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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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Generally speaking, you want the right hand side to be zero. By factoring a polynomial, you're looking to write it as $(x-a)(x-b)(x-c) = 0.$ Notice that there will be a term there $-abc$ that is not a power of $x$. So, don't move the 8 over -- it won't help. Cubics are hard to factor. You might not be able to find $a,b,c$ such that they are real numbers. There is a general formula, but it is quite complicated. I'm not sure this problem can be easily solved with simple operations. –  Arkamis Aug 21 '12 at 13:53
    
it would easy if you had +8 instead of having -8. Otherwise, Ross's answer is the final one. –  B. S. Aug 21 '12 at 13:54
    
2  
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. –  Mark Bennet Aug 21 '12 at 13:56
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3 Answers

up vote 3 down vote accepted

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 $-2/3$

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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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I didn't actually. It was + 8 instead of - 8. –  Andre Oseguera Aug 21 '12 at 13:54
    
@AndreOseguera: You wrote $-8$ in your primary question. Please take more attention writing the question here. :) –  B. S. Aug 21 '12 at 14:18
    
Yeah, sorry, I wrote it thinking I moved it over to the other side already and got all backwards. Won't happen again. :) –  Andre Oseguera Aug 21 '12 at 14:20
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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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For formatting, you could look at meta.math.stackexchange.com/questions/1773/… Basically you put $\LaTeX$ between dollar signs. I'm not sure what this answer contributes beyond what jasoncube wrote. –  Ross Millikan Aug 21 '12 at 14:36
    
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... –  SKT Aug 21 '12 at 14:43
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