# HELP...Polynomial Rings Problem...

• May 28th 2008, 08:52 PM
ginafara
HELP...Polynomial Rings Problem...
I am reviewing some problems from polynomial rings and hope that someone can tell me if I am on the right track..

PROBLEM: Let $f(x) \in R[x]$. If $f(a) = 0, f'(a)=0$show that $(x-a)^{2}$ divides $f(x)$.

What I have done:

$f(x) = g(x)(x-a)^{2}+r(x) g(x), r(x) \in R[x], deg r(x) < deg (x-a)^{2}$ or $deg r(x) = 0$

$f'(x) = 2g(x)(x-a) + g'(x)(x-a)^{2} + r'(x)$

$f(a) = g(a)(0) + r(a) \implies f(a) = r(a) = 0$

so $f(x) = g(x)(x-a)^{2}$

$f'(a) = 2g(a)(0) + g'(a)(0) +r'(a) \implies f'(a) = r'(a) = 0$

Thus $(x-a)^{2}$ is a factor of $f(x) \implies (x-a)^{2}|f(x)$

Is this even close???? Thanks in advance
• May 28th 2008, 11:33 PM
Moo
Hello,

In my opinion (it's not the field I prefer), there is a problem :

Quote:

Originally Posted by ginafara

$f(a) = g(a)(0) + r(a) \implies f(a) = r(a) = 0$

so $f(x) = g(x)(x-a)^{2}$

here, because from r(a)=0, you state that $r(x)=0 ,\ \forall x$

Plus, if you go this way, you can directly say that $r'(x)=0$ since it's the null function ~
• May 29th 2008, 01:39 AM
kalagota
i think you are actually done but you missed some reasons.

note that $deg \, r(x) < deg \, (x-a)^2$ or $deg \, r(x) = 0$, that is $deg \, r(x) = 0 \, or \, 1$

but the case $deg \, r(x) = 1$ cannot happen otherwise $r'(x)$ is a constant and that $f'(a) \neq 0$.

thus $deg \, r(x) = 0$, that is $r(x)$ is a constant function. but you have shown that $r(a) = 0$ and therefore $r(x) = 0, \, \forall \, x$.
• May 29th 2008, 08:15 AM
ThePerfectHacker
If $f(a) = 0$ then $f(x) = (x-a)g(x)$.
Thus, $f'(x) = g(x) + (x-a)g'(x)$ since $f'(a) = 0$,it follows $g(a) = 0$ so $g(x) = (x-a)h(x)$.
And thus $f(x) = (x-a)^2h(x)$.