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P1 Question..... watch

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    (Original post by Bhaal85)
    If the equation is x^3-19x-30, therefore:

    (X+2)(X^2-2X-15) which can be further factorised to give:

    (X+2)(X+3)(X-5)
    That is the answer, just done it. If you need someone to elaborate feel free to ask bono or Elpaw.
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    by algebraic juggling:


    (x³-19x-30)/(x+2)

    = (x²(x+2))/(x+2) -2x²/(x+2) -(19x+30)/(x+2)

    = x² - (2x(x+2))/(x+2) + 4x/(x+2) -(19x + 30)/(x+2)

    = x² - 2x - 15(x+2)/(x+2) + (30 - 30)/(x+2)

    = x² -2x -15

    = (x-5)(x+3)
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    (Original post by eastldn)
    it was on the P1 JAN 04 PAPER, 1ST QUESTION..... (x+2) is a factor
    f(x) = X^3 - 19x - 30
    f(-2) = 0
    im sure it can be factorised, bt how?
    you said f(x) = x^3 - 19x^2 - 30 originally. that can't be factorised by (x+2)
 
 
 
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