Question

Simplify:n squared + 8n + 16 over n squared minus 6n + 9 divided by n + 4 over n minus 3

Answer 1

\[\frac{n^2+8n+16}{n^2-6n+9}\times \frac{n-3}{n+4}\] is a start, then factor as
\[\frac{(n+4)(n+4)(n-3)}{(n-3)(n-3)(n+4)}\] and so some canceling

Answer 2

(n+4)(n+4)
-----------
(n-3)(n-3)

Answer 3

so this divided by (n+4)/(n-3)

Answer 4

so will get (n+4)/(n-3)

Answer 5

Thanks! :)

Answer 6

yw

good luck

bye