# Prove that a polynomial has no rational roots

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.

• If $a/b$ is a rational root, in lowest terms, then $P(x)=(bx-a)Q(x)$ where $Q$ also has integer coefficients. – Angina Seng Aug 4 at 8:16

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.
Suppose $$Q$$ is a polynomial over the integers s.t. $$Q(0)$$ and $$Q(1)$$ are odd. Then for all $$x \in \mathbb{Z}$$, $$Q(x)$$ is odd. Proof: just reduce modulo 2.
Now suppose $$P$$ has a rational root $$\frac{a}{b}$$, fully simplified. By the rational roots theorem, $$b$$ is a factor of $$P$$'s leading coefficient, which is odd. Let $$P$$ be of degree $$n$$. Then the polynomial $$Q(x) = b^n P(x / b)$$ has coefficients over the integers. Then $$Q \equiv P$$ mod 2 (since $$b \equiv 1$$ mod 2). Then $$Q(0)$$ and $$Q(1)$$ are both odd. Then $$Q(a)$$ is odd. But $$Q(a) = b^n P(a/b) = 0$$. Contradiction.
• @rishikesh Thanks for pointing out the error; it was a typo. See the edited version for the correct definition of $Q$. – Doctor Who Aug 5 at 5:23