Question

Simplify. Show work and explain.
13. n^5 over n - 6 * n^2 - 6n over n^8

Answer 1

Factor 3y^2 out of the numerator

Answer 2

Awesome!!

Answer 3

Remember to multiply by the reciprocal and factor n^2-6n by factoring out n

Answer 4

cancel out n - 6

Answer 5

yes

Answer 6

so now we have n^5 * n over n^8

Answer 7

no

Answer 8

what do you mean

Answer 9

\[\frac{n^5}{n-6}\times\frac{n(n-6)}{n^8}=\frac{1}{n^2}\]

Answer 10

@abb0t

Answer 11

?

Answer 12

I need someone to finish Mertsj's work

Answer 13

\[\large \frac{n^5}{n-6}\times \frac{n^2-6n}{n^8}\]

Hmm do you understand what he did so far? That was a bunch of steps all done at once, I can understand if it was a little confusing.

Answer 14

I understand that, but I need the explanation of how that = 1 over n^2

Answer 15

\[\large \frac{n^5}{n-6}\times \frac{\color{orangered}{n^2-6n}}{n^8}\]
For this orange part, factor an n out of each term.\[\large \frac{n^5}{n-6}\times \frac{\color{orangered}{n(n-6)}}{n^8}\]

Answer 16

From here we'll simply multiply across,
Put brackets around the n-6 on the bottom, so the multiplication is a little clearer.\[\large \frac{n^5}{(n-6)}\times \frac{n(n-6)}{n^8}\]
Then multiplying across gives us,\[\large \frac{n^5\times n(n-6)}{n^8(n-6)} \qquad = \qquad \frac{n^6(n-6)}{n^8(n-6)}\]Understand so far? :o

Answer 17

yes

Answer 18

The n-6's can divide out,\[\large \frac{n^6\cancel{(n-6)}}{n^8\cancel{(n-6)}} \qquad = \qquad \frac{n^6}{n^8}\]

Answer 19

From here, we want to remember our rules for exponents.
When we divide terms of the same base, we `subtract` the exponents.\[\large \frac{n^6}{n^8} \qquad = \qquad n^{6-8}\]

Answer 20

then's it's 1 over n^2

Answer 21

\[\large n^{-2}\qquad = \qquad \frac{1}{n^2}\]Yes :)

Answer 22

you can do that OR the biggest exponent is on the bottom <--- 8 - 6 = 2

Answer 23

substract