Let $f(.)$ be a continuous Lebesgue-density-function with compact support $\mathrm{supp}f=\{x:f(x)\neq0\}$. Let $p(.)$ be a polynomial with degree $k$.
Then the convolution $h:=f\star p$ is an integrable function and can be written as polynomial.
What I tried:
$f\star p(y)=\int_{-\infty}^{+\infty}f(x-y)p(y)dy=\int_{-\infty}^{+\infty}f(y)p(x-y)dy$.
$p=\sum_{m=0}^{k}\alpha_mt^m$. We only consider $p(t)=t^m \forall m=0,\ldots,k$.
So $f\star p(x) =\int_{-\infty}^{+\infty}f(y)(x-y)^m$ which is by the binomial theorem $\sum_{i=1}^{m}x^i{m\choose i}\int_{-\infty}^{\infty}f(y)(-y)^{m-i}$ Where the product of the terms after $x^i$ denote the polynomial coefficient. So $f \star p$ is a polynomial.
Now I want to show that $f \star p$ is integrable.
The support is compact so we can reach the maximum of p$\max_{x \in \mathrm{supp} f}p(x)=:K<\infty$
$\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}f(x-y)p(y)dydx=\int_{-\infty}^{\infty}\int_{supp f}f(x-y)p(y)dydx\le\int_{-\infty}^{+\infty}\int_{\mathrm{supp} f}f(x-y)Kdydx$.
Since $f$ is a density function
$\int_{-\infty}^{+\infty}\int_{\mathrm{supp} f}f(x-y)dydx\le1$ (I'm not sure about that)
So $\int_{-\infty}^{+\infty}(f\star p)(x)dx\le K<\infty$
No, $f*g$ need not be integrable. Let $f=1$ on $[0,1],$ $f=0$ elsewhere. Let $p(y)=y.$ Then
$$f*p(x) = \int_0^1 1\cdot(x-y)\,dy = x-1/2.$$
The function on the right is not integrable on $\mathbb R.$