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?
Best Answer
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.