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

Question

taibidzhl · Accepted Answer

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 \to a \times c = a c

multiply the two first terms together.

0 \to a \times d = a d

multiply the two outer terms together.

I \to b \times c = b c

multiply the two inner terms together.

L \to b \times d = b d

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:

2 \times 11 \times 5

. This can be done as:

(2 \times 11) \times 5 = 22 \times 5 = 110

2 \times (11 \times 5) = 2 \times 55 = 110

(2 \times 5) \times 11 = 10 \times 11 = 110 \leftarrow

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) (x + 2) (x - 3) \leftarrow

FOIL applies

= (x - 1) (x^{2} - 3 x + 2 x - 6)

= (x - 1) (x^{2} - x - 6) \leftarrow

FOIL no longer applies

(x - 1) (x^{2} - x - 6)

= x (x^{2} - x - 6) - 1 (x^{2} - x - 6)

= x^{3} - x^{2} - 6 x - x^{2} + x + 6

= x^{3} - 2 x^{2} - 5 x + 6

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

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