# How do I find all real and complex zeros of x^5+x+1?

Oct 17, 2015

Factor into the product of a quadratic and a cubic, then find the roots of those using the quadratic formula and Cardano's method.

#### Explanation:

${x}^{5} + x + 1 = \left({x}^{2} + x + 1\right) \left({x}^{3} - {x}^{2} + 1\right)$

${x}^{2} + x + 1 = 0$ has solutions $x = \textcolor{b l u e}{\frac{- 1 \pm i \sqrt{3}}{2}}$

To solve ${x}^{3} - {x}^{2} + 1 = 0$ first substitute $t = x - \frac{1}{3}$

Then

${t}^{3} = {\left(x - \frac{1}{3}\right)}^{3} = {x}^{3} - {x}^{2} + \frac{1}{3} x - \frac{1}{27}$

$- \frac{1}{3} t = - \frac{1}{3} \left(x - \frac{1}{3}\right) = - \frac{1}{3} x + \frac{1}{9}$

So

${t}^{3} - \frac{1}{3} t + \frac{25}{27} = {x}^{3} - {x}^{2} + \frac{1}{3} x - \frac{1}{27} - \frac{1}{3} x + \frac{1}{9} + \frac{25}{27}$

$= {x}^{3} - {x}^{2} + 1$

So our cubic becomes:

${t}^{3} - \frac{1}{3} t + \frac{25}{27} = 0$

Multiply through by $27$ to get:

$27 {t}^{3} - 9 t + 25 = 0$

Next let $t = u + v$ to get

$0 = 27 \left({u}^{3} + 3 {u}^{2} v + 3 u {v}^{2} + {v}^{3}\right) - 9 \left(u + v\right) + 25$

$= 27 {u}^{3} + 27 {v}^{3} + \left(81 u v - 9\right) \left(u + v\right) + 25$

Add the constraint that $v = \frac{1}{9 u}$

Then $81 u v - 9 = 9 \left(9 u v - 1\right) = 9 \left(1 - 1\right) = 0$, so our cubic becomes:

$0 = 27 {u}^{3} + 27 {\left(\frac{1}{9} u\right)}^{3} + 25 = 27 {u}^{3} + \frac{1}{27 {u}^{3}} + 25$

Multiply through by $27 {u}^{3}$ to get a quadratic in ${u}^{3}$:

$0 = 729 {\left({u}^{3}\right)}^{2} + 675 \left({u}^{3}\right) + 1$

So:

${u}^{3} = \frac{- 675 \pm \sqrt{{675}^{2} - 4 \cdot 729}}{2 \cdot 729}$

$= \frac{- 675 \pm 27 \sqrt{625 - 4}}{27 \cdot 54}$

$= \frac{- 25 \pm \sqrt{621}}{54}$

$= \frac{- 25 \pm \sqrt{9 \cdot 69}}{54}$

$= \frac{- 25 \pm 3 \sqrt{69}}{54}$

Since the derivation was symmetric in $u$ and $v$, the Real root is given by:

$t = \sqrt{\frac{- 25 + 3 \sqrt{69}}{54}} + \sqrt{\frac{- 25 - 3 \sqrt{69}}{54}}$

$= \frac{1}{3} \sqrt{\frac{- 25 + 3 \sqrt{69}}{2}} + \frac{1}{3} \sqrt{\frac{- 25 - 3 \sqrt{69}}{2}}$

and

$x = t + \frac{1}{3} = \textcolor{b l u e}{\frac{1}{3} + \frac{1}{3} \sqrt{\frac{- 25 + 3 \sqrt{69}}{2}} + \frac{1}{3} \sqrt{\frac{- 25 - 3 \sqrt{69}}{2}}}$

$\approx - 0.75487766625$

The Complex roots are found using $\omega = - \frac{1}{2} + i \frac{\sqrt{3}}{2}$ as:

$x = \textcolor{b l u e}{\frac{1}{3} + \frac{\omega}{3} \sqrt{\frac{- 25 + 3 \sqrt{69}}{2}} + {\omega}^{2} / 3 \sqrt{\frac{- 25 - 3 \sqrt{69}}{2}}}$

$x = \textcolor{b l u e}{\frac{1}{3} + {\omega}^{2} / 3 \sqrt{\frac{- 25 + 3 \sqrt{69}}{2}} + \frac{\omega}{3} \sqrt{\frac{- 25 - 3 \sqrt{69}}{2}}}$

That is

$x \approx 0.87744 + 0.74486 i$

$x \approx 0.87744 - 0.74486 i$

Postscript

Note that in general, quintic equations of the form ${x}^{5} + x + a = 0$ are not so easily (!) solved. They are normally not expressible in terms of ordinary square and cube roots or any kind of ordinary radical.

There is a special kind of radical called a Bring Radical, which represents the principal root of this quintic. That is:

$B R \left(a\right)$ is the principal root of ${x}^{5} + x + a = 0$

If $a \in \mathbb{R}$ then $B R \left(a\right) \in \mathbb{R}$ is the Real root.