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].

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- Nov 28th 2009, 07:41 AMjohnt4335Ring 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]. - Nov 28th 2009, 07:46 AMqmechA 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)

- Nov 28th 2009, 08:57 AMjohnt4335
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?

- Nov 28th 2009, 09:11 AMclic-clac
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: $\displaystyle (x-a)|g'(x)\Rightarrow (x-a)|p(x)\Rightarrow (x-a)^2|g(x)$