Question

Please Help for Fan and Medal!

Answer 1

Part A: Divide (4x^4y^3 + 8x^3y^2 - 12x^2y - 16x^2y^4) by -4x^2y. 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)
@mathmale @Hero @ganeshie8 @Luigi0210

Answer 2

Have you tried seeing if you can factor anything out?

Answer 3

Yes this is my answer: \[4x ^{2}y(x^2y^2- 4y^3 + 2xy - 3)\]

Answer 4

Is it correct @Luigi0210

Answer 5

@Hero Am I correct?

Answer 6

@Broskishelleh you mind showing the work you did to get that result?

Answer 7

Okay, give me a sec to type it out

Answer 8

Sorry for taking so long :(
First make two groups
(4x^4y^3 + 8x^3y^2) - (12x^2y - 16x^2y^4)
Next find like terms
(4x^4y^3 + 8x^3y^2) = 4x^3y^2(2xy)
(12x^2y - 16x^2y^4) =

Answer 9

(12x^2y - 16x^2y^4) = 4x^2y(7xy^3)

Answer 10

@Hero Are you still here?

Answer 11

What is the purpose of creating two groups? That isn't necessary in this case. At least not for the first step. The first step is to find what is common to each term (Greatest Common Factor)

Answer 12

Oh this was how I was tutored, sorry for me being dumb :(

Answer 13

There are different approaches for different kinds of problems. For one problem type, factoring by grouping may be appropriate. But not for this one. At least not yet.

Answer 14

@tHe_FiZiCx99 Hi

Answer 15

Ok then

Answer 16

@Hero Ah I get it so what shall we do next?

Answer 17

Have you figured out the GCF yet?

Answer 18

They all have 4 as a GCF

Answer 19

Find the GCF of 4x^4y^3 + 8x^3y^2 - 12x^2y - 16x^2y^4

Answer 20

You have to find the GCF of the variables as well

Answer 21

(4x^4y^3 + 8x^3y^2) - (12x^2y - 16x^2y^4)
4(x^4y^3 - 2x^3y^2) - (3x^2y - 4x^2y^4)

Answer 22

4(x^4y^3 - 2x^3y^2) - (3x^2y + 4x^2y^4)

Answer 23

I'm am certain that you have confused one method with another here. All you have to do is factor out the GCF, not group terms.

Answer 24

So how can I do it correctly if you may? I'm sorry for the stress and hassle I know I am not strong in Math

Answer 25

@Hero

Answer 26

@paki

Answer 27

To find GCF of 4x^4y^3 + 8x^3y^2 - 12x^2y - 16x^2y^4
look at the constants, x variables and y variables separately and find the GCF of each.
GCF of 4, 8, -12, -16 is 4
GCF of x^4, x^3, x^2, x^2 is x^2
GCF of y^3, y^2, y, y^4 is y

Therefore, the GCF of 4x^4y^3 + 8x^3y^2 - 12x^2y - 16x^2y^4 is 4x^2y

Answer 28

Maybe this video can help you:
https://www.youtube.com/watch?v=lX3eni0luro

Answer 29

Thank you @aum and @Hero

Answer 30

Is that Part 1?

Answer 31

Factoring is more than just knowing what the GCF is. You still have to factor out the GCF. Watch the video to learn how to do that.

Answer 32

Can i maybe have an explanation videos usually don't stick well. I am not that kind of learner. Though, I will still view it, just in case.

Answer 33

It won't hurt to look at the video. It should help more than you think.

Answer 34

Okay, then in a bit can I open Part 2 in a different question?

Answer 35

Parts B and C are dependent on answer to Part A. You might as well finish the whole question here.

Answer 36

Okay I am in the end of the video

Answer 37

What did you learn?

Answer 38

It helped so much! I saw what you meant @Hero

Answer 39

Let's go back, I guess

Answer 40

4x^4y^3 + 8x^3y^2 - 12x^2y - 16x^2y^4
I see they all have 4, and x^2, and y
So like @aum said we have 4x^2y

Answer 41

So, I guess next we divide 4x^2y by -4x^2y

Answer 42

Before dividing you still have to factor the numerator.
Numerator is: 4x^4y^3 + 8x^3y^2 - 12x^2y - 16x^2y^4 = 4x^2y( ...... ? ......)

Answer 43

Hmm, so like I said before it is 4x^2y(x^2y^2−4y^3+2xy−3)

Answer 44

There's a couple ways to approach it. You can either factor out the numerator like what @aum is suggesting or you can split the fraction as follows:

\[\frac{4x^4 y^3}{-4x^2y} + \frac{8x^3y^2}{-4x^2y} - \frac{12x^2y}{-4x^2y} - \frac{16x^2y^4}{-4x^2y}\] then simplify each fraction.

Answer 45

That is correct.
Now cancel the 4x^2y in the numerator and the denominator. Don't forget the minus sign in the denominator.

Answer 46

Oh man, my mother needs me to go to sleep. Can we continue tomorrow? I have to fast tomorrow so I need to get off in about 5 more minutes.

Answer 47

Make sure you double-check your factorization.

Answer 48

Alright. Bump the question up tomorrow and someone who is logged on at that time may help you.

Answer 49

Can I tag you guys?

Answer 50

No permission needed for tagging! But not everyone may be logged in at that time you are.

Answer 51

If you factor something out, it has to multiply back out to get the same thing you started with.

Answer 52

Okay @Hero and @aum Thanks and I will hopefully continue about 1PM

Answer 53

Good night and good luck.

Answer 54

4x^2y(x^2y^2−4y^3+2xy−3) = 4x^4y^3 - 16x^2y^4 + 8x^3y^2 - 12x^2y
which agrees with the original expression (just a couple of terms switched around) and so your factorization of the numerator is correct.

Answer 55

@Hero @LearningIsAwesome HELP!

Answer 56

This looks a bit difficult but I'll try

Answer 57

I helped you, so hopefully you could do the same back :-)

Answer 58

Re-write as \[\frac{4x^4 y^3}{-4x^2y} + \frac{8x^3y^2}{-4x^2y} - \frac{12x^2y}{-4x^2y} - \frac{16x^2y^4}{-4x^2y}\]

then simplify each fraction

Answer 59

Okay give me a sec

Answer 60

My answer was \[-x^2y^2+4y^3-2xy+3\] for all of the fractions

Answer 61

I need to leave for prayers in a bit so may we please go on to the next step

Answer 62

Was that my answer?

Answer 63

I need to leave for prayers so bye I guess.

Answer 64

Looks right