Derivative and Ring Question

let f(x) be an element of R[x]. If f(a)=0 and f'(a)=0 [f'(a) is the derivative of f(x) at a], show that (x-a)^2 divides f(x).

I set f(x)=(x-a)g(x) for some g(x) in R[x]. Took the derivative and got an answer but i'm just curious if my proof matches up with someone elses. Any help would be great!

By hypothesis, since f(a) = 0, we have (x - a) | f(x).
So, we can write f(x) = (x - a) g(x) for some g(x) in R[x].

Using the product rule, we have
f '(x) = g(x) + (x - a) g'(x).

Since f '(a) = 0, we have f '(a) = 0 = g(a) + (a - a) g'(a) ==> g(a) = 0.
Hence, (x - a) | g(x) and thus g(x) = (x - a) h(x) for some h(x) in R[x].

Therefore, f(x) = (x - a) * (x - a) h(x) = (x - a)^2 h(x) for some h(x) in R[x].
==> (x - a)^2 | f(x).

I hope this helps!