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

- anonymous

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- schrodinger

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- e.mccormick

Well, what did you write?

- e.mccormick

Do you know what a GCF is?

- anonymous

Yes, greatest common factor

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- anonymous

I just don't know where to start..

- e.mccormick

Well, I would work it in reverse.

- e.mccormick

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.

- anonymous

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.

- e.mccormick

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.

- e.mccormick

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.

- anonymous

Do you think you could explain to me how i could factor out the variables?

- e.mccormick

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?

- anonymous

3y^2 and 9x^2. Correct?

- e.mccormick

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?

- anonymous

-6x^4y^5-15x^3+9^3?

- e.mccormick

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?

- anonymous

Yes, please. Hahaha, sorry I'm not getting this right away.

- e.mccormick

\(5x^3y^2+3xy^3 \)
\(xy^2(5x^{3-1}y^{2-2}+3x^{1-1}y^{3-2}) \)
\(xy^2(5x^2+3y) \)

- e.mccormick

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.

- anonymous

How do you know what to subtract the exponents by?

- e.mccormick

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.

- anonymous

Oh, i see

- e.mccormick

So try \(−6x^4y^5 − 15x^3y^2 + 9x^2y^3 \) again. =)

- e.mccormick

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.

- anonymous

Okay, I will be right back!

- anonymous

Would it be this? 3x^2y^2(-2x^2y^3-5x+3y)?

- e.mccormick

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

- e.mccormick

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.

- anonymous

Oh, thank you so much :D So, I sorta just solved the whole thing but throughout my steps there where the forms?

- e.mccormick

Exactly!

- anonymous

I get it now, do you think you could help with one more? I feel as if it is much simpler

- e.mccormick

Sure.

- anonymous

15x^4 + 9x^3 +8x

- e.mccormick

OK. So, you need to factor this, right?

- anonymous

Yes, and i think you would take out the x , correct? since its x^1?

- e.mccormick

Yep!

- anonymous

I need help with this next step

- e.mccormick

Which next step?

- anonymous

When you subtract from the exponents

- e.mccormick

Well, because it is just one x, you remove one from each of the exponents of x. The x alone just disappears.

- anonymous

So, it would now be 15x^4-^1 +9x^3-^1

- e.mccormick

Almost. The 8 still stays there... just the x goes away.
\(x(15x^{4-1} + 9x^{3-1} +8x^{1-1})\)

- anonymous

Oh, haha i forgot about the 8

- e.mccormick

=)
And after the subtraction, that is:
\(x(15x^3 + 9x^2 +8)\)
So, anything else you can factor?

- anonymous

x^2

- e.mccormick

You already factored the x out of the 8, so you can't really factor x more.

- anonymous

Then what can i factor? The constant?

- e.mccormick

Well, are there any common factors in the constants?

- anonymous

I would say no, right?

- e.mccormick

Correct! Now, I do not know if you are doing things like (v+2)(v-5), but it will not factor that way.

- anonymous

I messed something up actually, the 9x^3 is supposed to be 9y^3.

- e.mccormick

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.

- e.mccormick

Oops.... Hmmm.... well... that would make it different.

- e.mccormick

So it is?
\(15x^4 + 9y^3 +8x\)

- anonymous

correct

- e.mccormick

Hmmm... well, there is no GCF at all in that.

- anonymous

Since I made this equation I could keep the variable x

- anonymous

Sorry, I got disconnected

- e.mccormick

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.

- anonymous

What would I multiply again?

- e.mccormick

Well, it works the opposite way. Instead of subtracting from exponents, you add. Let me show with just the y:
\(2x^2y(15x^4 + 9y^3 +8x)\)
\(2x^2(15x^4y^{0+1} + 9y^{3+1} +8xy^{0+1})\)
\(2x^2(15x^4y^{1} + 9y^{4} +8xy^{1})\)
\(2x^2(15x^4y + 9y^{4} +8xy)\)

- e.mccormick

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.

- anonymous

So, the y then goes next to the other variables except the one in front?

- e.mccormick

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.

- e.mccormick

And after I multiply it, it is no longer out front....

- anonymous

Oh okay, I see! Thanks for helping me =)

- e.mccormick

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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