Divisibility of Quadratic Equation

Let us write the above quadratic equation with $x = -2$ as one of its roots:

$(x + 2)(x + A) = 0 \Rightarrow 2 + A = 5k, 2A = k^2 + 5$ .

Solving for $A$ in terms of $k$ in the first linear relation and substituting it into the second quadratic relation yields:

$2(5k - 2) = k^2 + 5 \Rightarrow 0 = k^2 - 10 k + 9 \Rightarrow 0 = (k-1)(k-9) \Rightarrow k = 1,9.$

Now, checking each of these roots for $k$ in the original quadratic equation finally produces:

$k = 1 \Rightarrow x^2 + 5x + 6 = (x+2)(x+3)$ and $k = 9 \Rightarrow x^2 + 45x + 86 = (x+2)(x+43)$ .

which is only satisfied by $\boxed{k = 9}.$