## Example of the Solution of a Quartic Polynomial

Here’s a worked example of the solution of a quartic polynomial, as described in my earlier post on solving polynomials algebraically.

This example is artificial, but the principle would be the same for any other quartic with real or complex coefficients. In this case, the roots are all real.

$0 = 16 x^4 + 96 x^3 - 24 x^2 - 664 x - 315$.

First, we divide through by 16 to make this monic:

$x^4 + 6 x^3 - \frac{3}{2} x^2 - \frac{83}{2} x - \frac{315}{16}$.

Depress this by substituting
$x = y - \frac{6}{4} = y - \frac{3}{2}$:

$0 = y^4 + c y^2 +b y + a$.

where
$c = -15$;
$b = -10$;
$a = 24$;

that is:

$0 = y^4 - 15 y^2 - 10 y + 24$.

To solve this, we wish to convert this to the biquadratic,

$0 = (y^2 + p y + s)(y^2 + q y + t)$,

which means that we must solve the cubic in $p^2$:

$0 = p^6 + 2c p^4 + (c^2 - 4a) p^2 - b^2$

which is

$0 = p^6 - 30 p^4 + 129 p^2 - 100$

Depress this with
$p^2 = g - \frac{-30}{3} = g + 10$:

$0 = g^3 - 171 g - 810$.

Now make the ‘magic’ substitution

$g = f + \dfrac{171}{3f} =f + \dfrac{57}{f}$,

and then multiply by $f^3$:

$0 = f^6 - 810 f^3 +185193$.

That’s just a quadratic in $f^3$, and so we have:

$f^3 = (810 \pm \sqrt{810^2 - 4 \times 185193}) /2 = 405 \pm 84 \sqrt{3} i$,

hence

$f = \sqrt[3]{405 \pm 84 \sqrt{3} i} * \omega^n = (\frac{15}{2} \pm \frac{\sqrt{3}}{2}i) * \omega^n$

where $1 \ne \omega = \sqrt[3]{1} = -\frac{1}{2} + \frac{\sqrt{3}}{2}i$.

(That is a fortunate reduction I discovered numerically, and then checked algebraically; you might find that you must work either numerically, or with a more complicated expression in other examples.)

So we have the six values $f \in \left\{\frac{15}{2} \pm \frac{\sqrt{3}}{2}i, -\frac{9}{2} \pm \frac{7\sqrt{3}}{2}i, -3 \pm 4\sqrt{3}i \right\}$.

Now, as it happens, it doesn’t matter which one of these is chosen. Let’s look at them all!

Back-substituting, we get the three values (each occurring twice) $g \in {-9, -6, 15}$.

Back-substituting again, we get the six values $p \in {\pm 1, \pm 2, \pm 5}$.

Now,

$q = -p$;
$s = \alpha - \beta$;
$t = \alpha + \beta$;

where

$\alpha = (c + p^2)/2$;
$\beta = b / 2p$;

We’ll take some values of $p$ in turn to see what happens.

To solve the biquadratic $0 = (y^2 + p y + s)(y^2 + q y + t)$, we may solve the two factors separately:

$0 = (y^2 + p y + s)$;
$0 = (y^2 + q y + t)$.

First, $p=1 \Rightarrow \alpha=-7, \beta=-5 \Rightarrow q=-1, s=-2, t=-12$:

$0 = (y^2 + y - 2) \Rightarrow y \in {-2 , 1}$;
$0 = (y^2 - y - 12) \Rightarrow y \in {-3 , 4}$.

Alternatively, $p=2 \Rightarrow \alpha=-\frac{11}{2}, \beta=-\frac{5}{2} \Rightarrow q=-2, s=-3, t=-8$:

$0 = (y^2 + 2 y - 3) \Rightarrow y \in {-3 , 1}$;
$0 = (y^2-2y-8) \Rightarrow y \in{-2,4}$.

Alternatively, $p=5 \Rightarrow \alpha=5, \beta=-1 \Rightarrow q=-5, s=6, t=4$:

$0 = (y^2 + 5 y + 6) \Rightarrow y \in {-3 , -2}$;
$0 = (y^2 - 5 y + 4) \Rightarrow y \in {1 , 4}$.

Alternatively, we can take $p=-1, -2, -5$; in those cases the roles of the variables are just reversed: $p \leftrightarrow q, s \leftrightarrow t$. Hence in all cases we have

$y \in {-3 , -2, 1, 4}$.

Hence, with the final back-substitution:

$x \in {-\frac{9}{2} , -\frac{7}{2}, -\frac{1}{2}, \frac{5}{2}}$.

In each case, it is easy to check that

$0 = 16 x^4 + 96 x^3 - 24 x^2 - 664 x-315$

and we are finished.