# Math Help - Ring of Polynomials

1. ## Ring of Polynomials

Assume that (x-a) | g(x) in the ring of polynomials R[x]. Prove that (x-a)^2 | g(x) if and only if (x-a) |
g' in R[x].

2. ## A start

If (x-a)^2 | g(x) then g(x) = (x-a)^2 p(x) for some polynomial p(x). Take the derivative of g(x) and see if (x-a) | g'(x)

3. Okay, I can see that, but I am still stuck on the "if and only if" part. If (x-a) | g', then g' = (x-a)q(x) for some polynomial q, but then how exactly would I integrate that? Would it just be g = ((x-a)^2)q(x)/2 + b, and if so, how would I deal with the "+b"? or do I have to treat the q(x) as more than a constant like I suspect, and if so, how?

4. Hi

You assumed $(x-a)|g(x),$ i.e. there is a $p(x)$ such that $g(x)=(x-a)p(x).$

We have $g'(x)=(x-a)p'(x)+p(x)$ hence $p(x)=g'(x)-(x-a)p'(x).$

Therefore: $(x-a)|g'(x)\Rightarrow (x-a)|p(x)\Rightarrow (x-a)^2|g(x)$