If $(x-a)^2|f(x)$, then $(x-a)|f'(x)$ and if $(x-a)|f(x)$ and $(x-a)|f'(x)$, then $(x-a)^2|f(x)$

Question

Let $R$ a commutative ring and let $f(x) \in R[x]$.

(i) Prove that if $(x-a)^2|f(x)$, then $(x-a)|f'(x)$.

(ii)Prove that if $(x-a)|f(x)$ and $(x-a)|f'(x)$, then $(x-a)^2|f(x)$.

Any help in order to prove the last statements? I have been trying to prove these straight forward but I cannot see how to attack this exercises, thanks.

Answer 1

Write $f(x)=(x-a)^kg(x)$ and apply the product rule.

Answer 2

Hint for $(ii)$

Since $(x - \alpha) | f(x)$ one has $f(x)=g(x)(x-\alpha)$ and $f'(x) = g'(x)(x-\alpha) + g(x)$.

On the other side $f'(x) = (x- \alpha) \cdot h(x)$ simillary.

Hence: $g'(x)(x-\alpha) + g(x) = (x-\alpha) \cdot h(x)$ $\iff (x - \alpha)\left(g'(x) + \frac{g(x)}{(x-\alpha)}\right) = (x - \alpha)h(x)$

Can you continue from here?