Factoring Quadratics

To factor a trinomial with a leading coefficient of $1,$ we need to find two numbers whose product is the independent term. $$\begin{array}{c}{x}^2-7x-8\end{array}$$ In this case, we have that the constant term is $\text{-}8.$ This is a negative number, so for a product to be negative, the factors must have opposite signs (one positive and one negative).

Factor Constants	Product of Constants
$1$ and $\text{-}8$	$\text{-}8$
$\text{-}1$ and $8$	$\text{-}8$
$2$ and $\text{-}4$	$\text{-}8$
$\text{-}2$ and $4$	$\text{-}8$

Next, let's consider the coefficient of the linear term. $$\begin{array}{c}{x}^2-7x-8\end{array}$$ In this case, since the linear coefficient is $\text{-}7,$ we need the sum of the factors to be $\text{-}7.$

Factors	Sum of Factors
$1$ and $\text{-}8$	$\text{-}7$
$\text{-}1$ and $8$	$7$
$2$ and $\text{-}4$	$\text{-}2$
$\text{-}2$ and $4$	$2$

We found two numbers whose product is $\text{-}8$ and whose sum is $\text{-}7.$ These numbers are $1$ and $\text{-}8.$ With this we can factor the given trinomial.