# How do you find all the zeros of f(x)=x^4-x^2-3x+3?

Jul 23, 2016

Real zeros: $1$, $\frac{1}{3} \left(1 + \sqrt[3]{\frac{83 + 9 \sqrt{85}}{2}} + \sqrt[3]{\frac{83 - 9 \sqrt{85}}{2}}\right)$

and related Complex zeros.

#### Explanation:

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

I suspect there may be a typo here. Was ${x}^{4}$ supposed to be ${x}^{3}$? I will proceed assuming that it is correct as given...

First note that since the sum of the coefficients of $f \left(x\right)$ is zero, $f \left(1\right) = 0$ and $\left(x - 1\right)$ is a factor:

${x}^{4} - {x}^{2} - 3 x + 3 = \left(x - 1\right) \left({x}^{3} + {x}^{2} - 3\right)$

The discriminant $\Delta$ of a cubic polynomial in the form $a {x}^{3} + b {x}^{2} + c x + d$ is given by the formula:

$\Delta = {b}^{2} {c}^{2} - 4 a {c}^{3} - 4 {b}^{3} d - 27 {a}^{2} {d}^{2} + 18 a b c d$

In our case $a = 1$, $b = 1$, $c = 0$ and $d = - 3$, so we find:

$\Delta = 0 + 0 + 12 - 243 + 0 = - 231 < 0$

Since $\Delta < 0$ this cubic has one Real zero and a Complex conjugate pair of non-Real zeros.

We could use the substitution $t = \frac{1}{x}$ to get a simplified cubic, but on this occasion, let's multiply $\left({x}^{3} + {x}^{2} - 3\right)$ by $27$ and use $t = 3 x - 1$. (That will make it slightly easier to find the roots in $a + b i$ form.)

$27 \left({x}^{3} - {x}^{2} - 3\right)$

$= 27 {x}^{3} - 27 {x}^{2} - 81$

$= {\left(3 x\right)}^{3} - 3 {\left(3 x\right)}^{2} - 81$

$= {\left(3 x - 1\right)}^{3} - 3 \left(3 x - 1\right) - 83$

$= {t}^{3} - 3 t - 83$

So we want to solve ${t}^{3} - 3 t - 83 = 0$

Use Cardano's method:

Let $t = u + v$

${u}^{3} + {v}^{3} + 3 \left(u v - 1\right) \left(u + v\right) - 83 = 0$

Eliminate the $\left(u + v\right)$ term by letting $v = \frac{1}{u}$ to find:

${u}^{3} + \frac{1}{u} ^ 3 - 83 = 0$

Multiply through by ${u}^{3}$ to get:

${\left({u}^{3}\right)}^{2} - 83 \left({u}^{3}\right) + 1 = 0$

${u}^{3} = \frac{83 \pm \sqrt{{83}^{2} - 4 \left(1\right) \left(1\right)}}{2 \cdot 1}$

$= \frac{83 \pm \sqrt{6889 - 4}}{2}$

$= \frac{83 \pm \sqrt{6885}}{2}$

$= \frac{83 \pm 9 \sqrt{85}}{2}$

Since this is Real valued and our derivation was symmetric in $u$ and $v$, we can use one of these roots for ${u}^{3}$ and the other for ${v}^{3}$ to deduce the roots of our cubic in $t$.

Then we can use $x = \frac{1}{3} \left(t + 1\right)$ to find the Real zero of our original cubic:

${x}_{1} = \frac{1}{3} \left(1 + \sqrt[3]{\frac{83 + 9 \sqrt{85}}{2}} + \sqrt[3]{\frac{83 - 9 \sqrt{85}}{2}}\right)$

and related Complex zeros:

${x}_{2} = \frac{1}{3} \left(1 + \omega \sqrt[3]{\frac{83 + 9 \sqrt{85}}{2}} + {\omega}^{2} \sqrt[3]{\frac{83 - 9 \sqrt{85}}{2}}\right)$

${x}_{3} = \frac{1}{3} \left(1 + {\omega}^{2} \sqrt[3]{\frac{83 + 9 \sqrt{85}}{2}} + \omega \sqrt[3]{\frac{83 - 9 \sqrt{85}}{2}}\right)$

where $\omega = - \frac{1}{2} + \frac{\sqrt{3}}{2} i$ is the primitive Complex cube root of $1$.