Using the FOIL method, what is #(4x+3)(x+2)#?

2 Answers
Jun 24, 2016

#(4x+3)(x+2) =4x^2+11x+6#

Explanation:

FOIL is short for First, Outside, Inside, Last, indicating the various combinations of terms from each of the binomial factors to multiply then add:

#(4x+3)(x+2) = overbrace((4x*x))^"First" + overbrace((4x*2))^"Outside" + overbrace((3*x))^"Inside" + overbrace((3*2))^"Last"#

#=4x^2+8x+3x+6#

#=4x^2+11x+6#

If we did not use FOIL, then we might do the calculation by breaking up each of the factors in turn using distributivity:

#(4x+3)(x+2) = 4x(x+2)+3(x+2)#

#= (4x*x)+(4x*2)+(3*x)+(3*2)#

#= 4x^2+8x+3x+6#

#= 4x^2+11x+6#

So for binomials, FOIL helps you avoid one step.

The main downside of FOIL is that it is limited to binomials.

Jun 24, 2016

#(4x+3)(x+2)=4x^2+11x+6#

Explanation:

Letters FOIL in FOIL method stand for First, Outer, Inner, Last and is used to multiply two binomials.

Here we are multiplying #(4x+3)# and #(x+2)#.

This means first multiply the terms which occur first in each binomial i.e. #4x# and #x# in above example. Outer means multiply the outermost terms in the product i.e. #4x# and #2#.

Inner means multiply the innermost two terms i.e. #3# and #x# and lastly multiply the terms which occur last in each binomial i.e. #3# and #2#.

Hence #(4x+3)(x+2)=4x xx x + 4x xx 2 + 3 xx x + 3 xx2#

= #4x^2+8x+3x+6#

= #4x^2+11x+6#