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

##### 1 Answer

#### Answer:

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^2+x+1 = 0# has solutions#x = color(blue)((-1+-isqrt(3))/2)#

To solve

Then

#t^3 = (x-1/3)^3 = x^3-x^2+1/3x-1/27#

#-1/3t = -1/3(x-1/3) = -1/3x+1/9#

So

#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#

So our cubic becomes:

#t^3-1/3t+25/27 = 0#

Multiply through by

#27t^3-9t+25 = 0#

Next let

#0 = 27(u^3+3u^2v+3uv^2+v^3)-9(u+v)+25#

#=27u^3+27v^3+(81uv-9)(u+v)+25#

Add the constraint that

Then

#0 = 27u^3+27(1/9u)^3+25 = 27u^3 + 1/(27u^3)+25#

Multiply through by

#0 = 729(u^3)^2+675(u^3)+1#

So:

#u^3 = (-675+-sqrt(675^2-4*729))/(2*729)#

#=(-675+-27sqrt(625-4))/(27*54)#

#=(-25+-sqrt(621))/54#

#=(-25+-sqrt(9*69))/54#

#=(-25+-3sqrt(69))/54#

Since the derivation was symmetric in

#t = root(3)((-25+3sqrt(69))/54) + root(3)((-25-3sqrt(69))/54)#

#= 1/3root(3)((-25+3sqrt(69))/2) + 1/3root(3)((-25-3sqrt(69))/2)#

and

#x = t+1/3 = color(blue)(1/3+1/3root(3)((-25+3sqrt(69))/2) + 1/3root(3)((-25-3sqrt(69))/2))#

The Complex roots are found using

#x = color(blue)(1/3+omega/3 root(3)((-25+3sqrt(69))/2) + omega^2/3 root(3)((-25-3sqrt(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~~0.87744+0.74486 i#

#x~~0.87744-0.74486 i#

**Postscript**

Note that in general, quintic equations of the form

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

#BR(a)# is the principal root of#x^5+x+a = 0#

If