# How do you factor the expression x^2 +7x-8?

May 12, 2016

${x}^{2} + 7 x - 8 = \left(x - 1\right) \left(x + 8\right)$

#### Explanation:

Given ${x}^{2} + 7 x - 8$

Note that the sum of the coefficients is $0$. That is $1 + 7 - 8 = 0$.

So $x = 1$ is a zero and $\left(x - 1\right)$ a factor:

${x}^{2} + 7 x - 8 = \left(x - 1\right) \left(x + 8\right)$

The linear factor $\left(x + 8\right)$ can be constructed by looking at the coefficients of ${x}^{2}$ and the constant term $- 8$ that we want in the product:

• Since the coefficient of ${x}^{2}$ is $1$ in the product, we require a coefficient $1$ for $x$.

• Since the constant term is $- 8$ in the product, we require a constant $+ 8$ in the factor which will become $- 8$ when multiplied by $- 1$.