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Math Help - Factoring a polynomial/simplifying

  1. #1
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    Factoring a polynomial/simplifying

    I have an test tomorrow. I studied everything..but i know my teacher will give us hard questions he did not even solve in the class..

    how to factor this polynomial:
    4y^2 - 81

    i know that a^2 - b^2 = (a-b) (a+b)


    Simplify:
    (m^2 n ^3)^-2 (m^-3 n^2)^-3

    we learned that if we have (m^2 m^3)^2 then the answer is m^2 n^3 x m^2 n^3 but how about if we have a -ve square or -3?

    Simplify a radical:
    √8 + √18

    according to my teacher the solution is..
    = √(2x2x2) + √(3x3x3)
    =2√2 + 3√2
    =5√3

    how is that possible?

    Please help
    Thanks in advance
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  2. #2
    Super Member malaygoel's Avatar
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    Quote Originally Posted by Cherry View Post

    Simplify a radical:
    √8 + √18

    according to my teacher the solution is..
    = √(2x2x2) + √(3x3x3)
    =2√2 + 3√2
    =5√3

    how is that possible?
    the solution is
    \sqrt{8}+\sqrt{18}
    =\sqrt{2*2*2}+\sqrt{3*3*2}
    =2\sqrt{2}+3\sqrt{2}
    =5\sqrt{2}

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  3. #3
    Super Member malaygoel's Avatar
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    Quote Originally Posted by Cherry;28279
    how to factor this polynomial:
    [B
    4y^2 - 81[/B]

    i know that a^2 - b^2 = (a-b) (a+b)

    2y=a
    9=b

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  4. #4
    Super Member malaygoel's Avatar
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    Quote Originally Posted by Cherry View Post


    Simplify:
    (m^2 n ^3)^-2 (m^-3 n^2)^-3

    we learned that if we have (m^2 m^3)^2 then the answer is m^2 n^3 x m^2 n^3 but how about if we have a -ve square or -3?
    do you mean
    \frac{1}{({m^2n^3})^2}* \frac{1}{({m^{-3}}{n^2})^3}
    it is the same with negative powers
    (m^{-2})^2=m^{-2}*m^{-2}=m^{(-2)+(-2)}=m^{-4}
    hence your expression simplifies to
    \frac{1}{{m^4n^6}}* \frac{1}{{m^{-9}}{n^6}}
    =\frac{1}{{m^4m^{-9}}}* \frac{1}{{n^6n^6}}
    =\frac{1}{{m^{4+(-9)}}}* \frac{1}{{n^{6+6}}}
    =\frac{1}{{m^{-5}}}* \frac{1}{{n^{12}}}
    =\frac{m^{5}}{n^{12}}
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    Last edited by malaygoel; November 21st 2006 at 03:44 AM.
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  5. #5
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    Quote Originally Posted by malaygoel View Post
    do you mean
    \frac{1}{({m^2n^3})^2}* \frac{1}{({m^{-3}}{n^2})^3}

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    Nope.
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  6. #6
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    Quote Originally Posted by malaygoel View Post
    do you mean
    \frac{1}{({m^2n^3})^2}* \frac{1}{({m^{-3}}{n^2})^3}
    it is the same with negative powers
    (m^{-2})^2=m^{-2}*m^{-2}=m^{(-2)+(-2)}=m^{-4}
    hence your expression simplifies to
    \frac{1}{{m^4n^6}}* \frac{1}{{m^{-9}}{n^6}}
    =\frac{1}{{m^4m^{-9}}}* \frac{1}{{n^6n^6}}
    =\frac{1}{{m^{4+(-9)}}}* \frac{1}{{n^{6+6}}}
    =\frac{1}{{m^{-5}}}* \frac{1}{{n^{12}}}
    =\frac{m^{5}}{n^{12}}
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    Ahhh I see

    Thanks for helping me. I really appreciate it!
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