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: $(n^{2} + 1)(n + 1)(n-1) = (n + i)(n - i)(n + 1)(n-1)$