solving quintics up to a number of degree

Question

Let's say I have the following fifth degree polynomial (the coefficients are left simple for clarity) $$\mathcal{P}(x)=2x^5+3x^4-2x^3-x^2-x+1$$

I know that there exists a real number $\eta$ such that $\mathcal{P}(\eta)=0.$

So we can write $\mathcal{P}(x)$ as: $$\mathcal{P(x)}=(x-\eta)\times\mathcal{O}(x)$$ where $\mathcal{O}(x)$ is a polynomial of fourth order. Now my mission is to find out $\mathcal{O}(x)$ and then solve $$(x-\eta)(\mathcal{O}(x))=0$$

If I have a problem I can factorize $\mathcal{O}(x)$ to two other polynomials and so I can solve them. What do you think of my work friends?

I'm not sure I follow... are you saying that you can solve every quintic? If my math is right, one solution is $x\in[-3,-2]$... — abiessu, Feb 04 '14 at 19:22
It's like solving a quintic using the fundamental theorem of algebra. — user126208, Feb 04 '14 at 19:23
I don't think anyone has said that it is not possible to solve a given quintic, just that there is no general formula for all solutions to all quintics. — abiessu, Feb 04 '14 at 19:24
Use rational root theorem. This one splits over $\Bbb Z[x]$ in a linear-quartic form. — Balarka Sen, Feb 13 '14 at 13:37

score 0 · Accepted Answer · answered Feb 04 '14 at 19:31

0

Yes, this is absolutely correct. Unfortunately there's the tricky step of finding $\eta$... how would you go about this? Once you've found it you can, of course, solve the quartic by means of the general formula.

answered Feb 04 '14 at 19:31

GPerez

6,766

solving quintics up to a number of degree

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