<s>[INST] Finding the real roots and complex roots to a polynomial I have tried solving and finding the real roots to the polynomial
$ {x^5 - 5x + 3} $ 
by saying that one of the solutions has to be a factor of the $ +3 $, so the factors might be 
$ \pm 1, or \pm 3  $
and when I place the supposed factors into the polynomial, I get that none of them zero out, so I guess it means no real numbers have the zeroes for this function. So how do I go about finding the complex roots if I don't have the real roots?
 [/INST] Suppose that
$$x^5-5x+3=(x^2+a x+b)(x^3+c x^2+d x+e)$$ Expand and group to get
$$(b e-3)+x (a e+b d+5)+x^2 (a d+b c+e)+x^3 (a c+b+d)+x^4
   (a+c)=0$$ n which all coefficients must be equal to $0$.
Then, successively, $c=-a$, $b+d=a^2\implies d=a^2-b$, $e=-a^3+2ab$ make that we are left with
$$-3 - a^3 b + 2 a b^2=0 \qquad \text{and} \qquad 5 - a^4 + 3 a^2 b - b^2=0$$ where $a=1$, $b=-1$ seem to be "obvious" solutions.
So, back to $c,d,e$
$$x^5-5x+3=(x^2+x-1)(x^3-x^2+2x-3)$$ The first term has two real roots and the second has  then the equation has one real root and two non-real complex conjugate roots since $\Delta=-175$.
Now, have look here for the solution of the cubic equation.
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