# Further Maths - Cubics

> Source: https://ollybritton.com/notes/a-level/further-maths/topics/cubics/ · Updated: 2025-04-26 · Tags: further-maths, school

Here, using $z$ instead of $x$ means that the variable is complex. $w$ is also sometimes used.

The first step is to find the one real solution. Since it's a cubic, there will be three solutions and by examining the graph you can see that there always must be at least one real solution (cubics always cross the $y$-axis at least once).

For $z = 1$:

$$
z^3 + 9z^2 + 33z + 25
1^3 + 9\times1^2 + 33\times1 + 25 \neq 0
$$

For $z = -1$:

$$
(-1)^3 + 9\times(-1)^2 + 33\times-1 + 25 = 0
-1 + 9 - 33 + 25 = 0
$$

So we have one bracket, $(z + 1)$. We can now write out the cubic like so:

$$
(z+1)(Az^2 + Bz + C)
$$

We can work out $A, B$ and $C$ by inspection:

* $A$ must be $1$ since the final result of multiplying everything out has a

---
Olly Britton — https://ollybritton.com. Machine-readable index: https://ollybritton.com/llms.txt