Question 94670
Factor:
{{{n^4-1}}} Notice that {{{n^4 = (n^2)^2}}}, so now you can write:
{{{(n^2)^2 - 1^2}}} The difference of two squares can be factored:{{{A^2-B^2 = (A+B)(A-B)}}} Applying this to your problem:
{{{(n^2)^2 -1^2 = (n^2+1)(n^2-1)}}} but again, we have the difference of two squares in the second factor, {{{n^2-1 = (n+1)(n-1)}}}so...
{{{n^4-1 = (n^2+1)(n+1)(n-1)}}} we would normally stop here, because the factors of {{{n^2+1}}}involve the imaginary number {{{sqrt(-1) = i}}}
If you are interested though:
{{{n^2+1 = (n+i)(n-i)}}} but don't use this if you have not yet come to imaginary numbers.