How do you multiply #(x+3)(x-3)+(x+7)(x-3)#?

1 Answer
Feb 13, 2016

Answer:

#color(blue)(2x^2+4x-30)#

Explanation:

We can use the FOIL Method,
(First, Outer, Inner, and Last)

like this,

#(a+b)(c+d) = ac + ad + bc + cd#

Solving the problem,

#(x+3)(x-3) + (x+7)(x-3)#

we can simplify the 1st term, #(x+3)(x-3)#

#(x+3)(x-3) = x^2 -3x+3x-9 #

#x^2 -3x+3x-9#, combine all like terms, to simplify, if there is any, like #-3x# and #+3x#, we can combine this.

Since #a + (-a) = 0# (Additive Inverse Property), we can cancel the like terms giving,

#= x^2 - 9#

now apply FOIL method, to the 2nd term.

# (x+7)(x-3) = x^2-3x+7x-21#

simplify, by combining like terms,

#= x^2+4x-21#

Plug it all,

#x^2 - 9 + x^2+4x-21#

combine like terms to simplify again, we must simplify the answer as long as possible.

#= 2x^2+4x-30#