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

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

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Answer 1 · 2015-10-17T18:28:27

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:

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

$x^{2} + x + 1 = 0$ $x^2+x+1 = 0$ has solutions $x = \frac{- 1 \pm i \sqrt{3}}{2}$ $x = color(blue)((-1+-isqrt(3))/2)$

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

Then

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

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

So

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

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

So our cubic becomes:

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

Multiply through by $27$ $27$ to get:

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

Next let $t = u + v$ $t = u + v$ to get

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

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

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

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

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

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

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

So:

$u^{3} = \frac{- 675 \pm \sqrt{675^{2} - 4 \cdot 729}}{2 \cdot 729}$ $u^3 = (-675+-sqrt(675^2-4*729))/(2*729)$

$= \frac{- 675 \pm 27 \sqrt{625 - 4}}{27 \cdot 54}$ $=(-675+-27sqrt(625-4))/(27*54)$

$= \frac{- 25 \pm \sqrt{621}}{54}$ $=(-25+-sqrt(621))/54$

$= \frac{- 25 \pm \sqrt{9 \cdot 69}}{54}$ $=(-25+-sqrt(9*69))/54$

$= \frac{- 25 \pm 3 \sqrt{69}}{54}$ $=(-25+-3sqrt(69))/54$

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

$t = \sqrt[3]{\frac{- 25 + 3 \sqrt{69}}{54}} + \sqrt[3]{\frac{- 25 - 3 \sqrt{69}}{54}}$ $t = root(3)((-25+3sqrt(69))/54) + root(3)((-25-3sqrt(69))/54)$

$= \frac{1}{3} \sqrt[3]{\frac{- 25 + 3 \sqrt{69}}{2}} + \frac{1}{3} \sqrt[3]{\frac{- 25 - 3 \sqrt{69}}{2}}$ $= 1/3root(3)((-25+3sqrt(69))/2) + 1/3root(3)((-25-3sqrt(69))/2)$

and

$x = t + \frac{1}{3} = \frac{1}{3} + \frac{1}{3} \sqrt[3]{\frac{- 25 + 3 \sqrt{69}}{2}} + \frac{1}{3} \sqrt[3]{\frac{- 25 - 3 \sqrt{69}}{2}}$ $x = t+1/3 = color(blue)(1/3+1/3root(3)((-25+3sqrt(69))/2) + 1/3root(3)((-25-3sqrt(69))/2))$

$\approx - 0.75487766625$ $~~ -0.75487766625$

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

$x = \frac{1}{3} + \frac{ω}{3} \sqrt[3]{\frac{- 25 + 3 \sqrt{69}}{2}} + \frac{ω^{2}}{3} \sqrt[3]{\frac{- 25 - 3 \sqrt{69}}{2}}$ $x = color(blue)(1/3+omega/3 root(3)((-25+3sqrt(69))/2) + omega^2/3 root(3)((-25-3sqrt(69))/2))$

$x = \frac{1}{3} + \frac{ω^{2}}{3} \sqrt[3]{\frac{- 25 + 3 \sqrt{69}}{2}} + \frac{ω}{3} \sqrt[3]{\frac{- 25 - 3 \sqrt{69}}{2}}$ $x = color(blue)(1/3+omega^2/3 root(3)((-25+3sqrt(69))/2) + omega/3 root(3)((-25-3sqrt(69))/2))$

That is

$x \approx 0.87744 + 0.74486 i$ $x~~0.87744+0.74486 i$

$x \approx 0.87744 - 0.74486 i$ $x~~0.87744-0.74486 i$

Postscript

Note that in general, quintic equations of the form $x^{5} + x + a = 0$ $x^5+x+a = 0$ 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:

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

If $a in RR$ then $BR(a) in RR$ is the Real root.

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