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anonymous
 one year ago
Create your own factorable polynomial with a GCF. Rewrite that polynomial in two other equivalent forms. Explain how each form was created.
anonymous
 one year ago
Create your own factorable polynomial with a GCF. Rewrite that polynomial in two other equivalent forms. Explain how each form was created.

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e.mccormick
 one year ago
Best ResponseYou've already chosen the best response.1Well, what did you write?

e.mccormick
 one year ago
Best ResponseYou've already chosen the best response.1Do you know what a GCF is?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes, greatest common factor

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I just don't know where to start..

e.mccormick
 one year ago
Best ResponseYou've already chosen the best response.1Well, I would work it in reverse.

e.mccormick
 one year ago
Best ResponseYou've already chosen the best response.1As in, Start with something that has say three factors, one of them being a GCF, and then multiply it out. The different stages of it being ultiplied will give you the different forms.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay, so I would need a whole expression, with three factors, do you think this;−6x4y5 − 15x3y2 + 9x2y3 would be good enough? I am not very good at factoring and such.

e.mccormick
 one year ago
Best ResponseYou've already chosen the best response.1That would work. It does not nessisarily need to be that complex, but there is nothing really wrong with that. You can just factor it one step at a time, and that would give you the different forms.

e.mccormick
 one year ago
Best ResponseYou've already chosen the best response.1So, if you start with −6x^4y^5 − 15x^3y^2 + 9x^2y^3, you can factor out the constants then varisables, and doing so will get you three different versions and a method to describe.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Do you think you could explain to me how i could factor out the variables?

e.mccormick
 one year ago
Best ResponseYou've already chosen the best response.1Well, you can only factor out the number of them that is the SMALLEST exponent. x^2+x < In that, the smallest exponent is 1, x alone, so all I could factor out is x. x(x+1) x^999+x^2 < the big, 999 as the first enponent does not matter for GCF. Only the 2. x^2(x^997+1) So on yours, −6x^4y^5 − 15x^3y^2 + 9x^2y^3 what is the smallest exponents for the variables?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.03y^2 and 9x^2. Correct?

e.mccormick
 one year ago
Best ResponseYou've already chosen the best response.1The 3 and 9 are not part of te variables. They are constants. So just the y^2 and x^2. So, if you factor out x^2y^2, what do you get?

e.mccormick
 one year ago
Best ResponseYou've already chosen the best response.1You need to take the 2 off of every exponent, since you are taking out 2 x and 2 y. Then you put x^2y^2 out front and the rest in ( ) 5x^3y^2+3xy^3 becomes: xy^2(5x^2+3y) Want me to show it step by step for that example?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes, please. Hahaha, sorry I'm not getting this right away.

e.mccormick
 one year ago
Best ResponseYou've already chosen the best response.1\(5x^3y^2+3xy^3 \) \(xy^2(5x^{31}y^{22}+3x^{11}y^{32}) \) \(xy^2(5x^2+3y) \)

e.mccormick
 one year ago
Best ResponseYou've already chosen the best response.1See how in the middle step I subtract out an x and y^2 from every exponent? If the exponent becomes 0, the variable is completey factored out and disappears.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0How do you know what to subtract the exponents by?

e.mccormick
 one year ago
Best ResponseYou've already chosen the best response.1Another name for factoring is unmultiplying. That is really what you are doing. Putting things in a form where they are not multiplied. They are factored. I am taking out the smallest number of exponents. The just x is really x^1. So I subtract out 1 for it. My y is y^2 that I can take out, so I subtract out 2. Then, in front I have xy^2.

e.mccormick
 one year ago
Best ResponseYou've already chosen the best response.1So try \(−6x^4y^5 − 15x^3y^2 + 9x^2y^3 \) again. =)

e.mccormick
 one year ago
Best ResponseYou've already chosen the best response.1It will become: x^2y^2(something) You already know this because as you correctly said, x^2 and y^2 are the smallest exponents of x and y.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay, I will be right back!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Would it be this? 3x^2y^2(2x^2y^35x+3y)?

e.mccormick
 one year ago
Best ResponseYou've already chosen the best response.1That is the final form. You need a total of 3 forms, so I would do the numebrs seperate. \(−6x^4y^5 − 15x^3y^2 + 9x^2y^3\) < Form 1 \(−6x^{42}y^{52} − 15x^{32}y^{22} + 9x^{22}y^{32}\) \(x^2y^2(−6x^2y^3 − 15x + 9y)\) < Form 2 \(x^2y^2(−(3\cdot 2)x^2y^3 − (3\cdot 5)x + (3\cdot 3)y)\) \(3x^2y^2(−2x^2y^3 − 5x + 3y)\) < Form 3

e.mccormick
 one year ago
Best ResponseYou've already chosen the best response.1That is for the "Rewrite that polynomial in two other equivalent forms. Explain how each form was created." part of the instructions. Form 1 is original. 2 is variables factored out. 3 is you took form 2 and also factored out the constants.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh, thank you so much :D So, I sorta just solved the whole thing but throughout my steps there where the forms?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I get it now, do you think you could help with one more? I feel as if it is much simpler

e.mccormick
 one year ago
Best ResponseYou've already chosen the best response.1OK. So, you need to factor this, right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes, and i think you would take out the x , correct? since its x^1?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I need help with this next step

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0When you subtract from the exponents

e.mccormick
 one year ago
Best ResponseYou've already chosen the best response.1Well, because it is just one x, you remove one from each of the exponents of x. The x alone just disappears.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So, it would now be 15x^4^1 +9x^3^1

e.mccormick
 one year ago
Best ResponseYou've already chosen the best response.1Almost. The 8 still stays there... just the x goes away. \(x(15x^{41} + 9x^{31} +8x^{11})\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh, haha i forgot about the 8

e.mccormick
 one year ago
Best ResponseYou've already chosen the best response.1=) And after the subtraction, that is: \(x(15x^3 + 9x^2 +8)\) So, anything else you can factor?

e.mccormick
 one year ago
Best ResponseYou've already chosen the best response.1You already factored the x out of the 8, so you can't really factor x more.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Then what can i factor? The constant?

e.mccormick
 one year ago
Best ResponseYou've already chosen the best response.1Well, are there any common factors in the constants?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I would say no, right?

e.mccormick
 one year ago
Best ResponseYou've already chosen the best response.1Correct! Now, I do not know if you are doing things like (v+2)(v5), but it will not factor that way.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I messed something up actually, the 9x^3 is supposed to be 9y^3.

e.mccormick
 one year ago
Best ResponseYou've already chosen the best response.1So really: \(15x^4 + 9x^3 +8x\) \(x(15x^{41} + 9x^{31} +8x^{11})\) \(x(15x^3 + 9x^2 +8)\) That is it... no common constants, no \((x\pm h ) (x\pm k ) \), etc.

e.mccormick
 one year ago
Best ResponseYou've already chosen the best response.1Oops.... Hmmm.... well... that would make it different.

e.mccormick
 one year ago
Best ResponseYou've already chosen the best response.1So it is? \(15x^4 + 9y^3 +8x\)

e.mccormick
 one year ago
Best ResponseYou've already chosen the best response.1Hmmm... well, there is no GCF at all in that.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Since I made this equation I could keep the variable x

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Sorry, I got disconnected

e.mccormick
 one year ago
Best ResponseYou've already chosen the best response.1Yah, we all did. Well: \(15x^4 + 9y^3 +8x\) \(3(5x^4 + 3y^3) +8x\) \(15x^4+8x + 9y^3 \) \(x(15x^3+8) + 9y^3 \) None of that makes for anything that would really factor out of the whole thing. BUT, you could do that whole work it in reverse thing I talked about. What if I take \(15x^4 + 9y^3 +8x\) and do this: \(2x^2y(15x^4 + 9y^3 +8x)\) Now you can just multiply the things in to get the other forms.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0What would I multiply again?

e.mccormick
 one year ago
Best ResponseYou've already chosen the best response.1Well, it works the opposite way. Instead of subtracting from exponents, you add. Let me show with just the y: \(2x^2y(15x^4 + 9y^3 +8x)\) \(2x^2(15x^4y^{0+1} + 9y^{3+1} +8xy^{0+1})\) \(2x^2(15x^4y^{1} + 9y^{4} +8xy^{1})\) \(2x^2(15x^4y + 9y^{4} +8xy)\)

e.mccormick
 one year ago
Best ResponseYou've already chosen the best response.1Why I did y^[0+1] os because originally there is no y, so 0, and I am adding 1. If you get the concept, you can just add a y where there was no y before. I just wanted to show it with all possible steps so you would get the concept.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So, the y then goes next to the other variables except the one in front?

e.mccormick
 one year ago
Best ResponseYou've already chosen the best response.1It starts out in front. Then I multiply it through all the terms. If the term has a y alread (middle term) I add to the exponent. If it does not have a y already, it gets a y.

e.mccormick
 one year ago
Best ResponseYou've already chosen the best response.1And after I multiply it, it is no longer out front....

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh okay, I see! Thanks for helping me =)

e.mccormick
 one year ago
Best ResponseYou've already chosen the best response.1Nasty lag at the moment... You can do the same basic thing with the x^2 and then the 2. Each time it is less and less factored, but has the same value/meaning because you are just multiplying.
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