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?

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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?

Mathematics
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@phi please help
someone
  • phi
you divide each "term" by the denominator can you do the first term \[ \frac{8x^4y^3 }{-2x^2y}\]

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Other answers:

  • phi
first do the numbers 8/(-2) then x*x*x*x/x*x then y*y*y/y
what?
  • phi
Divide (8x4y3 + 4x3y2 - 2x2y - 12x2y4) by -2x2y. that means you divide each term by -2x2y the first term is 8x4y3 divided by -2x2y
  • phi
as you know, \[ \frac{8x^4y^3 }{-2x^2y}= \frac{8}{-2} \cdot \frac{x^4}{x^2} \cdot \frac{y^3}{y} \]
ok so first is 8/-2 which equals -4
  • phi
or, if we expand the exponents for x \[ \frac{8}{-2} \cdot \frac{x^4}{x^2} \cdot \frac{y^3}{y} \\ \frac{8}{-2} \cdot \frac{x\cdot x \cdot x \cdot x}{x\cdot x} \cdot \frac{y^3}{y} \]
  • 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.
4*256*27
is this right
  • phi
letters stay letters
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
  • phi
we use the idea that anything divided by itself is 1 \[ \frac{x \cdot x}{x \cdot x} = 1 \]
  • phi
so \[ \frac{x^4}{x^2} = \frac{x\cdot x \cdot x\cdot x}{x\cdot x}= x^2 \]
  • phi
now do the y^3/y
y3
  • phi
\[ \frac{y \cdot y \cdot y}{y} \] you have one y/y pair that you can cancel
  • phi
you should get \[ \frac{8}{-2} \cdot \frac{x^4}{x^2} \cdot \frac{y^3}{y} \\ -4 x^2y^2\]
  • phi
now do the next term 4x3y2 divided by -2x2y
-2 x^1.5 y^2
@phi is the answer -4^2y^2-2x^1.5y^26xy^4
@DarronW do you know if this is the answer

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