How do you foil #(9x + 7)(6x + 4)#?

1 Answer
Aug 10, 2015

Answer:

#54x^2+78x+28#

Explanation:

FOIL stands for: First (times First), Outside (times Outside), Inside (times Inside), Last (times Last).

In the problem to expand #(9x+7)(6x+4)#, First-times-First is #9x * 6x = 54x^2#, Outside-times-Outside is #9x * 4 = 36x#, Inside-times-Inside is #7 * 6x=42x#, and Last-times-Last is #7 * 4=28#.

Combining these gives #(9x+7)(6x+4)=54x^2+36x+42x+28#. The #36x# and #42x# are "like terms" and combine to give #36x+42x=78x#. Therefore, the answer is:

#(9x+7)(6x+4)=54x^2+78x+28#

The reason this works is the distributive property: #a*(b+c)=a*b+a*c#, and the commutative property: #a*b=b*a#

#(a+b)*(c+d)=(a+b)*c+(a+b)*d#

#=a*c+b*c+a*d+b*d#

#=a*c+a*d+b*c+b*d# (which is what you'd get by using "FOIL" on the original product)