# How do you solve x^2+15x+56=0?

Mar 27, 2016

$x = - 7$ or $x = - 8$

#### Explanation:

Find a pair of factors of $56$ with sum $15$.

The pair $7 , 8$ works in that $7 \cdot 8 = 56$ and $7 + 8 = 15$

Hence:

$0 = {x}^{2} + 15 x + 56 = \left(x + 7\right) \left(x + 8\right)$

So:

$x = - 7$ or $x = - 8$

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Alternative method

I will use the difference of squares identity:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

with $a = \left(2 x + 15\right)$ and $b = 1$ as follows:

First multiply the whole equation by $4$ (to cut down on the use of fractions) to get:

$0 = 4 {x}^{2} + 60 x + 224$

$= {\left(2 x + 15\right)}^{2} - 225 + 224$

$= {\left(2 x + 15\right)}^{2} - {1}^{2}$

$= \left(\left(2 x + 15\right) - 1\right) \left(\left(2 x + 15\right) + 1\right)$

$= \left(2 x + 14\right) \left(2 x + 16\right)$

$= \left(2 \left(x + 7\right)\right) \left(2 \left(x + 8\right)\right)$

$= 4 \left(x + 7\right) \left(x + 8\right)$

Hence $x = - 7$ or $x = - 8$