Question

I'm horrible with algebra...
I got a metal though!(:

Answer 1

Part A: Divide (4x4y3 + 8x3y2 - 12x2y - 16x2y4) by -4x2y. Show your work, and justify each step. (6 points)

Part B: How would your answer in Part A be affected if the x2 variable in the denominator was just an x? (2 points)

Part C: What is the degree and classification of the polynomial you got in Part A? (2 points)

Answer 2

@sleepyjess @confluxepic @TheSmartOne @THEHELPER123

Help?):

Answer 3

@ganeshie8

Answer 4

\[\large \dfrac{4x^4y^3 + 8x^3y^2 - 12x^2y - 16x^2y^4}{-4x^2y}\]

Answer 5

so divide the top by the bottom right?

Answer 6

Yes.. for partA, factor out \(-4x^2y\) from the numerator and cancel it wid denominator :

\[\large \dfrac{-4x^2y(-x^2y^2 - 2xy + 3x + 4y^3)}{-4x^2y}\]

Answer 7

holy cow.. okay. I'll write it out.

Answer 8

You get

\[\require{cancel}\dfrac{\cancel{-4x^2y}(-x^2y^2 - 2xy + 3 + 4y^3)}{\cancel{-4x^2y}} = -x^2y^2 - 2xy + 3+ 4y^3\]

Answer 9

so would the part on the end be the product?

Answer 10

yes thats the end of part a

Answer 11

and then for part a the question would look like this \[ (4x4y3 + 8x3y2 - 12x2y - 16x2y4) \div -4x2y\]

Answer 12

I don't know how to get the fraction/divide bar you used...

Answer 13

you may copy paste the existing latex code by right clicking on latex, then click
"Show Math As" ---> "TeX Commands"

Answer 14

but we're done with part A right ?

Answer 15

yes. and thank you! Would you be able to help with part b also? If you don't want to its okay!(:

Answer 16

for part b :
If the denominator was just an \(x\), then one \(x\) wont be cancelled in the numerator increasing the degree of result by 1 compared to having \(x^2\) in the denominator.

Answer 17

for part c, look at the answer from part A and see if you can figure out the "degree"

Answer 18

soo would it be \[x^3y^2-2xy +3+4y^3\]

Answer 19

is that what you get for part b ?

Answer 20

yes... is it correct?

Answer 21

\[\dfrac{4x^4y^3 + 8x^3y^2 - 12x^2y - 16x^2y^4}{-4xy}\]

\[\dfrac{-4xy(-x^3y^2 -2x^2y + 3x + 4xy^3)}{-4xy}\]

\[\require{cancel}\dfrac{\cancel{-4xy}(-x^3y^2 -2x^2y + 3x + 4xy^3)}{\cancel{-4xy}}\]

\[-x^3y^2 -2x^2y + 3x + 4xy^3\]

Answer 22

ohh.. alrighty then. Thank you thank you! I can do part C(:

Answer 23

sounds good :) yw!