so we know that the factors of (x-6) and (x+2) will yield a remainder of -14, thus by the remainder theorem we can say that the values of x = 6 and x = -2 will yield -14, that is for our function f(6) = f(-2) = -14, so let's plug those two values and see what we get for our "k" and "f"
![\boxed{x=6}\hspace{5em}f(6)=-x^3+fx^2+kx-62\\\\\\ -14=-(6)^3+f(6)^2+k(6)-62\implies -14=36f+6k-278 \\\\\\ 264=36f+6k\implies 264=6(6f+k)\implies \cfrac{264}{6}=6f+k \\\\\\ 44=6f+k\implies 44-6f=k \\\\[-0.35em] ~\dotfill\\\\ \boxed{x=-2}\hspace{5em} f(-2)=-x^3+fx^2+kx-62\\\\\\ -14=-(-2)^3+f(2)^2-k(2)-62\implies -14=8+4f-2k-62 \\\\\\ -14=4f-2k-54\implies 40=4f-2k\implies 40=2(2f-k)](https://img.qammunity.org/2024/formulas/mathematics/high-school/hvcpd82ubrojzeizeh7r3pxapdds2fhqmb.png)
