Let $I$ be the set of polynomials $f(x)$ for with $f(2) = f'(2) = f''(2) = 0$. Prove this set forms an ideal (a) Let $I \subset \mathbb {R}[x]$ be the set of polynomials $f(x)$ for with $f(2) = f'(2) = f''(2) = 0$. Prove that this set forms an ideal, and find a polynomial $g(x)$ for which $I = g(x)R[x]$
(b) Let $J \subset \mathbb {R}[x]$ be the set of polynomials $f(x)$ for which $f(2) = f'(3) = 0$. Show that $J$ is not an ideal. 
I have no idea about how to approach this problem, the only thing I sure is $f(x) = (x-2)g(x)$, but I don't know how to solve it. Please help me to do this! Thank you!
 A: Use Taylor's expansion at $x=2$, which is a exact formula for polynomials. It follows from the hypotheses that this expansion begins with $\dfrac{f'''(2)}6(x-2)^3$ In other words, the ideal is generated by $(x-2)^3$.
A: Hint for the first part of (a): try directly checking the conditions for being an ideal.  For instance, suppose that you know the polynomial $p(x)$ is in $I$.  Then how would you show that $p(x)q(x)$ is in $I$?  Let $r(x)=p(x)q(x)$; what do you know about $r(2)$?  What do you know about $r'(x)$ in general, and how does that let you calculate $r'(2)$?
Hint for the second part of (a): as suggested by Bernard's answer, the Taylor series provides a big hint here: if $p(x)\in I$, then you know that $p(x)$ has a triple root at $2$, and thus that it's divisible by $(x-2)^3$.  Contrariwise, if $p(x)$ is divisible by $(x-2)^3$ - i.e., if $p(x)=(x-2)^3q(x)$ - then can you show that $p(x)$ is in $I$?
Hint for (b): consider your answer for the first part of (a).  What piece of that breaks down when you're calculating $p(x)$ and $p'(x)$ at different points?  Can you try and use that to find polynomials $p, q$ so that $p(x)\in J$ but $p(x)q(x)$ isn't?
