## anonymous one year ago Part A: Divide (8x4y3 + 4x3y2 - 2x2y - 12x2y4) by -2x2y. Show your work, and justify each step. Part B: How would your answer in Part A be affected if the x2 variable in the denominator was just an x? Part C: What is the degree and classification of the polynomial you got in Part A?

1. anonymous

2. anonymous

someone

3. phi

you divide each "term" by the denominator can you do the first term $\frac{8x^4y^3 }{-2x^2y}$

4. phi

first do the numbers 8/(-2) then x*x*x*x/x*x then y*y*y/y

5. anonymous

what?

6. phi

Divide (8x4y3 + 4x3y2 - 2x2y - 12x2y4) by -2x2y. that means you divide each term by -2x2y the first term is 8x4y3 divided by -2x2y

7. phi

as you know, $\frac{8x^4y^3 }{-2x^2y}= \frac{8}{-2} \cdot \frac{x^4}{x^2} \cdot \frac{y^3}{y}$

8. anonymous

ok so first is 8/-2 which equals -4

9. phi

or, if we expand the exponents for x $\frac{8}{-2} \cdot \frac{x^4}{x^2} \cdot \frac{y^3}{y} \\ \frac{8}{-2} \cdot \frac{x\cdot x \cdot x \cdot x}{x\cdot x} \cdot \frac{y^3}{y}$

10. phi

if you have the "same thing" in the "top" and "bottom" they divide out (give you 1) so every x/x can be cancelled.

11. anonymous

4*256*27

12. anonymous

is this right

13. phi

letters stay letters

14. anonymous

I have 10 mins and about 20 secs left can you try and help a little faster sorry for rushing I just don't understand

15. phi

we use the idea that anything divided by itself is 1 $\frac{x \cdot x}{x \cdot x} = 1$

16. phi

so $\frac{x^4}{x^2} = \frac{x\cdot x \cdot x\cdot x}{x\cdot x}= x^2$

17. phi

now do the y^3/y

18. anonymous

y3

19. phi

$\frac{y \cdot y \cdot y}{y}$ you have one y/y pair that you can cancel

20. phi

you should get $\frac{8}{-2} \cdot \frac{x^4}{x^2} \cdot \frac{y^3}{y} \\ -4 x^2y^2$

21. phi

now do the next term 4x3y2 divided by -2x2y

22. anonymous

-2 x^1.5 y^2

23. anonymous

@phi

24. anonymous