$P(x)$ is a polynomial such that when $P(x)$ is divided by $x-13$ the remainder is $19$, and when $P(x)$ is divided by $x+17$ the remainder is $-11$. What is the remainder when $P(x)$ is divided by $(x-13)(x+17)$?
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,
$$\begin{aligned} P(13) &= 13a+b = 19 \\
P(-17) &= -17a+b = -11 \end{aligned}$$.
Solving this system of equations gives us $a=1$ and $b=6$. So the remainder is $\boxed{x+6}$.