How do you factor #x^2 + 7x + 8x + 56# by grouping?

2 Answers
Mar 9, 2018

#x=-8# & #x=-7#

Explanation:

Let's look at the #x^2+7x# and the #8x+56# separately.

#color(red)(x^2+7x)+color(blue)(8x+56)=0#

We can factor an #x# out of the red term, and an #8# out of the blue term. We get:

#color(red)(x(x+7))+color(blue)(8(x+7))=0#

Since #x+7# is multiplying two terms, we can rewrite this as:

#(x+8)(x+7)=0#

Setting the factors equal to zero, we get:

#x=-8# & #x=-7#

Mar 9, 2018

#(x+8)(x+7)#

Explanation:

This expression is already in the perfect form for grouping. Note that #x^2+7x = x(x+7)# and #8x+56 = 8(x+7)# so that both pairs of terms in #color(red)(x^2+7x)+color(blue)(8x+56)# have a common factor :

#color(red)(x^2+7x)+color(blue)(8x+56) = color(red)(x(x+7))+color(blue)(8(x+7)) #

You can now use the law #bcolor (red)a+c color(red)a = (b+c)color(red)a# to write this as

#(x+8)(x+7)#