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    math help

    factor completely

    16x2-2x-3
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    Quote Originally Posted by fancyface View Post
    factor completely

    16x2-2x-3
    (8x + 3)(2x - 1)
    I cannot really give a good explanation to how I factored. It was more of a guess.
    I just saw that 3 was prime and only has 2 factors: 3 and 1.
    Thus, they must appear in the end of the factor.
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    Quote Originally Posted by fancyface View Post
    factor completely

    16x2-2x-3
    Multiply the coefficient of the leading term by the constant term:
    16 x -3 = -48
    Now find two factors of -48 that add to the coefficient of the linear term: -2

    Factors......Sum
    1, -48........-47
    2, -24........-22
    3, -16........-13
    etc.

    I get that 6 x -8 = -48 and 6 + -8 = -2

    So:
    16x^2-2x-3 = 16x^2 + 6x - 8x - 3

     = (16x^2 + 6x) - (8x + 3)

     = 2x \cdot (8x + 3) - 1 \cdot (8x + 3)

     = (2x - 1)(8x + 3)

    If such a pair of factors cannot be found, the quadratic does not factor over the rational numbers.

    -Dan
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