anonymous
  • anonymous
Create your own factorable polynomial with a GCF. Rewrite that polynomial in two other equivalent forms. Explain how each form was created.
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
e.mccormick
  • e.mccormick
Well, what did you write?
e.mccormick
  • e.mccormick
Do you know what a GCF is?
anonymous
  • anonymous
Yes, greatest common factor

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
I just don't know where to start..
e.mccormick
  • e.mccormick
Well, I would work it in reverse.
e.mccormick
  • 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
  • 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
  • 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
  • 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
  • anonymous
Do you think you could explain to me how i could factor out the variables?
e.mccormick
  • 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
  • anonymous
3y^2 and 9x^2. Correct?
e.mccormick
  • 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
  • anonymous
-6x^4y^5-15x^3+9^3?
e.mccormick
  • 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
  • anonymous
Yes, please. Hahaha, sorry I'm not getting this right away.
e.mccormick
  • 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
  • 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
  • anonymous
How do you know what to subtract the exponents by?
e.mccormick
  • 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
  • anonymous
Oh, i see
e.mccormick
  • e.mccormick
So try \(−6x^4y^5 − 15x^3y^2 + 9x^2y^3 \) again. =)
e.mccormick
  • 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
  • anonymous
Okay, I will be right back!
anonymous
  • anonymous
Would it be this? 3x^2y^2(-2x^2y^3-5x+3y)?
e.mccormick
  • 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
  • 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
  • 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
  • e.mccormick
Exactly!
anonymous
  • anonymous
I get it now, do you think you could help with one more? I feel as if it is much simpler
e.mccormick
  • e.mccormick
Sure.
anonymous
  • anonymous
15x^4 + 9x^3 +8x
e.mccormick
  • e.mccormick
OK. So, you need to factor this, right?
anonymous
  • anonymous
Yes, and i think you would take out the x , correct? since its x^1?
e.mccormick
  • e.mccormick
Yep!
anonymous
  • anonymous
I need help with this next step
e.mccormick
  • e.mccormick
Which next step?
anonymous
  • anonymous
When you subtract from the exponents
e.mccormick
  • 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
  • anonymous
So, it would now be 15x^4-^1 +9x^3-^1
e.mccormick
  • e.mccormick
Almost. The 8 still stays there... just the x goes away. \(x(15x^{4-1} + 9x^{3-1} +8x^{1-1})\)
anonymous
  • anonymous
Oh, haha i forgot about the 8
e.mccormick
  • e.mccormick
=) And after the subtraction, that is: \(x(15x^3 + 9x^2 +8)\) So, anything else you can factor?
anonymous
  • anonymous
x^2
e.mccormick
  • e.mccormick
You already factored the x out of the 8, so you can't really factor x more.
anonymous
  • anonymous
Then what can i factor? The constant?
e.mccormick
  • e.mccormick
Well, are there any common factors in the constants?
anonymous
  • anonymous
I would say no, right?
e.mccormick
  • 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
  • anonymous
I messed something up actually, the 9x^3 is supposed to be 9y^3.
e.mccormick
  • 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
  • e.mccormick
Oops.... Hmmm.... well... that would make it different.
e.mccormick
  • e.mccormick
So it is? \(15x^4 + 9y^3 +8x\)
anonymous
  • anonymous
correct
e.mccormick
  • e.mccormick
Hmmm... well, there is no GCF at all in that.
anonymous
  • anonymous
Since I made this equation I could keep the variable x
anonymous
  • anonymous
Sorry, I got disconnected
e.mccormick
  • 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
  • anonymous
What would I multiply again?
e.mccormick
  • 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
  • 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
  • anonymous
So, the y then goes next to the other variables except the one in front?
e.mccormick
  • 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
  • e.mccormick
And after I multiply it, it is no longer out front....
anonymous
  • anonymous
Oh okay, I see! Thanks for helping me =)
e.mccormick
  • 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.

Looking for something else?

Not the answer you are looking for? Search for more explanations.