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(2n + 2)(2n – 2)
_____________
A.) 4n2 – 4

B) 4n2 – 4n – 4

C.) 4n2 + 4n – 4

D.) 4n2 + 2n – 4


Sagot :

[tex](2n+2)(2n-2)\\\\=(2n)^2 -2^2 ~~~~~~;[a^2 -b^2 =(a-b)(a+b)]\\\\=4n^2 -4[/tex]

Answer:

A) 4n²- 4

Step-by-step explanation:

use the FOIL method:

F = first, meaning multiply the 1st terms in each factor

O = outer, meaning multiply the 1st term in first factor and last term in second

I = inner, meaning multiply the 2nd term in first factor and 1st term in second

L = last, meaning multiply the last terms in each factor

F = 2n · 2n = 4n²

O = 2n(-2) = -4n

I = 2(2n) = 4n

L = 2(-2) = -4

you have 4n² - 4n + 4n - 4

this can be simplified by adding the two middle terms (which will zero out):

4n²- 4