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Math Help - Really Really Urgent

  1. #1
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    Really Really Urgent

    I REALLY NEED HELP WITH THESE TWO PROBLEMS, please someone help solveeeeee i'm begginggg you, I keep getting them wrong everytime I do them

    How would I multiply this problem? the small numbers are twos
    http://img397.imageshack.us/img397/7599/math13ep.jpg



    How would I perform the indicated operation on this problem?
    http://img397.imageshack.us/img397/9062/math20me.jpg
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  2. #2
    Eater of Worlds
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    For number 2:

    It's just a division/cancellation problem.

    \frac{a^{2}-a-30}{a+5}

    Factor the numerator and you will quickly see the cancellations.

    If you're having a rough time factoring, ask yourself, what 2 numbers when multiplied equal -1 and when added equal -30

    How about -6 and 5

    (-6)(5)=-30 and -6+5=-1

    a^{2}-6a+5a-30

    (a^{2}-6a)+(5a-30)

    a(a-6)+5(a-6)

    (a+5)(a-6)

    Now, you finish?.
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  3. #3
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    Quote Originally Posted by galactus
    For number 2:

    It's just a division/cancellation problem.

    \frac{a^{2}-a-30}{a+5}

    Factor the numerator and you will quickly see the cancellations.

    If you're having a rough time factoring, ask yourself, what 2 numbers when multiplied equal -1 and when added equal -30

    How about -6 and 5

    (-6)(5)=-30 and -6+5=-1

    a^{2}-6a+5a-30

    (a^{2}-6a)+(5a-30)

    a(a-6)+5(a-6)

    (a+5)(a-6)

    Now, you finish?.
    how would I finish? I'm not trying to act stupid
    would it be A - 6 for the answer?
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  4. #4
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    You ought to go see your teacher if you're still lost from this point on.

    Just some friendly advice.

    Here's all is left:

    \frac{(a-6)(a+5)}{a+5}

    Now cancel what needs cancelled and that's it.
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  5. #5
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    Quote Originally Posted by galactus
    You ought to go see your teacher if you're still lost from this point on.

    Just some friendly advice.

    Here's all is left:

    \frac{(a-6)(a+5)}{a+5}

    Now cancel what needs cancelled and that's it.
    Its homeschool and its self-taught, I'm going into a normal school next year, I Just need these two problems and im finished with this grade
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  6. #6
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    Quote Originally Posted by happyboy

    how would I finish? I'm not trying to act stupid
    would it be A - 6 for the answer?

    Yes, you got it! . You updated your post while I was writing mine.

    Anytime you see a division problem set up like the one you have, rewrite it the way I did and see whether or not anything factors. If it does, they'll more than likely be some cancellations and you'll be on your way.
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  7. #7
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    Quote Originally Posted by galactus
    Yes, you got it! . You updated your post while I was writing mine.

    Anytime you see a division problem set up like the one you have, rewrite it the way I did and see whether or not anything factors. If it does, they'll more than likely be some cancellations and you'll be on your way.

    I just have one more problem and i'm done, I don't get how to do the other one at all tho
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  8. #8
    MHF Contributor Quick's Avatar
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    Quote Originally Posted by happyboy

    I just have one more problem and i'm done, I don't get how to do the other one at all tho
    I don't understand question #1 do you want us to factor 3x^2-2xy+2y^2?
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  9. #9
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    Quote Originally Posted by Quick
    I don't understand question #1 do you want us to factor 3x^2-2xy+2y^2?
    I don't understand it either, the x - y under it is apart of the problem too. and all it says is to multiply
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  10. #10
    MHF Contributor Quick's Avatar
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    Quote Originally Posted by happyboy
    I don't understand it either, the x - y under it is apart of the problem too. and all it says is to multiply
    So you want us to do \left(3x^2-2xy+2y^2\right)\times\left(x-y\right)?

    Alright, write it out...

    \left(x-y\right)\left(3x^2-2xy+2y^2\right)

    use the distributive property:
    \left(3x^2\left(x-y\right)-2xy\left(x-y\right)+2y^2\left(x-y\right)\right)

    use the distibutive property again: 3x^2\left(x\right)-3x^2\left(y\right)-2xy\left(x\right)+2xy\left(y\right)+2y^2\left(x<br />
\right)-2y^2\left(y\right)

    multiply: 3x^3-3x^2y-2x^2y+2y^2x+2y^2x-2y^3

    group like terms: 3x^3-(3x^2y+2x^2y)+(2y^2x+2y^2x)-2y^3

    add/subtract: \boxed{3x^3-5x^2y+4y^2x-2y^3}

    I think that's the answer you want (if I understand the question correctly)
    Last edited by Quick; July 13th 2006 at 12:15 PM.
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  11. #11
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    Quote Originally Posted by Quick
    So you want us to do \left(3x^2-2xy+2y^2\right)\times\left(x-y\right)?
    I'm guessing that is it, all the problem said was multiply and x-y was under it. this is a really confusing problem
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