Question

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

Answer 1

Well, what did you write?

Answer 2

Do you know what a GCF is?

Answer 3

Yes, greatest common factor

Answer 4

I just don't know where to start..

Answer 5

Well, I would work it in reverse.

Answer 6

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.

Answer 7

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.

Answer 8

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.

Answer 9

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.

Answer 10

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

Answer 11

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?

Answer 12

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

Answer 13

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?

Answer 14

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

Answer 15

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?

Answer 16

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

Answer 17

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

Answer 18

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.

Answer 19

How do you know what to subtract the exponents by?

Answer 20

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.

Answer 21

Oh, i see

Answer 22

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

Answer 23

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.

Answer 24

Okay, I will be right back!

Answer 25

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

Answer 26

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

Answer 27

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.

Answer 28

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

Answer 29

Exactly!

Answer 30

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

Answer 31

Sure.

Answer 32

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

Answer 33

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

Answer 34

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

Answer 35

Yep!

Answer 36

I need help with this next step

Answer 37

Which next step?

Answer 38

When you subtract from the exponents

Answer 39

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

Answer 40

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

Answer 41

Almost. The 8 still stays there... just the x goes away.

\(x(15x^{4-1} + 9x^{3-1} +8x^{1-1})\)

Answer 42

Oh, haha i forgot about the 8

Answer 43

=)

And after the subtraction, that is:
\(x(15x^3 + 9x^2 +8)\)

So, anything else you can factor?

Answer 44

x^2

Answer 45

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

Answer 46

Then what can i factor? The constant?

Answer 47

Well, are there any common factors in the constants?

Answer 48

I would say no, right?

Answer 49

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

Answer 50

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

Answer 51

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.

Answer 52

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

Answer 53

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

Answer 54

correct

Answer 55

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

Answer 56

Since I made this equation I could keep the variable x

Answer 57

Sorry, I got disconnected

Answer 58

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.

Answer 59

What would I multiply again?

Answer 60

Well, it works the opposite way. Instead of subtracting from exponents, you add. Let me show with just the y:

\(2x^2y(15x^4 + 9y^3 +8x)\)

\(2x^2(15x^4y^{0+1} + 9y^{3+1} +8xy^{0+1})\)

\(2x^2(15x^4y^{1} + 9y^{4} +8xy^{1})\)

\(2x^2(15x^4y + 9y^{4} +8xy)\)

Answer 61

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.

Answer 62

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

Answer 63

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.

Answer 64

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

Answer 65

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

Answer 66

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.