Question

For what value of $k,-4$ is a zero of the polynomial $x^2-x-(2k+2)$?

Answer 1

In the given polynomial $f\left(x\right)=x^2-x-(2k+2)$, substitute $x=-4$ as shown below:

$f(-4)=(-4{)}^2-(-4)-(2k+2)=16+4-2k-2=18-2k$

Therefore, $f(-4)=18-2k$.

Since it is given that $x=-4$ is a zero of the polynomial $f\left(x\right)=x^2-x-(2k+2)$, therefore $f(-4)=0$ that is:

$18-2k=0\Rightarrow 2k=18\Rightarrow k=\frac{18}{2}\Rightarrow k=9$

Hence $k=9$.