jmc

Since we are dividing by a quadratic, the remainder will have degree less than 2. So the remainder $r(x)=ax+b$ for some constants $a$ and $b$. We have, $$P(x)=(x-13)(x+17)Q(x)+ax+b$$where $Q(x)$ is the quotient of the division.

Then, using the Remainder Theorem, $$\begin{array}{rl}P(13)& =13a+b=19\\ P(-17)& =-17a+b=-11\end{array}$$. Solving this system of equations gives us $a=1$ and $b=6$. So the remainder is $\boxed{x+6}$.