Originally Posted by

**oblixps** I assume for contradiction that q(x) is reducible or that q(x) = k(x)g(x). Then i know that q(1/x) = k(1/x)g(1/x) and thus $\displaystyle x^n q(\frac{1}{x}) = x^n k(\frac{1}{x})g(\frac{1}{x}) $ and i know this is a polynomial since the original polynomial was of degree n and the lowest "degree" after substituting 1/x will be "-n" but multiplying by x^n will make sure all the exponents are positive. I also know that $\displaystyle x^n q(1/x) = p(x) $ so $\displaystyle x^n k(\frac{1}{x})g(\frac{1}{x}) = p(x) $ which means that p(x) is the product of 2 polynomials of smaller degree and thus p(x) is reducible giving a contradiction.

Is this proof correct, or is it maybe missing details, has wrong arguments, etc.?