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Math Help - f(x) again :)

  1. #1
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    f(x) again :)

    f(x)=x^3+9x^2+20x+12

    divide f(x) by x=6

    thanks in advance
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  2. #2
    A riddle wrapped in an enigma
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    Quote Originally Posted by coyoteflare View Post
    f(x)=x^3+9x^2+20x+12

    divide f(x) by x + 6

    thanks in advance
    Hello coyote,

    Use synthetic division:

    Code:
     
     
    -6 | 1  9  20  12
           -6 -18 -12
        --------------
         1  3   2   0
    (x^3+9x^2+20x+12) \div (x+6)=x^2+3x+2

    Sorry coyote. I should've asked if you knew synthetic division.

    Code:
            x^2 + 3x   + 2
          __________________________
    x + 6 | x^3 + 9x^2 + 20x + 12
          -(x^3 + 6x^2)
           --------------
                  3x^2 + 20x
                -(3x^2 + 18x)
                  ------------
                          2x + 12
                        -(2x + 12)
                          =======
                                0
    Last edited by masters; January 8th 2009 at 11:43 AM.
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  3. #3
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    is this the only way of doing it because i dont know synthetic division
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  4. #4
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    No, you can just use "regular" division:
    x+6) x^3+ 9x^2+ 20x+ 12

    x divides into x^3 x^2 times so the quotient starts "x^2+ ..."
    Multiplying x^2 by x+ 6 gives
    x^3+ 9x^2+ 20x+ 12
    x^3+ 6x^2
    subtracting: 3x^2+ 20x+ 12

    Now x divides into 3x^2 3x times so we have "x^2+ 3x+ ..."
    Multiplying 3x by x+ 6 gives
    3x^2+ 20x+ 12
    3x^2+ 18x
    subtracting: 2x+ 12

    Finally, x divides into 2x twice so we have "x^2+ 15x+ 2"
    Multiplying 2 by x+ 6,
    2x+ 12
    2x+ 12
    which has remainder 0.
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  5. #5
    A riddle wrapped in an enigma
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    Quote Originally Posted by coyoteflare View Post
    is this the only way of doing it because i dont know synthetic division
    Coyote, I've edited my post to show the long division.
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  6. #6
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    no problem

    Quote Originally Posted by masters View Post
    Coyote, I've edited my post to show the long division.
    i thanked you also as yours was explained clearer although ive got the hang of it now anyway thanks though, i would like to learn synthetic division but not until after my exams if i can get by without it xD
    thanks again
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