## Firejay5 2 years ago Simplify. Show work and explain. 13. n^5 over n - 6 * n^2 - 6n over n^8

1. Mertsj

Factor 3y^2 out of the numerator

2. Mertsj

Awesome!!

3. Mertsj

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

4. Firejay5

cancel out n - 6

5. Mertsj

yes

6. Firejay5

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

7. Mertsj

no

8. Firejay5

what do you mean

9. Mertsj

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

10. Firejay5

@abb0t

11. abb0t

?

12. Firejay5

I need someone to finish Mertsj's work

13. zepdrix

$\large \frac{n^5}{n-6}\times \frac{n^2-6n}{n^8}$ 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.

14. Firejay5

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

15. zepdrix

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

16. zepdrix

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

17. Firejay5

yes

18. zepdrix

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

19. zepdrix

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

20. Firejay5

then's it's 1 over n^2

21. zepdrix

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

22. Firejay5

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

23. Firejay5

substract