= Solution
Assume the necessary <derivatives> decay enough for all boundary terms to vanish. One <integration by parts> gives
$$
\widetilde{f'}(k)=\int_{\mathbb R}f'(x)e^{-ikx}\,dx=ik\widetilde f(k).
$$
Repeating proves the <Fourier transform of a derivative> rule $\boxed{\widetilde{f^{(m)}}=(ik)^m\widetilde f}$. On the other hand, differentiation under the <integral> gives
$$
\frac{d^m\widetilde f}{dk^m}=\int_{\mathbb R}(-ix)^mf(x)e^{-ikx}\,dx,
$$
and therefore $\boxed{\widetilde{x^mf}=i^m\widetilde f^{(m)}}$.
Write $p_n(x)=\sum_{j=0}^na_jx^j$, with $a_n\ne0$. Applying the second rule and part (a) gives
$$
\mathcal F[p_n(x)g(x)](k)=\sqrt{2\pi}\sum_{j=0}^na_ji^jg^{(j)}(k)=q_n(k)e^{-k^2/2},
$$
where
$$
\boxed{q_n(k)=\sqrt{2\pi}\sum_{j=0}^na_ji^jr_j(k).}
$$
Only the $j=n$ term can contribute degree $n$. Its leading coefficient is $\sqrt{2\pi}a_n(-i)^n\ne0$, so $q_n$ has exactly the same degree $n$, with no possible leading-term cancellation. The coefficients of this transformed <polynomial> can be complex even when those of $p_n$ are real.
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