# Ring of Polynomials

• Nov 28th 2009, 07:41 AM
johnt4335
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].
• Nov 28th 2009, 07:46 AM
qmech
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)
• Nov 28th 2009, 08:57 AM
johnt4335
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 AM
clic-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)\$