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Mathematics 22 Online
OpenStudy (redbirdd):

Divide the following polynomial, then place the answer in the proper location on the grid. Write your answer in order of descending powers of x. (x^3 + y^3)/(x - y)

OpenStudy (will.h):

i don't want to just solve it for you there are 2 ways to solve this. Long division and synthetic division you may review your lesson or you may check khan academy they provide great explanations

OpenStudy (will.h):

i'll help you

OpenStudy (redbirdd):

I don't know how to do any of this because my teacher doesn't do anything but sit on his butt and get paid. I'm getting frustrated

OpenStudy (will.h):

okay i want you to watch this video and then we will solve this together. https://www.khanacademy.org/math/algebra2/arithmetic-with-polynomials/synthetic-division-of-polynomials/v/synthetic-division

OpenStudy (will.h):

i'll be right here for you just let me know when we start to solve

OpenStudy (redbirdd):

Ok so I think I get it a little, but I still need help

OpenStudy (will.h):

No problem but anyway here.. we won't use synthetic division why? because our divisor isn't something like x+1 or x-3 here we have x-y (2 variables) the best way to divide here is long division and i will walk you through this

OpenStudy (redbirdd):

Ok thank you so much

OpenStudy (will.h):

|dw:1474031241532:dw| i hope that helps but since they said descending order i don't think it would matter here why? because both terms has a degree of 2. Bingo that's all you may ask for my help when you need me again :)

OpenStudy (redbirdd):

Thank you. I think when I have a problem that I can do the synthetic divisiojn with, ill try that.

OpenStudy (will.h):

do you remember the 1st problem we solved? go ahead and solve it yourself and i may have forgot to apply descending order so you may check that

OpenStudy (redbirdd):

Ok

OpenStudy (will.h):

no we solved it correctly just the last term edit it and put the divisor as denominator (just for the last term)

OpenStudy (will.h):

My notifications don't work well so if you need me you should tag me like this @will.H bye ...:)

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