If $a$ divides $b$, then $a$ divides $3b^3-b^2+5b$. 
Prove: Suppose $a$ and $b$ are integers. If $a\mid b$, then $a\mid3b^3-b^2+5b$.

I think I have an idea of how to prove this, but I'm not entirely sure.
I can prove that each individual term in the polynomial $3b^3-b^2+5b$ is a multiple of $a$. 


*

*$3b^3 = a(3b^2c) = ad$

*$-b^2 = a(-ac^2) = ae$

*$5b = a(5c) =af$.


Adding together the equalities (1)-(3) we obtain $3b^3-b^2+5b=ad+ae+af =a(d + e + f) $ where $(d + e + f) \in \mathbb{Z}$, demonstrating that $3b^3-b^2+5b$ is a multiple of $a$ and that $a\mid3b^3-b^2+5b$.
This is not my formal proof, I just want to make sure the reasoning is correct.

if $a|b$ then $b=a.t$ for some integer $t$, so $3b^3-b^2+5b=3(at)^3-(at)^2+5(at)=a(3a^2t^3-at^2+5t)$, and thus clearly $a|3b^3-b^2+5b$
