Solve the Cubic $x^3+24x^2+6x-4$

Question

I'm having trouble solving this cubic: $x^3+24x^2+6x-4$. Is anyone able to help explain how to get the values of $x$?

Answer 1

There's an algorithm, and I'll outline only part of it.

1. Depress the cubic

Substitute $y$ where $x=y-8$

$(y-8)^3+24(y-8)^2+6(y-8)-4=0$

$y^3-512-24y(y-8)+6(y-8)(4(y-8)+1)-4=0$

$y^3-512-24y(y-8)+(6y-48)(4y-23)-4=0$

$y^3-24y^2+24y^2+192y-138y-192y+48 \cdot 23 - 512-4=0$

$y^3-138y-588=0$

How$\textit{(exercise)}$ and why did I depress? The answer will be clear soon.

2. Wizardry from Hogwarts itself

Express $y$ as the sum of two variables, say $p$ and $q$. Don't worry, just do it.

$(p+q)^3-138(p+q)+588=0$

$p^3+q^3+3pq(p+q)-138(p+q)+588=0$

$p^3+q^3+(3pq-138)(p+q)+588=0$

Now, obviously $p$ and $q$ can vary and now we have a weird factor on our hands. But let's say there were particular values of both variables that we're looking for. Particularly, one that abolishes that nasty factor. Here's what I'm looking at.

$3pq-138=0$

Or, will be of use later:

$pq=46 \iff q=\dfrac{46}p \tag 1$

Now if we set this extra condition on the variables, they will have distinct solutions. Solutions that we now can find, now that we've decimated that factor. Moving on.

$p^3+q^3+588=0$

$p^3+\left(\dfrac{46}p \right)^3+588=0$

$p^3+\dfrac{46^3}{p^3}+588=0$

One more substitution(last one, I promise):

$z=p^3$

$z+\dfrac{46^3}z+588=0$

$z^2+588z+46^3=0$

$\begin{align} z&=\dfrac{-588 \pm \sqrt{-43600}}2 \\ &= -294 \pm 10\sqrt{109} i \end{align}$

3. Run backwards

So now to cube root $z$ and get $p$. Turns out you only need to pick one of the values for $z$ because of some complex conjugates when calculating $p$. For this I picked $z=-294+10\sqrt{109}$

With Euler's formula and the miracle of computers we get:

$p_1 \approx 4.035888724+5.450835001 i$

$p_2\approx -6.738505945+0.7697646609 i$

$p_3 \approx 2.702617221 - 6.22059962 i$

Longest possible string given to stay as true as possible to the answer.

From there you just use $(1)$ to get $q$ for each of those values and then calculate $x$ from there. Simple stuff.

And this is why we graph cubics instead.

Answer 2

As @Dietrich Burde commented,by the rational root theorem, there are no rational roots.

If you follow the steps given here and use the trignonometric for three real roots, you should find that the roots are given by $$x_k=-8+2 \sqrt{62} \cos \left(\frac{2 \pi k}{3}-\frac{1}{3} \cos ^{-1}\left(-\frac{243}{31 \sqrt{62}}\right)\right)\qquad \text{with} \qquad k=0,1,2$$