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Math Help - Solving trinomial

  1. #1
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    Solving trinomial

    Help!

    Solve algebraically: 10x^3+x^2-69x+54 = 0

    How do i solve this?
    How do i find one root so i can find the other 2?

    Thanks so much for any help,

    mike
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  2. #2
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    Hello, Mike!

    Solve algebraically: . 10x^3+x^2-69x+54 \:= \:0
    We have the Rational Roots Theorem.

    If a polynomial P(x) has a rational root, it 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.

    The constant term is 54 with factors: . \pm1,\:\pm2,\:\pm3,\:\pm6,\:\pm9,\:\pm18,\:\pm27,\  :\pm54

    The leading coefficient is 10 with factors: . \pm1,\:\pm2,\:\pm5,\:\pm10

    That's a lot of fractions to test, but we may get lucky . . .


    Try P(1) \:=\:10\!\cdot\!1^3 + 1^2 = 60\!\cdot\!1 + 54 \:=\:-4 . . . no

    Try P(2) \;=\;10\!\cdot\!2^3 + 2^2 - 69\!\cdot\!2 + 54 \:=\:0 . . . YES!


    Since x = 2 is a zero of P(x), then (x-2) is a factor.

    Using long division, we have: . 10x^3 + x^2 - 69x + 54 \;=\;(x-2)(10x^2 + 21x - 27)

    . . And we find that the quadratic also factors.


    Therefore: . (x-2)(x+3)(10x-9) \;=\;0\quad\Rightarrow\quad x \;=\;2,\,-3,\,\frac{9}{10}

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  3. #3
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    We can also try a little factoring.

    Rewrite as

    10x^{3}-9x^{2}+10x^{2}-9x-60x+54

    Now group:

    (10x^{3}+10x^{2}-60x)-(9x^{2}+9x-54)

    Now, factor:

    10x(x^{2}+x-6)-9(x^{2}+x-6)

    (10x-9)(x^{2}+x-6)

    But the quadratic factor further:

    (10x-9)(x-2)(x+3)

    And you can easily see the roots now.
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  4. #4
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    Nice. Thanks! that's a lot of help.
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