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

##### 1 Answer

Mar 27, 2016

#x = -7# or#x = -8#

#### Explanation:

Find a pair of factors of

The pair

Hence:

#0 = x^2+15x+56 = (x+7)(x+8)#

So:

#x = -7# or#x = -8#

**Alternative method**

I will use the difference of squares identity:

#a^2-b^2 = (a-b)(a+b)#

with

First multiply the whole equation by

#0 = 4x^2+60x+224#

#=(2x+15)^2-225+224#

#=(2x+15)^2-1^2#

#=((2x+15)-1)((2x+15)+1)#

#=(2x+14)(2x+16)#

#=(2(x+7))(2(x+8))#

#=4(x+7)(x+8)#

Hence