Question

Can someone please help me solve my last problem?! Kinda desperate

Answer 1

Yeah Becca. Post your problem

Answer 2

(9x^4+3x^3y-5x^2y^2+xy^3) divided by (3x^3+2x^2y-xy^2)

Answer 3

We have
\[\frac{(9x^4+3x^3y-5x^2y^2+xy^3)}{(3x^3+2x^2y-xy^2) }\]
Divide numerator and denominator by x, we get
\[\frac{(9x^3+3x^2y-5xy^2+y^3)}{(3x^2+2xy-y^2) }\]
Now Let's divide this

__3x______________________________
3x^2+2xy-y^2 | 9x^3+3x^2y-5xy^2+y^3
-( 9x^3+6x^2y-3xy^2)
0-3x^2y-2xy^2+y^3



__3x-y______________________________
3x^2+2xy-y^2 | 9x^3+3x^2y-5xy^2+y^3
-( 9x^3+6x^2y-3xy^2)
0-3x^2y-2xy^2+y^3
-(-3x^y-2xy^2+y^3)
0
so the quotient is 3x-y
\[\frac{(9x^4+3x^3y-5x^2y^2+xy^3)}{(3x^3+2x^2y-xy^2)}=3x-y\]

Answer 4

woww..thank you "master" haha
I really,really appreciate it :D

Answer 5

Welcome, did u understand?

Answer 6

most of it. :) You did a really great job writing everything out, I'm just not really a math person.

Answer 7

@becca94 practice it and you'll be soon a math person . If you ever need help just let me know

Answer 8

thanks so much! I will! :)