The division algorithm says we can find the quotient and remainder:

x^3+5x^2+6x+11 = q(x)(x+k)+3

Since these are equal, their evaluations are equal as well.

Let us evaluate this at x=-k.

Thus,

(-k)^3+5(-k)^2+6(-k)+11=q(-k)(0)+3

-k^3+5k-6k+11=3

k^3-5k+6k-8=0

Use the rational root theorem and see that k=4 is a solution. (In fact the only one).

Thus, k=4.