How do you foil #2 (x+2) (x+5)#?

1 Answer
Aug 10, 2015

Answer:

FOIL can only be applied to the 2 binomials; the result of that multiplication can then be multiplied by #2# to get:
#color(white)("XXXX")##2x^2+14x+20#

Explanation:

Evaluating the product of two binomials using FOIL:

Sum
#color(white)("XXXX")#product of First terms
#color(white)("XXXX")#product of Outside terms
#color(white)("XXXX")#product of Inside terms
#color(white)("XXXX")#product of Last terms

For the given example: #2(x+2)(x+5)#:

We can apply FOIL to #(x+2)(x+5)# as
#color(white)("XXXX")#First terms: #x, x##color(white)("XXXX")#Product #= x^2#
#color(white)("XXXX")#Outside terms: #x, +5##color(white)("XXXX")#Product #= +5x#
#color(white)("XXXX")#Inside terms: #+2, x##color(white)("XXXX")#Product #= +2x#
#color(white)("XXXX")#Last terms: #+2, +5##color(white)("XXXX")#Product #= +10#

Sum
#color(white)("XXXX")##= x^2+5x+2x+10#

#color(white)("XXXX")##= x^2+7x+10#

So
#color(white)("XXXX")##(x+2)(x+5) = x^2+7x+10#

and
#color(white)("XXXX")##2(x+2)(x+7) = 2x^2+14x+20#