The question is as follows:
Let $f$ be a polynomial with non-zero constant term. Let $h$ be a polynomial of degree $d$ ($d$ is a fixed non-negative number). Show that there exists a unique polynomial $g$ of degree at most $d$ such that $f(x)g(x)=h(x)+x^{d+1}k(x)$. Hint: This problem may most easily be solved by regarding it as a problem involving quotient vector spaces.
My idea is to consider the vector space $P_d$ (the set of all polynomials with degree at most $d$) as a subspace of $P$, and consider the quotient space $P/P_d$. Then $f$ must be equivalent to a polynomial of the form $x^{d+1}v(x)$, and $f(x)=u(x)+x^{d+1}v(x)$. Since $u\in P_d$, I can write $h(x)=u(x)q(x)+r(x)$. Then $f(x)q(x)=h(x)+x^{d+1}v(x)+r(x)$. However, I am stuck here and not sure how to get rid of $r(x)$.
Another method would be to set up a system of linear equation about coefficients of g. The system is lower triangular (and invertible, since all diagonal entries are nonzero) so we can definitely find a unique solution. However, this method seems to not using any quotient vector stuff at all.
Any hint? Thanks.