# How do you find all the real and complex roots of  P(x) = 3x^4 + x^3 + 2x^2 - 2?

May 5, 2016

It gets rather complicated, but it's doable...

#### Explanation:

$P \left(x\right) = 3 {x}^{4} + {x}^{3} + 2 {x}^{2} - 2$

You might try looking for rational zeros first, but you would not find any. This is a full blown quartic with $2$ Real roots and $2$ Complex roots, all irrational.

Normally with a quartic of the form $a {x}^{4} + b {x}^{3} + c {x}^{2} + \mathrm{dx} + e$, the first thing you would do would be to substitute something like $t = x + \frac{b}{4 a}$ in order to get a quartic in $t$ with no ${t}^{3}$ term.

In our example, there's an easier method by substituting $t = \frac{1}{x}$ and multiplying by $- 2 {t}^{4}$ to get:

$0 = 4 {t}^{4} - 4 {t}^{2} - 2 t - 6$

$= \left(2 {t}^{2} - a t + b\right) \left(2 {t}^{2} + a t + c\right)$

$= 4 {t}^{4} + \left(2 b + 2 c - {a}^{2}\right) {t}^{2} + a \left(b - c\right) t + b c$

Equating coefficients and rearranging a little, we get:

$\left\{\begin{matrix}b + c = \frac{{a}^{2} - 4}{2} \\ b - c = - \frac{2}{a} \\ b c = - 6\end{matrix}\right.$

Then:

${\left(\frac{{a}^{2} - 4}{2}\right)}^{2} = {\left(b + c\right)}^{2} = {\left(b - c\right)}^{2} + 4 b c = {\left(- \frac{2}{a}\right)}^{2} - 24$

Hence:

${\left({a}^{2}\right)}^{3} - 8 {\left({a}^{2}\right)}^{2} + 112 \left({a}^{2}\right) - 16 = 0$

Multiply through by ${3}^{3} = 27$ to get:

$0 = 27 {\left({a}^{2}\right)}^{3} - 216 {\left({a}^{2}\right)}^{2} + 3024 \left({a}^{2}\right) - 432$

$= {\left(3 \left({a}^{2}\right) - 8\right)}^{3} + 816 \left(3 \left({a}^{2}\right) - 8\right) + 6608$

Substitute $w = 3 {a}^{2} - 8$ to get:

${w}^{3} + 816 w + 6608 = 0$

Let $w = u + v$ to get:

${u}^{3} + {v}^{3} + 3 \left(u v + 272\right) \left(u + v\right) + 6608 = 0$

Add the constraint $v = - \frac{272}{u}$ to eliminate the term in $\left(u + v\right)$ and multiply through by ${u}^{3}$ to get:

${\left({u}^{3}\right)}^{2} + 6608 \left({u}^{3}\right) - 20123648 = 0$

Since this derivation is symmetric in $u$ and $v$, we can use the quadratic formula and hence the two Real roots of this quadratic as ${u}^{3}$ and ${v}^{3}$, to provide a Real value for $w$ and hence ${a}^{2}$...

${u}^{3} = \frac{- 6608 \pm \sqrt{{6608}^{2} + 4 \cdot 20123648}}{2}$

$= \frac{- 6608 \pm \sqrt{124160256}}{2}$

$= \frac{- 6608 \pm 48 \sqrt{53889}}{2}$

$= \frac{- 6608 \pm 48 \sqrt{53889}}{2}$

$= 8 \left(- 413 \pm 3 \sqrt{53889}\right)$

Hence:

$3 {a}^{2} - 8 = w = u + v = 2 \left(\sqrt[3]{- 413 + 3 \sqrt{53889}} + \sqrt[3]{- 413 - 3 \sqrt{53889}}\right)$

Hence, choosing the positive root:

$a = \sqrt{\frac{2}{3} \left(4 + \sqrt[3]{- 413 + 3 \sqrt{53889}} + \sqrt[3]{- 413 - 3 \sqrt{53889}}\right)}$

Then $b = \frac{{a}^{3} - 4 a - 4}{4 a}$ and $c = \frac{{a}^{3} - 4 a + 4}{4 a}$.

Hence quadratics in $t$ to solve.

Then the zeros of the original polynomial are the reciprocals of the roots of these quadratics.

The resulting expressions are rather complicated, but tractable.