Compute $a(x)b(x)+c(x)$ in $\mathrm{GF}(2^4)$ where the irreducible generator polynomial $x^4+x+1$. 
Let the coefficients of
  $$a(x) = x^3+x^2+1,$$
  $$b(x) = x^2+x+1,$$
  $$c(x) = x^3+x^2+x+1$$
  be in $\mathrm{GF}(2)$.
  Compute $a(x)b(x)+c(x)$ in $\mathrm{GF}(2^4)$ where the irreducible generator polynomial $x^4+x+1$.

This is the exact question in a criptography lesson midterm.
I have calculated the $a(x)b(x)+c(x) = x^5 + 2x^4 + 3x^3 + 3x^2 + 2x + 1$. As I know in a $\mathrm{GF}(2^4)$, coefficients can't be greater than 15. So it is ok for a intermediate result. Now what to do with $x^4+x+1$?
Can you help me?

I assume that Pablo's interpretation of your question is correct, and answer accordingly.
I just added a self-answered question giving a useful discrete logarithm table. I rely heavily on that.
Below $\gamma$ will denote the coset $x+I$ in the quotient ring $GF(2)[x]/I$, where the ideal $I=\langle x^4+x+1\rangle$. That table then tells us that
$$
\begin{aligned}
a(x)+I&=x^3+x^2+1+I=\gamma^3+\gamma^2+1=\gamma^{13},\\
b(x)+I&=x^2+x+1+I=\gamma^2+\gamma+1=\gamma^{10},\\
c(x)+I&=x^3+x^2+x+1+I=\gamma^3+\gamma^2+\gamma+1=\gamma^{12}.
\end{aligned}
$$
Thus
$$
a(x)b(x)+I=\gamma^{13}\cdot\gamma^{10}=\gamma^{23}=\gamma^{15+8}=\gamma^8=\gamma^2+1.
$$
Therefore
$$
a(x)b(x)+c(x)+I=(\gamma^3+\gamma^2+\gamma+1)+(\gamma^2+1)=x^3+x+I.
$$
