How to solve $x+x^2+x^3+x^4+x^5=62$?

Question

$$x+x^2+x^3+x^4+x^5=62$$

The roots in the set of Real numbers is $x=2$, but how solve it?

Answer 1

Apply the rational roots theorem:

$$62\implies1,2,31,62$$

Thus, it is easy enough to check if $x=\pm1,\pm2,\pm31,\pm62$ are roots. Indeed, if we check, we find that $x=2$ is the only rational root. The remaining roots can be deduced to be complex since after factoring out, we find it has only one real relative minimum at $x=-1.55$ such that $\left(x^4+3 x^3+7 x^2+15 x+31\right)>0$ for all $x$.

Answer 2

By the Fundamental Theorem of Algebra you know it must have $5$ complex roots. Now from one of the comments you see that one is real and the other are complex. Mathematica gives: