Question

Factor completely. n ^4 - 1

Answer 1

(n^4 - 1) = (n^2 - 1) (n^2 +1) = (n - 1) (n + 1) (n^2 + 1)

Answer 2

Answer: Hello there!

here we have the equation n^4 - 1

using the relation:
`(a^(2) - b^(2)) = (a+b)*(a-b)`

we can write our equation as:

`(n^(4) - 1) = ((n^(2) )^(2) -1^(2) ) = (n^(2) + 1)(n^(2) - 1) = (n^(2) + 1)(n + 1)(n-1)`

and (n^2 + 1) has only complex roots, i and - i, then we can factorize this as (n -i)(n + i) = n*n + ni - ni (+i)*(-i) = (n^2 + 1)

then our equation is:
`(n^(2) + 1)(n + 1)(n-1) =  (n + i)(n - i)(n + 1)(n-1)`