I've got a question which shows short answers and no method so I'm trying to find a hand performed method of solving the cubic polynomial for the roots:
$$f(x) = 2x^3 + 8x^2 + 10x - 6$$
From the answers, I know the roots are: x = $0.4334, -2.2167+1.4170i, -2.2167-1.4170i$
The best I can do is factor out the $2$ then guess a real integer root and long divide, rinse/repeat until you find one that works. However that won't work in this example given no root is real and rational.
Thank you for any help!
You can solve cubics using a similar idea to 'completing the square'. The first step is to note that $(x+y)^3=x^3+3x^2y+3xy^2+y^3$ and use this to remove the quadratic factor. For your example, lets just factorize $x^3+4x^2+5x-3$. Once we do this we can multiply by $2$ to get the original equation. We write the equation as:
$$ \begin{split} x^3+4x^2+5x-3&=x^3+3\cdot\left(\frac{4}{3}\right)x^2+3\cdot\left(\frac{4}{3}\right)^2x+\left(\frac{4}{3}\right)^3-\frac{1}{3}x-\frac{145}{27} \\ &= \left( x+\frac{4}{3}\right)^3-\frac{1}{3}x-\frac{145}{27} \end{split} $$
We would now like to set a new variable $y=\left(x+\frac{4}{3}\right)$ so that the equation ends up looking like $y^3+ay+b$ for some $a,b\in\mathbb{R}$. So we write the equation as:
$$ \left(x+\frac{4}{3}\right)^3-\frac{1}{3}\left(x+\frac{4}{3}\right)-\frac{133}{27}=y^3-\frac{1}{3}y-\frac{133}{27} $$
Now we need to use a trick due to Cardano. We set $y=u+v$ in our equation above and we get:
$$ \begin{split} (u+v)^3-\frac{1}{3}(u+v)-\frac{133}{27}&=u^3+3u^2v+3uv^2+v^3-\frac{1}{3}(u+v)-\frac{133}{27} \\&= u^3+v^3+3uv(u+v)-\frac{1}{3}(u+v)-\frac{133}{27}\\ &=u^3+v^3+(u+v)\left(3uv-\frac{1}{3}\right)-\frac{133}{27} \end{split} $$
and now here is the point of making $y=u+v$: we have some extra freedom to choose our $u$ and $v$ and a convenient choice for us is to make $3uv-\frac{1}{3}=0$ so that the middle term cancels and we get:
$$ u^3+v^3=\frac{133}{27} $$
Now the quadratic:
$$ (z-u^3)(z-v^3)=z^2-(u^3+v^3)z+u^3v^3=z^2-\frac{133}{27}z+\left(\frac{1}{9}\right)^3 $$
has roots $u^3$ and $v^3$. So we can solve this quadratic and then the cube roots of the solutions will give us $u$ and $v$. We can then substitute these values in to get $y$ and finally $x$. This is a lot of work though! This process can be used to solve any cubic however.