How do you find the product of #(8x + 7)(x + 2) #?

1 Answer
Sep 15, 2017

Answer:

#8x^2+23x+14#

Explanation:

#(8x+7)(x+2)#
First, we will distribute parentheses/brackets
from #(8x+7)(x+2)# to
#8x*x+8x*2+7*x+7*2#
Example:
#(w+x)(y+z)=wy+wz+xy+xz#
In this case, #w# would be #8x#, #x# would be #7#, #y# would be #x# and #z# would be #2#.

#8x*x+8x*2+7*x+7*2# is the same as
#8x x+8x*2+7x+7*2#
Now we will do:
#8x^2+8x*2+7x+7*2#
Why did #8x x# change to #8x^2#?
It's because of this:
#aa = a^(1+1)=a^2#

So, now you have #8x^2+8x*2+7x+7*2#
You want to solve bit by bit. First we'll start with #8x*2#. #8*2=16# So we can replace #8x*2# with #16x#.
#8x^2+16x+7x+7*2#
#16x+7x=23x#
#8x^2+23x+7*2#
#7*2=14#

#8x^2+23x+14#
That's your answer! ^