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Math Help - polynomials

  1. #1
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    polynomials

    1- divide using long division (9x^3-9x^2-x-12) / (3x-5)

    2- simplify 3/x^2-9 + 4/x-3

    3- given the polynomial x^3-5x^2+2x+8 where x-4 is a factor. find another factor

    4- simplify 4x^5y^-3 / 20x^-2y

    *****
    sorry if the math is messy, but I don't know the code to make it nice.
    Last edited by Leah<3; May 22nd 2007 at 08:28 PM.
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  2. #2
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    for #2, i got:

    3(x-3)/(x+3)(x-3)(x-3) + 4(x+3)(x-3)/(x-3)(x+3)

    3(x-3) + 4(x^2-9)/x^2-9(x^2-9)


    is that wrong?
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Leah<3 View Post
    1- divide using long division (9x^3-9x^2-x-12) / (3x-5)
    i don't know how to type up the long division, so try it and come up with an answer and i'll tell you if its right.


    2- simplify 3/x^2-9 + 4/x-3
    you don't have to know the codes (though they are not that hard, i started learning them just today, and i'm doing pretty well i think) but you can use parenthesis to simplify. is this what you mean?

    Simplify \frac {3}{x^2 - 9} + \frac {4}{x - 3}


    3- given the polynomial x^3-5x^2+2x+8 where x-4 is a factor. find another factor
    think of other factors of 8 and plug them into the formula, if the result is zero, then x - the thing you plugged in is a factor.


    4- simplify 4x^5y^-3 / 20x^-2y
    again, use parenthesis

    should it be, simplify \frac {4x^5 y^{-3}}{20x^{-2}y} ?
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  4. #4
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    Quote Originally Posted by Jhevon View Post
    i don't know how to type up the long division, so try it and come up with an answer and i'll tell you if its right.
    i'm not sure how to do it, that's why i asked...

    Quote Originally Posted by Jhevon View Post
    you don't have to know the codes (though they are not that hard, i started learning them just today, and i'm doing pretty well i think) but you can use parenthesis to simplify. is this what you mean?

    Simplify \frac {3}{x^2 - 9} + \frac {4}{x - 3}
    yes. i tried it and got \frac {3(x-3) + 4(x^2-9)}{x^2-9(x^2-9)}

    Quote Originally Posted by Jhevon View Post

    think of other factors of 8 and plug them into the formula, if the result is zero, then x - the thing you plugged in is a factor.
    ....huh?

    Quote Originally Posted by Jhevon View Post

    should it be, simplify \frac {4x^5 y^{-3}}{20x^{-2}y} ?
    yes. i started working on it but got thrown off because of the negatives.
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  5. #5
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    the codes take up so much time!!
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  6. #6
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    Quote Originally Posted by Jhevon View Post


    think of other factors of 8 and plug them into the formula, if the result is zero, then x - the thing you plugged in is a factor.
    hm.. x-2? although i did try it and got -2.
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  7. #7
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    Quote Originally Posted by Leah<3 View Post
    1- divide using long division (9x^3-9x^2-x-12) / (3x-5)

    2- simplify 3/x^2-9 + 4/x-3

    3- given the polynomial x^3-5x^2+2x+8 where x-4 is a factor. find another factor

    4- simplify 4x^5y^-3 / 20x^-2y

    ...
    Hello,

    to #1:
    Code:
    (9x - 9x - x - 12)  (3x - 5) = 3x + 2x + 3 
    (3x - 15x)
    -------------
            6x - x
            6x - 10x
            ---------
                  9x - 12
                  9x - 15
                  --------
                         3   ←  remainder
    to #2:
    \frac{3}{x^2-9} + \frac{4}{ x-3}=\frac{3}{ (x+3)(x-3)}+\frac{4(x+3)}{(x+3)(x-3)} = \frac{4x+15}{x^2-9}

    to #3:

    a) use long division
    b) use synthetic division
    Code:
          | 1   -5   2   8
          | 0   4   -4   -8
    ------|----------------
    x = 4 | 1   -1   -2   0
    That means:  x^3 -5x^2 + 2x + 8 = (x-4)(x^2 - x - 2)
    The last factor can be factorized again so that you finally have:
     x^3 -5x^2 + 2x + 8 = (x-4)(x- 2)(x+1)


    to #4:
    \frac{4x^5 y^{-3}}{20x^{-2} y} = \frac{1}{5} x^{5-(-2)} y^{-3 - 1} = \frac{1}{5} x^7 y^{-4} = \frac{x^7}{5y^4}
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  8. #8
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Leah<3 View Post
    1- divide using long division (9x^3-9x^2-x-12) / (3x-5)
    It's going to be kind of hard to explain this one. i'll do the problem by hand and try to scan a copy and hopefully you'll be able to figure it out from that.

    2- simplify 3/x^2-9 + 4/x-3
    \frac {3}{x^2 - 9} + \frac {4}{x - 3} = \frac {3}{(x + 3)(x - 3)} + \frac {4}{x - 3} ....the LCD is (x + 3)(x - 3)
    ....................... = \frac {3 + 4(x + 3)}{(x + 3)(x - 3)}
    ....................... = \frac {4x + 15}{(x + 3)(x - 3)}


    3- given the polynomial x^3-5x^2+2x+8 where x-4 is a factor. find another factor
    By the rational roots theorem, the possible rational roots are given by the factors of 8 are: \pm 1, \pm 2, \pm 4, \pm 8

    start with \pm 1 and start plugging in each of the factors of 8 into the formula, if we get zero as the result, we have a root.

    we realize that -1 works, since, (-1)^3 - 5(-1)^2 + 2(-1) + 8 = 0

    since x = -1 is a root, it means x + 1 is a factor by the factor theorem




    4- simplify 4x^5y^-3 / 20x^-2y
    \frac {4x^5y^{-3}}{20x^{-2}y} = \frac {4}{20} \cdot \frac {x^5}{x^{-2}} \cdot \frac {y^{-3}}{y}
    ............. = \frac {1}{5} \cdot x^{5 - (-2)} \cdot y^{-3 - 1}
    ............. = \frac {1}{5} \cdot x^{7} \cdot y^{-4}
    ............. = \frac {x^7}{5y^4}


    EDIT: Thanks a lot earboth. anyway, i guess it's my own fault for going around asking a lot of questions before doing the problem
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