How do you FOIL where three factors are used #(x-1)(x+2)(x-3)#?

1 Answer
May 27, 2018

Answer:

#=x^3-2x^2-5x+6#

Explanation:

FOIL is an acronym standing for Firsts, Outers, Inners, Lasts

It refers specifically to the multiplication of two binomials: like

#(a+b)(c+d)# and is a way for students to remember how to multiply them:

#F rarr a xx c = ac" " # multiply the two first terms together.
#O rarr a xx d = ad" " # multiply the two outer terms together.
#I rarr b xx c = bc" " # multiply the two inner terms together.
#L rarr b xxd = bd" " # multiply the two last terms together.

To get #ac +ad +bc +bd#

Any product which is not of this form cannot be done by applying the #FOIL# principle.

I believe it is of more value if students realise that EACH term in the first bracket is multiplied by EACH term of the second bracket - regardless of how many terms there are in each.

Then #FOIL# is not needed.

In the case where three factors are multiplied, they can be multiplied in any order, but TWO have to be multiplied first and then that product is multiplied by the third.

Consider: #2xx11xx5#. This can be done as:
#(2 xx 11) xx 5 = 22xx5 = 110#
#2xx(11xx5)= 2 xx 55 =110#
#(2xx5) xx11= 10 xx 11 = 110" "larr# probably the easiest.

In #(x-1)(x+2)(x-3)# FOIL can only be applied for the first multiplication ....... it does not matter which factors are multiplied first.

#(x-1)color(blue)((x+2)(x-3))" "larr" "FOIL# applies

#=(x-1)color(blue)((x^2-3x+2x-6))#

#=(x-1)color(blue)((x^2-x-6))" "larr FOIL# no longer applies

#(color(red)(x)color(lime)(-1))color(blue)((x^2-x-6)#

#=color(red)(x)color(blue)((x^2-x-6))color(lime)(-1)color(blue) ((x^2-x-6)#

#=color(red)(x^3-x^2-6x" "color(lime)(-x^2+x+6)#

#=x^3-2x^2-5x+6#