David W. answered 04/28/17
Tutor
4.7
(90)
Experienced Prof
Almost always, the Distributive Property is written as: a(b+c) = ab + ac.
So, in this problem:
(x+1)(x-3) = (x+1)(x) + (x+1)(-3)
Now, you may apply the Commutative Property (ab = ba) then apply the Distributive Property again twice more:
(x+1)(x) = (x)(x+1)
(x)(x+1) = x^2 + x and
(x+1)(-3) = (-3)(x+1)
< (-3)(x+1) = -3x+ -3
That gives:
(x+1)(x-3) = x^2 + x - 3x - 3 [This is F-I-OL, not F-O-I-L]
(x+1)(x-3) = x^2 - 3x + x - 3 [Commute; now is F-O-I-L]
< (x+1)(x-3) = x^2 -2x - 3 [combine like terms]
So, in this problem:
(x+1)(x-3) = (x+1)(x) + (x+1)(-3)
Now, you may apply the Commutative Property (ab = ba) then apply the Distributive Property again twice more:
(x+1)(x) = (x)(x+1)
(x)(x+1) = x^2 + x and
(x+1)(-3) = (-3)(x+1)
< (-3)(x+1) = -3x+ -3
That gives:
(x+1)(x-3) = x^2 + x - 3x - 3 [This is F-I-OL, not F-O-I-L]
(x+1)(x-3) = x^2 - 3x + x - 3 [Commute; now is F-O-I-L]
< (x+1)(x-3) = x^2 -2x - 3 [combine like terms]
Andrew M.
04/28/17