Question

simplify by removing factors of 1
n^2-16/(n+4)^4

Answer 1

Hint: n^2-16 factors to (n-4)(n+4) using the difference of squares rule

Answer 2

so what does that mean?

Answer 3

huh i dont understand

Answer 4

If I have n+4 up top and a n+4 down on the bottom, what will happen?

Answer 5

that it would be n-4 right

Answer 6

it would then be n+1

Answer 7

yes, the n+4 up top cancels with one of the n+4 terms down on the bottom

Answer 8

or n+8

Answer 9

not sure where you're getting n+1

Answer 10

so my answer then would be n+4

Answer 11

not quite, let me draw it out for you

Answer 12

one sec

Answer 13

alright

Answer 14

\[\Large \frac{n^2-16}{(n+4)^4}\]


Note: in step 2, i'm factoring using the difference of squares law
\[\Large \frac{(n-4)(n+4)}{(n+4)^4}\]


\[\Large \frac{(n-4)(n+4)}{(n+4)(n+4)(n+4)(n+4)}\]



\[\Large \frac{(n-4)\cancel{(n+4)}}{\cancel{(n+4)}(n+4)(n+4)(n+4)}\]



\[\Large \frac{n-4}{(n+4)(n+4)(n+4)}\]



\[\Large \frac{n-4}{(n+4)^3}\]



So \[\Large \frac{n^2-16}{(n+4)^4}\] completely simplifies to \[\Large \frac{n-4}{(n+4)^3}\]


In other words, \[\Large \frac{n^2-16}{(n+4)^4} = \frac{n-4}{(n+4)^3}\]

Answer 15

let me know if you have questions

Answer 16

so then i would also subtract as well
and so my answer then would be n-4/(n+4)^3 right

Answer 17

yes n-4 all over (n+4)^3