Finding irrational and complex roots of a cubic polynomial

Question

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!

Answer 1

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.

Answer 2

Let $a=10,b=8,c=2,f=6$.

One of many ways get roots:

$X=\dfrac{-(a b+9 c f)+\sqrt{(a b+9 c f)^2-4 (b^2-3 a c) (a^2+3 b f)}}{2 (b^2-3 a c)}=(-47+\sqrt{1965})/2$

$A=a+2 b X+3 c X^2=5895-133 \sqrt{1965}$

$F=a X+b X^2+c X^3-f=655 (-133+3\sqrt{1965})$

$G=\sqrt[3]{A^3-27 c F^2}=\sqrt[3]{-7860 (-5895+133\sqrt{1965})}$

$Q=\left\{G,-(-1)^{1/3} G,(-1)^{2/3} G\right\}=\left\{2^{2/3} \sqrt[3]{-1965 \left(133 \sqrt{1965}-5895\right)},-(-2)^{2/3} \sqrt[3]{1965 \left(133 \sqrt{1965}-5895\right)},-2^{2/3} \sqrt[3]{1965 \left(133 \sqrt{1965}-5895\right)}\right\}$

$x=X+\dfrac{3 F}{Q-A}=\left\{\frac{1}{2} \left(\sqrt{1965}+\frac{3930 \left(3 \sqrt{1965}-133\right)}{2^{2/3} \sqrt[3]{-1965 \left(133 \sqrt{1965}-5895\right)}+133 \sqrt{1965}-5895}-47\right),\frac{1}{2} \left(\sqrt{1965}+\frac{3930 \left(3 \sqrt{1965}-133\right)}{-(-2)^{2/3} \sqrt[3]{1965 \left(133 \sqrt{1965}-5895\right)}+133 \sqrt{1965}-5895}-47\right),\frac{1}{2} \left(\sqrt{1965}+\frac{3930 \left(3 \sqrt{1965}-133\right)}{-2^{2/3} \sqrt[3]{1965 \left(133 \sqrt{1965}-5895\right)}+133 \sqrt{1965}-5895}-47\right)\right\}$

Verifing in pari/gp:

? X=(-47+sqrt(1965))/2;
?
? A=5895-133*sqrt(1965);
?
? F=655*(-133+3*sqrt(1965));
?
? G=(-7860*(-5895+133*sqrt(1965)))^(1/3);
?
? Q=[G,-(-1)^(1/3)*G,(-1)^(2/3)*G];
?
? x=[X+3*F/(Q[1]-A),X+3*F/(Q[2]-A),X+3*F/(Q[3]-A)]
%1 = [-2.2167138319319099042883646224845123271 + 1.4169509457209434022679297010446478619*I,
-2.2167138319319099042883646224845123271 - 1.4169509457209434022679297010446478620*I,
0.43342766386381980857672924496902465415 - 2.4904167685289165540 E-39*I]