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

Mathematics
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Well, what did you write?
Do you know what a GCF is?
Yes, greatest common factor

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I just don't know where to start..
Well, I would work it in reverse.
As 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.
Okay, 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.
That 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.
So, 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.
Do you think you could explain to me how i could factor out the variables?
Well, 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?
3y^2 and 9x^2. Correct?
The 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?
-6x^4y^5-15x^3+9^3?
You 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?
Yes, please. Hahaha, sorry I'm not getting this right away.
\(5x^3y^2+3xy^3 \) \(xy^2(5x^{3-1}y^{2-2}+3x^{1-1}y^{3-2}) \) \(xy^2(5x^2+3y) \)
See 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.
How do you know what to subtract the exponents by?
Another name for factoring is un-multiplying. 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.
Oh, i see
So try \(−6x^4y^5 − 15x^3y^2 + 9x^2y^3 \) again. =)
It 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.
Okay, I will be right back!
Would it be this? 3x^2y^2(-2x^2y^3-5x+3y)?
That 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 \(−6x^{4-2}y^{5-2} − 15x^{3-2}y^{2-2} + 9x^{2-2}y^{3-2}\) \(x^2y^2(−6x^2y^3 − 15x + 9y)\) <-- 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
That 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.
Oh, thank you so much :D So, I sorta just solved the whole thing but throughout my steps there where the forms?
Exactly!
I get it now, do you think you could help with one more? I feel as if it is much simpler
Sure.
15x^4 + 9x^3 +8x
OK. So, you need to factor this, right?
Yes, and i think you would take out the x , correct? since its x^1?
Yep!
I need help with this next step
Which next step?
When you subtract from the exponents
Well, because it is just one x, you remove one from each of the exponents of x. The x alone just disappears.
So, it would now be 15x^4-^1 +9x^3-^1
Almost. The 8 still stays there... just the x goes away. \(x(15x^{4-1} + 9x^{3-1} +8x^{1-1})\)
Oh, haha i forgot about the 8
=) And after the subtraction, that is: \(x(15x^3 + 9x^2 +8)\) So, anything else you can factor?
x^2
You already factored the x out of the 8, so you can't really factor x more.
Then what can i factor? The constant?
Well, are there any common factors in the constants?
I would say no, right?
Correct! Now, I do not know if you are doing things like (v+2)(v-5), but it will not factor that way.
I messed something up actually, the 9x^3 is supposed to be 9y^3.
So really: \(15x^4 + 9x^3 +8x\) \(x(15x^{4-1} + 9x^{3-1} +8x^{1-1})\) \(x(15x^3 + 9x^2 +8)\) That is it... no common constants, no \((x\pm h ) (x\pm k ) \), etc.
Oops.... Hmmm.... well... that would make it different.
So it is? \(15x^4 + 9y^3 +8x\)
correct
Hmmm... well, there is no GCF at all in that.
Since I made this equation I could keep the variable x
Sorry, I got disconnected
Yah, 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.
What would I multiply again?
Well, it works the opposite way. Instead of subtracting from exponents, you add. Let me show with just the y: \(2x^2y(15x^4 + 9y^3 +8x)\) \(2x^2(15x^4y^{0+1} + 9y^{3+1} +8xy^{0+1})\) \(2x^2(15x^4y^{1} + 9y^{4} +8xy^{1})\) \(2x^2(15x^4y + 9y^{4} +8xy)\)
Why 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.
So, the y then goes next to the other variables except the one in front?
It 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.
And after I multiply it, it is no longer out front....
Oh okay, I see! Thanks for helping me =)
Nasty 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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