p(x) is a polynomial with real coefficients and derivative r(x) = p'(x). For some positive integers a, b, r^{a}(x) divides p^{b}(x). Prove that for some real numbers A and α and for some integer n, we have p(x) = A(x - α)^{n}.

**Solution**

Straightforward.

Write p(x) = A ∏ (x - α_{i})^{ni}.

Then r(x) = p(x) ∑ n_{i}/(x - α_{i}) = (A ∏ (x - α_{i})^{ni-1} ) q(x), where no α_{i} is a root of q(x). This is easily seen, because q(x) is a sum of terms, all but one of which has a factor (x - α_{i}). But q(x) divides p(x)^{b} which has no roots except the α_{i}. Hence q(x) must be a constant. But now the degree of r(x) is wrong unless there is just one α_{i}.

© John Scholes

jscholes@kalva.demon.co.uk

15 Sep 1999