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Math Help - Need help on Pre Calc Problem...Factoring

  1. #1
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    Need help on Pre Calc Problem...Factoring

    factor the polynomial as a product of linear and a factor of a constant or an irreducible polynimial


    6X^3-5X^2+3X-1


    Thanks guys
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  2. #2
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    Quote Originally Posted by victorfk06 View Post
    factor the polynomial as a product of linear and a factor of a constant or an irreducible polynimial


    6X^3-5X^2+3X-1


    Thanks guys
    i dont really have a good way of explaining how i factor, its just a run through of factors in my head to see what guesses might make sense...

    (3x^2-x+1)(2x-1)
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  3. #3
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    Hello, victorfk06!

    Factor the polynomial: . 6x^3-5x^2+3x-1
    Are you familiar with the Factor Theorem?

    Given a polynomial p(x), if p(a) = 0, then (x-a) is a factor of p(x).


    And the Rational Roots Theorem?

    A rational root of a polynomial equation is of the form: . \frac{n}{d}
    . . where n is a factor of the constant term,
    . . and d is a factor of the leading coefficient.


    We want a number which makes the polynomial equal zero.

    There are eight candidates.
    . . The numerator is a factor of 1: . \pm1
    . . The denominator is a factor of 6: . \pm1,\:\pm2,\:\pm3,\:\pm6
    The possible rational roots are: . \pm1,\;\pm\tfrac{1}{2},\:\pm\tfrac{1}{3},\:\pm\tfr  ac{1}{6}

    And we find that x = \tfrac{1}{2} works: .   6\left(\tfrac{1}{2}\right)^3 - 5\left(\tfrac{1}{2}\right)^2 + 3\left(\tfrac{1}{2}\right) - 1 \;=\;0

    Hence, \left(x-\tfrac{1}{2}\right)\;\hdots\;\text{ or }(2x-1) is a factor.


    Using long division, we find that:

    . . 6x^3 - 5x^2 + 3x - 1 \;=\;(2x-1)(3x^2-x+1) . . . as emathinstruction already pointed out.


    We find that the quadratic expression doesn't factor . . . so we're done!

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