## 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.

1. e.mccormick

Well, what did you write?

2. e.mccormick

Do you know what a GCF is?

3. anonymous

Yes, greatest common factor

4. anonymous

I just don't know where to start..

5. e.mccormick

Well, I would work it in reverse.

6. 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.

7. 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.

8. 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.

9. 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.

10. anonymous

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

11. 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?

12. anonymous

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

13. 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?

14. anonymous

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

15. 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?

16. anonymous

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

17. e.mccormick

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

18. 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.

19. anonymous

How do you know what to subtract the exponents by?

20. 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.

21. anonymous

Oh, i see

22. e.mccormick

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

23. 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.

24. anonymous

Okay, I will be right back!

25. anonymous

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

26. 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

27. 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.

28. anonymous

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

29. e.mccormick

Exactly!

30. anonymous

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

31. e.mccormick

Sure.

32. anonymous

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

33. e.mccormick

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

34. anonymous

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

35. e.mccormick

Yep!

36. anonymous

I need help with this next step

37. e.mccormick

Which next step?

38. anonymous

When you subtract from the exponents

39. e.mccormick

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

40. anonymous

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

41. e.mccormick

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

42. anonymous

Oh, haha i forgot about the 8

43. e.mccormick

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

44. anonymous

x^2

45. e.mccormick

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

46. anonymous

Then what can i factor? The constant?

47. e.mccormick

Well, are there any common factors in the constants?

48. anonymous

I would say no, right?

49. 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.

50. anonymous

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

51. 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.

52. e.mccormick

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

53. e.mccormick

So it is? $$15x^4 + 9y^3 +8x$$

54. anonymous

correct

55. e.mccormick

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

56. anonymous

Since I made this equation I could keep the variable x

57. anonymous

Sorry, I got disconnected

58. 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.

59. anonymous

What would I multiply again?

60. e.mccormick

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

61. 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.

62. anonymous

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

63. 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.

64. e.mccormick

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

65. anonymous

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

66. 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.