Question

factor (x^2-1)x^4n-(x^2y^8n-y^8n) completely.

Answer 1

\[-nx^2y^8+ny^8+nx^6-nx^4\]

Answer 2

Is the n of 4n part of the exponent or not?

Answer 3

yes

Answer 4

Is this the problem where 8n is an exponent?

\( (x^2-1)x^{4n} - (x^2y^{8n}-y^{8n}) \)

Answer 5

yes

Answer 6

\( (x^2-1)x^{4n} - (x^2y^{8n}-y^{8n})\)

Factor out y^(8n) from the second part.

\( =(x^2 - 1)x^{4n} - y^{8n} (x^2 - 1) \)

Now you have a commonn factor of x^2 - 1

Answer 7

\(=(x^2 - 1)(x^{4n} - y^{8n}) \)

Now use difference of squares on each factor.

Answer 8

so is it (x+1)(x-1)1^4n(x-y^2n) ?

Answer 9

\( (x + 1)(x - 1)(x^{2n} + y ^{4n}) (x^{2n} - y^{4n} ) \)

The last term above can be factored again using difference of squares.

Answer 10

okay thank you

Answer 11

\( = (x + 1)(x - 1)(x^{2n} + y ^{4n}) (x^n + y^{2n} ) (x^n - y^{2n} )\)

Answer 12

wlcm