Answer
$z^2-2z-48=(z-8)(z+6)$
Work Step by Step
We are trying to fill in the blank in the equation $z^2-2z-48=(z-8)(z+\square)$. In order to do so, we will factor the trinomial on the left side of the equation to determine the second factor.
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 $z^2-2z-48$, we are looking for two numbers whose product is $-48$ and whose sum is $-2$. The numbers $-8$ and $6$ meet these criteria, because $$6\times(-8)=-48\;\text{and}\;6+(-8)=-2$$When we insert these numbers into the blanks, we arrive at the factors: $(z-8)(z+6)$.
