Prove that a polynomial has no rational roots

contest-mathpolynomialsrational numbersroots

Let $P(x)$ be an integer polynomial whose leading coefficient is odd. Suppose that $P(0)$
and $P(1)$ are also odd.
Prove that $P(x)$ has no rational roots.

I have been able to prove that there are no integer roots (using the binomial theorem), and I'm stuck.

Best Answer

Let $P(x)$ be an integer polynomial whose leading coefficient is odd. $P(0)$ and $P(1)$ are also odd. Prove that $P(x)$ has no rational roots.

Let the polynomial $P(x)$ have a rational root $\frac ab\Rightarrow P(x)=(bx-a)Q(x)$, where $Q(x)$ is an integer polynomial. $b,a$ have to be odd since the leading coefficient, $P(0)$ are odd respectively. Now $P(1)$ is even since $b-a$ is even. Hence, a contradiction.

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