If a polynomial is divisible by \(x-a\), then \(a\) is a root of this polynomial.
Here (x-a) is equal to x+2:
\(x-a=x+2 \\
-a=2 \\
a=-2\)
So -2 is a root of the polynomial. That means x³-6x²+kx+10=0 for x=-2. Now substitute -2 for x in the equation and solve for k:
\((-2)^3-6 \times (-2)^2+ k \times (-2)+10=0 \\
-8-6 \times 4-2k+10=0 \\
-8-24-2k+10=0 \\
-2k-22=0 \\
-2k=22 \\
\boxed{k=-11}\)
