Question

v4-1 factor completeley

Answer 1

is that:
\[v ^{4}-1\]

Answer 2

yep

Answer 3

So far I have\[(v^{2}+1)(v+1)(v-1)\]

Answer 4

If so it factors to:
(v^2 + 1)(v^2 - 1)
(v^2 + 1)(v + 1)(v - 1)

Answer 5

The v^2 + 1 is the sum of perfect squares and therefor does not factor in the real number line (you could factor it using imaginary numbers though)

Answer 6

(v^2-1)(v^2+1)= (v+1)(v-1)(v^2+1)