Question

x^2-16 over x^2-8x+16

Answer 1

\[\frac{x^2-16}{x^2-8x+16}=\frac{(x+4)(x-4)}{(x-6)(x-2)}\]
Finish it from here?

Answer 2

*(x-4)(x-4)
For the bottom sorry

Answer 3

so its -1

Answer 4

Not exactly:
\[\frac{(x+4)(x-4)}{(x-4)(x-4)}=\frac{x+4}{x-4}\]

Answer 5

\[\frac{ x^2-16 }{ x^2-8x+16 } = \frac{ (x + 4)(x-4) }{ (x - 4)(x-4) }\]

Answer 6

so a (x-4) from the numerator and a (x-4) from the denominator "cancel out".

Answer 7

^

Answer 8

oh okay , but if it was -4 over 4 it would be -1 right?

Answer 9

\[\frac{ x^2-16 }{ x^2-8x+16 } = \frac{ (x + 4)(x-4) }{ (x - 4)(x-4) } = \frac{ (x+4) }{ (x-4) }\]

Answer 10

Nope, those can't cancel.. you have to leave them as they are

Answer 11

okay thank you (:

Answer 12

Anytime :)