# Number of rational roots

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?

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

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.