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
Backlinks
Metadata
date: 2020-09-03 18:21
tags:
- '@?further-maths'
- '@?school'
- '@?public'
title: Further Maths - Cubics