Further Maths - Cubics

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 $A, B \text{and}, C$ by inspection:

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