If $p(x)$ is a convex nonnegative polynomial in $x$, is $x^2 p(x)$ convex?
Its second derivative is $$ x^2 p''(x) + 4x p'(x) + 2 p(x). $$ We know $x^2 p''(x)$ is nonnegative by convexity of $p$, and $2p(x)$ is nonnegative by assumption. I don't know how to conclude anything about $x p'(x)$ though.
If the general case is not convex, what about the special case $p(x) = (a + bx)^{2k},\ k \in \mathbb{N},\ a, b \in \mathbb{R}$?
No: for instance, $f(x)=x^2(x-1)^2$ is not convex. A quick way to see this is that $f(0)=f(1)=0$ but $f$ is strictly positive in between.
More generally, if $p(r)=0$ for some $r\neq 0$ (and $p$ is not identically $0$), then $f(x)=x^2p(x)$ will not be convex since $f(0)=f(r)=0$ but $f$ is positive somewhere between $0$ and $r$.