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Math Help - factorising query

  1. #1
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    factorising query

    factorise: 13x^2 + 14x - 27 = 0

    (x-27)(13x+1) is this correct???
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by x-disturbed-x
    factorise: 13x^2 + 14x - 27 = 0

    (x-27)(13x+1) is this correct???
    (x-27)(13x+1) \ \  \ = \ x(13x+1) -27(13x+1)
     =\ 13x^2+x\ -351x-27
     =\ 13x^2 -350x-27

    So the answer is no.

    RonL
    Last edited by CaptainBlack; November 12th 2005 at 09:52 AM. Reason: stray ,
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  3. #3
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    Quote Originally Posted by x-disturbed-x
    factorise: 13x^2 + 14x - 27 = 0

    (x-27)(13x+1) is this correct???
    If you can't factor it just by spotting the zeroes, how else might you find the zeroes and put them in factored form?
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  4. #4
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    Re: 13x^2 + 14x -27 = 0

    The quadratic formula yields the roots of this degree 2 polynomial over the integers.

    x= +/- square root [14^2 - 4 (-27)13] - 14 / 13x2

    x= +/- 40 - 14 / 26

    x= 1 or x = -27/13

    Hence the polynomial factorises into (x-1)(x+ 27/13)=0
    or (x-1)(13x + 27)=0.

    QED (used for proofs but you get the picture).
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