How do you factor the expression $x^{2} + 7 x - 8$ $x^2 +7x-8$ ?

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How do you factor the expression $x^{2} + 7 x - 8$ $x^2 +7x-8$ ?

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Answer 1 · 2016-05-12T09:52:50

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

Explanation:

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

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

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

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

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

Since the coefficient of $x^{2}$ $x^2$ is $1$ $1$ in the product, we require a coefficient $1$ $1$ for $x$ $x$ .
Since the constant term is $- 8$ $-8$ in the product, we require a constant $+ 8$ $+8$ in the factor which will become $- 8$ $-8$ when multiplied by $- 1$ $-1$ .

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How do you factor the expression $x^{2} + 7 x - 8$ $x^2 +7x-8$ ?

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