Question

Factor completely: 49e^4-49

Answer 1

I tried setting is up like this: (7e^2)^2 - 7^2, then got (7e^2+7)(7e^2-7) but the answer returned incorrect.

Answer 2

try this :

\[49e^4 - 49 = 49(e^4-1) = 49(e^2+1)(e^2-1) = 49(e^2+1)(e+1)(e-1)\]

Answer 3

Thanks, that worked. Does x^2-1 always factor out to (x+1)(x-2)? Then is x^2+1 always prime?

Answer 4

sorry, (x+1)(x-1)

Answer 5

assume we're living in integers

then you're right, x^2+1 is a prime polynomial which cannot be factored

Answer 6

Notice below : \[x^2 - 1 = x^2-1^2\]

use the identity :
\[a^2-b^2 = (a+b)(a-b)\]

Answer 7

Thank you!

Answer 8

yw:)