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].
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?
Hi
You assumed $\displaystyle (x-a)|g(x),$ i.e. there is a $\displaystyle p(x)$ such that $\displaystyle g(x)=(x-a)p(x).$
We have $\displaystyle g'(x)=(x-a)p'(x)+p(x)$ hence $\displaystyle p(x)=g'(x)-(x-a)p'(x).$
Therefore: $\displaystyle (x-a)|g'(x)\Rightarrow (x-a)|p(x)\Rightarrow (x-a)^2|g(x)$