Question

Factor completely.



n4 - 1



(n2+ 1)(n - 1)
(n + 1)2(n - 1)(n + 1)
(n2+ 1)(n - 1)(n + 1) ******
I think it is the one with ******

Answer 1

well if you think about it as

\[(n^2)^2 - 1\]

you are looking at the difference of 2 squares

\[(n^2 -1)(n^2 + 1)\]

all you need to do now is factorise the 1st binomial which is also a difference of 2 squares.

Answer 2

yes, @Shannon20150 , the one with the stars, \((n^2+1)(n-1)(n+1)\), is correct :)

Answer 3

will u help me again