Question

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?

Answer 1

@phi please help

Answer 2

someone

Answer 3

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

Answer 4

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

Answer 5

what?

Answer 6

Divide (8x4y3 + 4x3y2 - 2x2y - 12x2y4) by -2x2y.

that means you divide each term by -2x2y

the first term is
8x4y3 divided by -2x2y

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

4*256*27

Answer 12

is this right

Answer 13

letters stay letters

Answer 14

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

Answer 15

we use the idea that anything divided by itself is 1

\[ \frac{x \cdot x}{x \cdot x} = 1 \]

Answer 16

so \[ \frac{x^4}{x^2} = \frac{x\cdot x \cdot x\cdot x}{x\cdot x}= x^2 \]

Answer 17

now do the y^3/y

Answer 18

y3

Answer 19

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

Answer 20

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

Answer 21

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

Answer 22

-2 x^1.5 y^2

Answer 23

@phi

Answer 24

@phi is the answer -4^2y^2-2x^1.5y^26xy^4

Answer 25

@DarronW do you know if this is the answer