Answer
$(w-8)(w+1)$
Work Step by Step
To factor a trinomial in the form $w^2+bw+c$, 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 $(w+\_)(w+\_)$.
In the case of $w^2-7w-8$, we are looking for two numbers whose product is $-8$ and whose sum is $-7$. The numbers $-8$ and $1$ meet these criteria because $$1\times(-8)=-8\;\text{and}\;1+(-8)=-7$$When we insert these numbers into the blanks, we arrive at the factors: $(w-8)(w+1)$.
