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Let $f(x) = a_0 + a_1 x + ...... + a_n x^n$ be a polynomial of degree n with integral coefficients. If $f(1), a_0, a_n$ are odd then number of rational roots are.

My Try:

Let $f(x)=(x-\alpha)g(x), \alpha \in \mathbb I$

$f(0)=(0-\alpha)g(x)$ is odd, therefore both $\alpha$ and $g(0)$ must be odd, hence $(1-\alpha)$ must be even but $f(1)$ is odd. Therefore it won't have any integral root. How to prove for rational root?

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    $\begingroup$ "number of rational roots are" what? $\endgroup$
    – user694818
    Oct 19, 2019 at 20:27
  • $\begingroup$ Not sure if this helps, but if $f$ has a rational root, $\frac{p}{q}$, then $q^nf$ has an integral root. $\endgroup$
    – rogerl
    Oct 19, 2019 at 20:27

1 Answer 1

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By the rational root theorem, any rational root must be of the form $\alpha=\frac uv$ with $u\mid a_0$ and $v\mid a_n$. In particular, both $u$ and $v$ are odd. Now $$v^nf(x)=(vx-u)g(x) $$ where $g$ has integer(!) coefficients. If we plug in $x=1$ the left hand side is odd, the right hand side is even, contradiction.

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  • $\begingroup$ I didn't know about rational root theorem, learnt something new. Thanks $\endgroup$
    – Zenix
    Oct 19, 2019 at 20:40

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