# How do you find all the zeros of f(x)=x^4-2x-1?

Aug 9, 2016

#### Explanation:

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

Since $f \left(x\right)$ has no term in ${x}^{3}$ it will factor as the product of a pair of quadratics with opposite middle terms:

${x}^{4} - 2 x - 1$

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

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

Equating coefficients and rearranging a little we get:

$\left\{\begin{matrix}b + c = {a}^{2} \\ b - c = - \frac{2}{a} \\ b c = - 1\end{matrix}\right.$

Then:

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

Multiply both ends by $\left({a}^{2}\right)$ and rearrange slightly to get:

${\left({a}^{2}\right)}^{3} + 4 \left({a}^{2}\right) - 4 = 0$

Use Cardano's method to solve this cubic in $\left({a}^{2}\right)$...

Let ${a}^{2} = u + v$

${u}^{3} + {v}^{2} + \left(3 u v + 4\right) \left(u + v\right) - 4 = 0$

Add the constraint $v = - \frac{4}{3 u}$ to eliminate the $\left(u + v\right)$ term:

${u}^{3} - \frac{64}{27 {u}^{3}} - 4 = 0$

Multiply through by $27 {u}^{3}$ and rearrange slightly to get:

$27 {\left({u}^{3}\right)}^{2} - 108 \left({u}^{3}\right) - 64 = 0$

Use the quadratic formula to find:

${u}^{3} = \frac{108 \pm \sqrt{{\left(- 108\right)}^{2} - 4 \left(27\right) \left(- 64\right)}}{2 \cdot 27}$

$= \frac{108 \pm \sqrt{11664 + 6912}}{54}$

$= \frac{108 \pm \sqrt{18576}}{54}$

$= \frac{108 \pm 12 \sqrt{129}}{54}$

$= \frac{54 \pm 6 \sqrt{129}}{27}$

Hence Real root:

${a}^{2} = \frac{1}{3} \left(\sqrt{54 + 6 \sqrt{129}} + \sqrt{54 - 6 \sqrt{129}}\right)$

We can use the positive root as the value of $a$:

$a = \sqrt{\frac{1}{3} \left(\sqrt{54 + 6 \sqrt{129}} + \sqrt{54 - 6 \sqrt{129}}\right)}$

Then:

$b = \frac{1}{2} \left({a}^{2} - \frac{2}{a}\right)$

$= \frac{1}{6} \left(\sqrt{54 + 6 \sqrt{129}} + \sqrt{54 - 6 \sqrt{129}}\right) - \frac{1}{\sqrt{\frac{1}{3} \left(\sqrt{54 + 6 \sqrt{129}} + \sqrt{54 - 6 \sqrt{129}}\right)}}$

$c = \frac{1}{2} \left({a}^{2} + \frac{2}{a}\right)$

$= \frac{1}{6} \left(\sqrt{54 + 6 \sqrt{129}} + \sqrt{54 - 6 \sqrt{129}}\right) + \frac{1}{\sqrt{\frac{1}{3} \left(\sqrt{54 + 6 \sqrt{129}} + \sqrt{54 - 6 \sqrt{129}}\right)}}$

This leaves us with two quadratics to solve:

${x}^{2} - \left(\sqrt{\frac{1}{3} \left(\sqrt{54 + 6 \sqrt{129}} + \sqrt{54 - 6 \sqrt{129}}\right)}\right) x + \left(\frac{1}{6} \left(\sqrt{54 + 6 \sqrt{129}} + \sqrt{54 - 6 \sqrt{129}}\right) - \frac{1}{\sqrt{\frac{1}{3} \left(\sqrt{54 + 6 \sqrt{129}} + \sqrt{54 - 6 \sqrt{129}}\right)}}\right) = 0$

${x}^{2} + \left(\sqrt{\frac{1}{3} \left(\sqrt{54 + 6 \sqrt{129}} + \sqrt{54 - 6 \sqrt{129}}\right)}\right) x + \left(\frac{1}{6} \left(\sqrt{54 + 6 \sqrt{129}} + \sqrt{54 - 6 \sqrt{129}}\right) + \frac{1}{\sqrt{\frac{1}{3} \left(\sqrt{54 + 6 \sqrt{129}} + \sqrt{54 - 6 \sqrt{129}}\right)}}\right) = 0$

We can find the roots of these quadratics using the quadratic formula, which are the zeros of the original quartic $f \left(x\right)$.