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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.

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  1. e.mccormick
    • one year ago
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    Well, what did you write?

  2. e.mccormick
    • one year ago
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    Do you know what a GCF is?

  3. anonymous
    • one year ago
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    Yes, greatest common factor

  4. anonymous
    • one year ago
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    I just don't know where to start..

  5. e.mccormick
    • one year ago
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    Well, I would work it in reverse.

  6. e.mccormick
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    Do you think you could explain to me how i could factor out the variables?

  11. e.mccormick
    • one year ago
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    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
    • one year ago
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    3y^2 and 9x^2. Correct?

  13. e.mccormick
    • one year ago
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    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
    • one year ago
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    -6x^4y^5-15x^3+9^3?

  15. e.mccormick
    • one year ago
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    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
    • one year ago
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    Yes, please. Hahaha, sorry I'm not getting this right away.

  17. e.mccormick
    • one year ago
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    \(5x^3y^2+3xy^3 \) \(xy^2(5x^{3-1}y^{2-2}+3x^{1-1}y^{3-2}) \) \(xy^2(5x^2+3y) \)

  18. e.mccormick
    • one year ago
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    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
    • one year ago
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    How do you know what to subtract the exponents by?

  20. e.mccormick
    • one year ago
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    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
    • one year ago
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    Oh, i see

  22. e.mccormick
    • one year ago
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    So try \(−6x^4y^5 − 15x^3y^2 + 9x^2y^3 \) again. =)

  23. e.mccormick
    • one year ago
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    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
    • one year ago
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    Okay, I will be right back!

  25. anonymous
    • one year ago
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    Would it be this? 3x^2y^2(-2x^2y^3-5x+3y)?

  26. e.mccormick
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    Exactly!

  30. anonymous
    • one year ago
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    I get it now, do you think you could help with one more? I feel as if it is much simpler

  31. e.mccormick
    • one year ago
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    Sure.

  32. anonymous
    • one year ago
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    15x^4 + 9x^3 +8x

  33. e.mccormick
    • one year ago
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    OK. So, you need to factor this, right?

  34. anonymous
    • one year ago
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    Yes, and i think you would take out the x , correct? since its x^1?

  35. e.mccormick
    • one year ago
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    Yep!

  36. anonymous
    • one year ago
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    I need help with this next step

  37. e.mccormick
    • one year ago
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    Which next step?

  38. anonymous
    • one year ago
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    When you subtract from the exponents

  39. e.mccormick
    • one year ago
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    Well, because it is just one x, you remove one from each of the exponents of x. The x alone just disappears.

  40. anonymous
    • one year ago
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    So, it would now be 15x^4-^1 +9x^3-^1

  41. e.mccormick
    • one year ago
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    Almost. The 8 still stays there... just the x goes away. \(x(15x^{4-1} + 9x^{3-1} +8x^{1-1})\)

  42. anonymous
    • one year ago
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    Oh, haha i forgot about the 8

  43. e.mccormick
    • one year ago
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    =) And after the subtraction, that is: \(x(15x^3 + 9x^2 +8)\) So, anything else you can factor?

  44. anonymous
    • one year ago
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    x^2

  45. e.mccormick
    • one year ago
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    You already factored the x out of the 8, so you can't really factor x more.

  46. anonymous
    • one year ago
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    Then what can i factor? The constant?

  47. e.mccormick
    • one year ago
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    Well, are there any common factors in the constants?

  48. anonymous
    • one year ago
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    I would say no, right?

  49. e.mccormick
    • one year ago
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    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
    • one year ago
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    I messed something up actually, the 9x^3 is supposed to be 9y^3.

  51. e.mccormick
    • one year ago
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    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
    • one year ago
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    Oops.... Hmmm.... well... that would make it different.

  53. e.mccormick
    • one year ago
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    So it is? \(15x^4 + 9y^3 +8x\)

  54. anonymous
    • one year ago
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    correct

  55. e.mccormick
    • one year ago
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    Hmmm... well, there is no GCF at all in that.

  56. anonymous
    • one year ago
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    Since I made this equation I could keep the variable x

  57. anonymous
    • one year ago
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    Sorry, I got disconnected

  58. e.mccormick
    • one year ago
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    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
    • one year ago
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    What would I multiply again?

  60. e.mccormick
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    So, the y then goes next to the other variables except the one in front?

  63. e.mccormick
    • one year ago
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    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
    • one year ago
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    And after I multiply it, it is no longer out front....

  65. anonymous
    • one year ago
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    Oh okay, I see! Thanks for helping me =)

  66. e.mccormick
    • one year ago
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    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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