jmc

Using the Remainder Theorem, we know that the remainder when $f(x)=k{x}^4+9x^3+k{x}^2+32x-11$ is divided by $x+5$ is $$f(-5)=k(625)+9(-125)+k(25)+32(-5)-11=650k-1296.$$So we have the remainder $$650k-1296=4$$Solving for $k$ gives us $k=\boxed{2}$.