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

    If a polynomial $\displaystyle P(x)$ has a rational root, it is of the form $\displaystyle \frac{n}{d}$
    . . where $\displaystyle n$ is a factor of the constant term
    . . and $\displaystyle d$ is a factor of the leading coefficient.

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

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

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


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

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


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

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

    . . And we find that the quadratic also factors.


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

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

    Rewrite as

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

    Now group:

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

    Now, factor:

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

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

    But the quadratic factor further:

    $\displaystyle (10x-9)(x-2)(x+3)$

    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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