Answer
$(z-2)(z+4)$
Work Step by Step
To factor a trinomial in the form $z^2+bz+c$, we must find two numbers whose product is $c$ and whose sum is $b$. We then insert these two numbers into the blanks of the factors $(z+\_)(z+\_)$.
In the case of $z^2+2z-8$, we are looking for two numbers whose product is $-8$ and whose sum is $2$. The numbers $-2$ and $4$ meet these criteria because: $$4\times(-2)=-8\;\text{and}\;4+(-2)=2$$When we insert these numbers into the blanks, we arrive at the factors: $(z-2)(z+4)$.
