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anonymous

  • one year ago

Part A: Factor 3x2y2 − 2xy2 − 8y2. Show your work. (4 points) Part B: Factor x2 + 10x + 25. Show your work. (3 points) Part C: Factor x2 − 36. Show your work. (3 points)

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  1. anonymous
    • one year ago
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    i need steps i have answers! Part A.) y^2(3x+4)(x-2) Part B.) (x-5)^2 Part C.) (x+6)(x-6)

  2. anonymous
    • one year ago
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    @zepdrix

  3. anonymous
    • one year ago
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    sorry I cant help u on this but I love the name :)

  4. anonymous
    • one year ago
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    ty :}

  5. anonymous
    • one year ago
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    yw and good luck on ur question!

  6. anonymous
    • one year ago
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    ty

  7. anonymous
    • one year ago
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    yw

  8. zepdrix
    • one year ago
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    \[\large\rm 3x^2y^2 − 2xy^2 − 8y^2\]What are we doing? :U Factoring this thing?

  9. zepdrix
    • one year ago
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    \[\large\rm 3x^2\color{orangered}{y^2} − 2x\color{orangered}{y^2} − 8\color{orangered}{y^2}\]Hmm there isn't a whole lot that we can do with this first one. Notice that the last term has no x's, so we can't pull out any factor out any amount of x. The last two terms have a 2 in them (8 is 4 times 2) but the first term does not, so we can't pull a 2 out. You can do something with the y's though, ya?

  10. anonymous
    • one year ago
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    Yes.

  11. zepdrix
    • one year ago
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    Oh I guess we can factor the x's after that though :3 hah

  12. zepdrix
    • one year ago
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    \[\large\rm \color{orangered}{y^2}(3x^2−2x−8)\]So you ok with that first step? :o

  13. anonymous
    • one year ago
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    Um (y2)(3x2+−2x+−8) =(y2)(3x2)+(y2)(−2x)+(y2)(−8) =3x2y2−2xy2−8y2 =3x2y2−2xy2−8y2

  14. zepdrix
    • one year ago
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    You just un-did the first step :) Not sure why hehe

  15. anonymous
    • one year ago
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    Opps >_<

  16. anonymous
    • one year ago
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    I am confused on this..

  17. zepdrix
    • one year ago
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    |dw:1440970256523:dw|So we're taking y^2 out of each term.

  18. zepdrix
    • one year ago
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    And from there, we're going to factor our x's. That first step giving you trouble though? :d

  19. anonymous
    • one year ago
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    y^2(3x+4)(x-2)

  20. zepdrix
    • one year ago
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    Yes, that's the answer. But we're not through the steps quite yet :) lol

  21. anonymous
    • one year ago
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    Ok lol :]

  22. zepdrix
    • one year ago
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    Since the x^2 has a coefficient, we'll have to try factoring by grouping for this one:\[\large\rm 3x^2-2x-8\qquad\to\qquad ax^2+bx+c\]Multiplying a and c gives us -24, ya? So we need factors of -24 that will add to -2. Example: \(\large\rm -12 * 2 = -24\) But \(\large\rm -12 + 2\ne-2\) Hmm so -12 and 2 are NOT the factors we're looking for.

  23. zepdrix
    • one year ago
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    \(\large\rm 3\cdot-8=-24\) But \(\large\rm 3+-8\ne-2\) Hmm looks like those factors don't work either :[

  24. zepdrix
    • one year ago
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    Any other ones you can try? :)

  25. anonymous
    • one year ago
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    Um... 2*-8=(-16)

  26. anonymous
    • one year ago
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    \[2\times-8= -16\]

  27. zepdrix
    • one year ago
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    No no, we want numbers that multiply to -24.

  28. anonymous
    • one year ago
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    OH OK.

  29. zepdrix
    • one year ago
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    3 and 8 didn't work. 12 and 2 didn't work. Any other combinations we can try? :o

  30. anonymous
    • one year ago
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    \[(-12) *2= -24 \] OR \[12*(-2)=-24\]

  31. zepdrix
    • one year ago
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    No we already tried 12 and 2. Gotta try some other factors of 24

  32. anonymous
    • one year ago
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    k

  33. anonymous
    • one year ago
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    \[6•4\]

  34. zepdrix
    • one year ago
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    \(\large\rm 6\cdot-4=-24\) Ok those work for multiplication. Let's try adding them. \(\large\rm 6+-4=2\) Hmm I got a 2 instead of a -2, that simply means I Put the negative on the wrong number. So the factors we want are -6 and 4. \(\large\rm -6\cdot4=-24\) \(\large\rm -6+4=-2\)

  35. anonymous
    • one year ago
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    ^0^

  36. zepdrix
    • one year ago
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    \[\large\rm 3x^2\color{orangered}{−2x}−8\]So what we want to do is, split up this -2x into -6x and 4x.\[\large\rm 3x^2\color{orangered}{−6x+4x}−8\]

  37. zepdrix
    • one year ago
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    And from there, we can do some grouping.\[\large\rm (3x^2−6x)+(4x−8)\]

  38. anonymous
    • one year ago
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    =)

  39. zepdrix
    • one year ago
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    So uh, factor some stuff :O what can you pull out of each bracket?

  40. anonymous
    • one year ago
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    3x2−2x−8

  41. anonymous
    • one year ago
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    \[3x^2−2x−8\]

  42. zepdrix
    • one year ago
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    ...?

  43. zepdrix
    • one year ago
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    factor some stuff out of this: (3x^2-6x)

  44. anonymous
    • one year ago
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    3x(x−2)

  45. zepdrix
    • one year ago
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    \[\large\rm (3x^2−6x)+(4x−8)\]\[\large\rm 3x(x−2)+(4x−8)\]Good good good :O Do something similar with the other set of brackets.

  46. anonymous
    • one year ago
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    \[(3x+4)(x−2)\]

  47. zepdrix
    • one year ago
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    Mmm k, so you get a 4 from the other brackets, ya?\[\large\rm 3x(x−2)+(4x−8)\]\[\large\rm 3x(x−2)+4(x−2)\]And you go a step further and pull the (x-2) from each term,\[\large\rm (x−2)(3x+4)\]Mmm k good, and ya, you get that. With the y^2 still in front.\[\large\rm y^2(3x+4)(x-2)\]

  48. anonymous
    • one year ago
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    k

  49. zepdrix
    • one year ago
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    For Part C recall your difference of squares formula:\[\large\rm a^2-b^2=(a-b)(a+b)\]Notice that in the formula, both numbers are squared on the left side of the equation. So in order to apply this rule to our equation,\[\large\rm x^2-36\]We'll first need to rewrite 36 as a square somehow.

  50. anonymous
    • one year ago
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    ok maybe \[6^2\]

  51. zepdrix
    • one year ago
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    ok great!\[\large\rm x^2-6^2\]Good. Now you can apply the formula:\[\large\rm x^2-6^2=(x-6)(x+6)\]Nice easy one step problem. It's really just about remembering your formula for that one.

  52. anonymous
    • one year ago
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    Can you write like Part A Part B And Part C I am really confused what is what >_< lol

  53. zepdrix
    • one year ago
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    We skipped Part B, maybe that's the confusion. I gotta go make a sammich, so I was trying to rush through it >.< lol sorry The big long explanation was Part A. And then these last two comments I posted were Part C.

  54. anonymous
    • one year ago
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    I really need part B too!

  55. zepdrix
    • one year ago
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    you -_- you and your ways lol

  56. zepdrix
    • one year ago
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    ok sec :3 lemme read it again

  57. anonymous
    • one year ago
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    Thank yyou!

  58. zepdrix
    • one year ago
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    \[\large\rm x^2 + 10x + 25\]So this is our typical factoring problem. Don't have a good grasp on these types yet? :o You look for `factors of 25` that will `sum to 10`. Lemme try a one and see what happens. \(\large\rm 25\cdot1=25\) \(\large\rm 25+1=26\) hmm, nope. they didn't add to 10.

  59. zepdrix
    • one year ago
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    Any other factors we can try?

  60. anonymous
    • one year ago
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    \[25+(-15)\]

  61. zepdrix
    • one year ago
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    `factors of 25` is what you're looking for. \(\large\rm 25\cdot(-15)\ne25\) Those are not factors of 25.

  62. zepdrix
    • one year ago
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    Example: If I was looking for factors of 24, i would list them like this maybe: 8 * 3 = 24 6 * 4 = 24 12 * 2 = 24 24 * 1 = 24 So those are all of the combinations I would want to try.

  63. anonymous
    • one year ago
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    There is not one.

  64. zepdrix
    • one year ago
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    25 ends in a 5, so it's 5 times something

  65. anonymous
    • one year ago
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    Yes 5*5...

  66. zepdrix
    • one year ago
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    \(\large\rm 5\cdot5=25\) \(\large\rm 5+5=10\) Ah there we go! That's the one we were looking for.

  67. anonymous
    • one year ago
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    Is that part B?

  68. anonymous
    • one year ago
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    (x-5)(x-5) or (x-5)^2

  69. anonymous
    • one year ago
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    \[(x-5)(x-5) or (x-5)^2\]

  70. zepdrix
    • one year ago
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    So for part B, you can either do factoring by grouping again:\[\large\rm x^2 + 10x + 25\]\[\large\rm x^2 + 5x+5x + 25\]\[\large\rm x(x + 5)+5(x + 5)\]\[\large\rm (x+5)(x + 5)\]\[\large\rm (x+5)^2\]Or just remember your shortcut for factoring. Yes. but not -5, +5 silly :)

  71. zepdrix
    • one year ago
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    The solution you listed at the start is incorrect. That's strange :P

  72. anonymous
    • one year ago
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    I am confused whats part a and wahts part c?

  73. anonymous
    • one year ago
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    *whats

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