Ok, so my question is relatively simple. I'm unsure how to factorise a cubic. Even simple ones i don't understand how to factorise them so i can eventually solve them.
I have two cubic functions in questions:
One is factorise: $2x^3 - 13x^2 + 22x - 8$
The other is solve: $x^3 -2x^2 -5x + 6x = 0$
If someone could explain in simple steps how to do these, and also if theres a certain process to factorising/solving cubics in general, it would be appreciated. Thanks
Here's the theorem: if a cubic $ax^3 + ... + d = 0$ with integer coefficients has a root, $r$ (or a factor $(x-r)$, which is the same thing) and that root is a rational number, then it's of the form $$ r = \frac{p}{q} $$ where $p$ is a divisor of $d$ and $q$ is a divisor of $a$. (Each can be positive or negative).
So in your first example, where $a = 2$ has positive divisors $1, 2$ and $d = 8$ has positive divisors $1, 2, 4, 8$, the only possible rational roots are \begin{align} \pm\frac{1}{1}\\ \pm\frac{2}{1}\\ \pm\frac{4}{1}\\ \pm\frac{8}{1}\\ \pm\frac{1}{2}\\ \pm\frac{2}{2}\\ \pm\frac{4}{2}\\ \pm\frac{8}{2}\\ \end{align}
If none of those are roots, you have to use Cardano's formula for the roots of a cubic, which is a pain in the neck.
To find whether any of those listed are roots...you just try each one. Fortunately, in that list there are several duplicates, which can save you some time.
