinct real roots | Graph crosses x-axis 3 times |
| 1 real root + 2 complex roots | Graph crosses x-axis once (the quadratic factor has $\Delta < 0$) |
| 3 real roots with a repeated root | Graph touches x-axis at the repeated root |
If the quadratic factor gives $\Delta < 0$ (no real roots), the cubic only has one x-intercept.
Worked Example: Full Exam Question#
Solve: $x^3 - x^2 - 8x + 12 = 0$
Step 1: Try $x = 2$:
$$ f(2) = 8 - 4 - 16 + 12 = 0 \checkmark $$Step 2: Synthetic division by $(x - 2)$:
Coefficients: $1, -1, -8, 12$
| Multiply by 2 | Add | |
|---|---|---|
| 1 | ||
| $1 \times 2 = 2$ | $-1 + 2 = 1$ | |
| $1 \times 2 = 2$ | $-8 + 2 = -6$ | |
| $-6 \times 2 = -12$ | $12 + (-12) = 0$ |
Quotient: $x^2 + x - 6$
Step 3: Factor: $x^2 + x - 6 = (x + 3)(x - 2)$
Step 4: Full factorisation:
$$ (x - 2)(x + 3)(x - 2) = (x - 2)^2(x + 3) = 0 $$Roots: $x = 2$ (repeated) or $x = -3$
The graph touches the x-axis at $x = 2$ and crosses at $x = -3$.
🚨 Common Mistakes#
- Giving up after trying $x = 1$ and $x = -1$: Keep trying! Go through $\pm 2, \pm 3, \pm 6$, and don’t forget fractions like $\pm\frac{1}{2}$ when the leading coefficient isn’t 1.
- Sign errors in synthetic division: Be extremely careful with negative signs. Write out each multiplication and addition step.
- Forgetting to check for repeated roots: If the quadratic factor gives a root that’s the same as the first factor, you have a repeated root. This affects the graph shape.
- Not using the Factor Theorem first: Some students try to factorise by grouping, which sometimes works but is unreliable for cubics. Always use Factor Theorem + division.
💡 Pro Tip: Start with Small Numbers#
Always test $x = 1$ and $x = -1$ first — they’re the easiest to calculate mentally. If neither works, try $x = 2$ and $x = -2$. Most exam cubics have at least one “nice” integer root.