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I have this problem:

$9x^3 - 18x^2 - 4x + 8 = 0$

However, I'm not sure how to find the values of $x$. I moved the 8 over and factor out an $x$, but the trinomial it created can't be factored. Could someone enlighten me?

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    You need the rational root theorem - there is one obvious (integer) root, and taking out this factor gives you a quadratic which is easy to factor.2012-08-21

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By trial and error method I'm solving this

Divide 9x3−18x2−4x+8 by x-2 reminder is 0 and quotient is 9x2-4x

x-2 = 0 and 9x2-4x = 0

Factorizing 9x2-4x = 0, we get (3x-2)(3x+2) = 0

ie. x = 2, x = 2/3, x = -2/3 are the solutions

(Sorry I'm new to this group and don't know how to write Square of a variable, But answer must be this)

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    Actually I didn't see the answer jason posted. I saw the question and just started solving it. Sorry if it was a duplicate. But thankyou for the link...2012-08-21
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The rational root theorem is your friend. It says all rational roots have numerators that are factors of the constant term and denominators that are factors of the leading term. Here the numerators can be $\pm 1, \pm 2, \pm 4, \pm 8$ and the denominators can be $1,3,9$. Not too many to try. When you find one, you can divide out that root to get a quadratic. It doesn't always work (as shown with the example with -8 for a constant term) but does sometimes, often in homework.

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    Yeah, sorry, I wrote it thinking I moved it over to the other side already and got all backwards. Won't happen again. :)2012-08-21
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Factor the function.

$9x^3-18x^2-4x+8=9x^2(x-2)-4(x-2)=(x-2)(9x^2-4)=(x-2)(3x-2)(3x+2)$

$(x-2)(3x-2)(3x+2)=0$

$x=2$ or $2/3$ or $-2/3$