# counting the real zeros of a polynomial and proving that it's irreducible over $\Bbb Q$

Let's consider the polynomial $$f\left( x \right) = \left( {x^2 + 2} \right)\prod\limits_{i = - k}^k {\left( {x - 2i} \right) + 2 \in {\Bbb Q}\left[ x \right]}$$ . Let's suppose that $p = 2k + 3 \geqslant 5$ is prime.

Prove the following:

$i)$ Prove that $f$ is irreducible and of degree $p$

$ii)$ Prove that p has exactly $p-2$ real zeros. I have no idea how to prove this, maybe with einsenstein and considering the derivate, but how? :/

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Can you prove that $f$ has degree $p$? – Gerry Myerson Nov 18 '12 at 22:49

Modulo $2$, $f(x)$ is just $x^p$; also, the constant term of $f(x)$ is 2; thus, you can apply Eisenstein with $p=2$.

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Ok , and how can I prove that has exactly p-2 real zeros? – Daniel Nov 18 '12 at 23:01
Maybe you can show it changes sign between $n-(1/2)$ and $n+(1/2)$ for $n=-k,\dots,k$. – Gerry Myerson Nov 18 '12 at 23:10
Well It's easy to see the following relation, let's call the polynomial $f_k$ so we can deduce the following $$f_k = \left( {f_{k - 1} - 2} \right)\left( {x - 2k} \right)\left( {x + 2k} \right) + 2$$ so derivating $f'_k = f'_{k-1} (x-2k)(x+2k)$ so, in each step I add two more critical points, and clearly $f'_1 = (x^2-2)(5x^2+4)$ thus $f'_k$ has exactly $2k$ critical points – Daniel Nov 18 '12 at 23:19
Not sure I follow your differentiation --- was expecting the Product Rule to make an appearance. – Gerry Myerson Nov 19 '12 at 1:12
My derivate is bad – Daniel Nov 20 '12 at 5:07