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-   -   Factoring in pre cal (https://www.askmehelpdesk.com/showthread.php?t=141272)

  • Oct 16, 2007, 07:19 AM
    robnfran5
    factoring in pre cal
    How do you factor ; x^4-7x^3+12x^2+4x-16
  • Oct 16, 2007, 07:54 AM
    terryg752
    Fundemental principle: If a polynomial is 0 when x = k, then (x - k) is a factor!

    Trial & error shows that the polynomial = 0 when x = 2

    Hence (x - 2) must be a factor

    So write as

    x^4 - 2 x^3 -5x^3 + 10x^2 + 2x^2 - 4x + 8x - 16

    = X^3 (x-2) - 5x^2(x -2) + 2x (x-2) +8 (x - 2)

    = (x -2) (X^3 - 5x^2 +2x + 8)

    Again X^3 - 5x^2 +2x + 8 = 0 when x = 4, hence it hax (x - 4) as factor!

    X^3 - 5x^2 +2x + 8 = x^3 - 4x^2 -x^2 + 4x - 2x + 8
    = x^2 (x - 4) - x (x - 4) -2 (x - 4)

    = (x^2 -x -2) (x-4)

    = (x -2) (x +1) (x -4)

    Hence the answer is (x -2) (x - 2) (x +1) (x - 4)
  • Oct 16, 2007, 08:49 AM
    robnfran5
    Thanks You, but I'm sorry I still don't understand:confused:
  • Oct 16, 2007, 08:57 PM
    terryg752
    Tell me which part you don't understand

    Terry

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