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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?

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    $\begingroup$ Look for a quadratic factor. $\endgroup$ – lulu May 31 '18 at 23:39
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    $\begingroup$ What you have done is use the "Rational Roots Theorem", and have excluded the possibility of rational roots. There may very well be real roots to this polynomial; in fact, since the degree is odd it has at least one real root. $\endgroup$ – Dave May 31 '18 at 23:41
  • $\begingroup$ Ok, Ill try looking for a quadratic factor. Thanks! $\endgroup$ – Guysudai1 May 31 '18 at 23:42
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    $\begingroup$ By examining the derivative $5(x^4-1)$ you will see a max at x=-1 and a min at x=1. At x=-1, the function = 7, while at x=1, the function = -1. This implies that the function has 3 real zeroes, one between -1 and 1, while the other two are (one each) $\gt 1$ and $\lt -1$. $\endgroup$ – herb steinberg May 31 '18 at 23:52
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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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  • $\begingroup$ Thanks for the detailed answer! $\endgroup$ – Guysudai1 Jun 1 '18 at 7:03
  • $\begingroup$ @fdsaddsa. You are very welcome ! Glad to help. $\endgroup$ – Claude Leibovici Jun 1 '18 at 7:12
  • $\begingroup$ I'm just getting complicated with the cubic equation's solution. I'll use this way of solution in my next problems $\endgroup$ – Guysudai1 Jun 1 '18 at 7:13
  • $\begingroup$ @fdsaddsa. It is not so complicated. Follow the steps given in the lnked page of Wikipedia. $\endgroup$ – Claude Leibovici Jun 1 '18 at 7:36
  • $\begingroup$ This was my slowly-arrived at method of solution, after I spent overmuch time trying to show irreducibility. Thanks for this! $\endgroup$ – Lubin Jun 1 '18 at 12:38

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