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    How can I solve the following for \theta without using trial and improvement to find a factor and then factorising the resulting quadratic? :
    2 \tan ^3 \theta - 33 \tan^2 \theta - 6 \tan \theta + 11 = 0
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    let tan(theta) = x?
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    (Original post by Jingers)
    let tan(theta) = x?
    Yes, but how can I find the factors in order to factorise this.
    also please note, I am NOT allowed to use a calculator,
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    Anyone thought of a way to do it?
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    any divisor of the form (x-a) has a|11 (can you see why).

    this narrows down what you have to guess considerably.
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    Change tan(theta) to x:

    

2 \tan ^3 \theta - 33 \tan^2 \theta - 6 \tan \theta + 11 = 0

    

2x^3-33x^2-6x+11=0

    Now, since the constant is 11, there are only four possible factors, 1, -1, 11 or -11

    EDIT: Are you sure you've written the question properly. Because you couldn't possibly do this in a non-calculator paper. The answers are \tan\theta = -0.66, 0.5 or16.66
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    Let x = \tan(\theta).

    First things first, it's evident that the only integer factors of 11 (\pm 1 & 11) are not roots themselves. So...

    The coefficient of x^3 isn't one, so let's divide through to make it easier to handle, albeit more clumsy-looking:

    \displaystyle x^3 - \frac{33}{2}x^2 - 3x + \frac{11}{2} = 0

    Now we're dealing with fractions. As usual, any roots will be found as a factor of the constant term. Let's try some:

    To get \frac{11}{2}, what are the combinations you would intuitively try? Remember we are looking for only one of the roots...

    Spoiler:
    Show
    Try x=\frac{1}{2}, the most 'obvious' choice of factor for the constant term.
    Substituting into the expression gives:

    \displaystyle \left( \frac{1}{2} \right)^3 - \frac{33}{2} \left( \frac{1}{2} \right)^2 - 3\left( \frac{1}{2} \right) + \frac{11}{2} = 0

    Hey presto. You can now reduce it to a quadratic, and use the formula to factorise that.


    That's how I'd approach it, anyway. :yep: Can't see an easier way at the moment. :no:

    What a horrible question. :p:
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    (Original post by james.h)
    Let x = \tan(\theta).

    First things first, it's evident that the only integer factors of 11 (\pm 1 & 11) are not roots themselves. So...

    The coefficient of x^3 isn't one, so let's divide through to make it easier to handle, albeit more clumsy-looking:

    \displaystyle x^3 - \frac{33}{2}x^2 - 3x + \frac{11}{2} = 0

    Now we're dealing with fractions. As usual, any roots will be found as a factor of the constant term. Let's try some:

    To get \frac{11}{2}, what are the combinations you would intuitively try? Remember we are looking for only one of the roots...

    Spoiler:
    Show
    Try x=\frac{1}{2}, the most 'obvious' choice of factor for the constant term.
    Substituting into the expression gives:

    \displaystyle \left( \frac{1}{2} \right)^3 - \frac{33}{2} \left( \frac{1}{2} \right)^2 - 3\left( \frac{1}{2} \right) + \frac{11}{2} = 0

    Hey presto. You can now reduce it to a quadratic, and use the formula to factorise that.


    That's how I'd approach it, anyway. :yep: Can't see an easier way at the moment. :no:

    What a horrible question. :p:
    Wow thanks for the help.
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    (Original post by Farhan.Hanif93)
    Wow thanks for the help.


    Don't know whether that'd work in all situations, though, so don't count on it. :p:
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    (Original post by innerhollow)
    Change tan(theta) to x:

    

2 \tan ^3 \theta - 33 \tan^2 \theta - 6 \tan \theta + 11 = 0

    

2x^3-33x^2-6x+11=0

    Now, since the constant is 11, there are only four possible factors, 1, -1, 11 or -11

    EDIT: Are you sure you've written the question properly. Because you couldn't possibly do this in a non-calculator paper. The answers are \tan\theta = -0.66, 0.5 or16.66
    Haha it's part of a STEP question so I'm not surprised it *seems* impossible. But the fact of the matter is that after finding x=0.5, I can get a quadratic to factorise and the formula does that for me
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    (Original post by Farhan.Hanif93)
    How can I solve the following for \theta without using trial and improvement to find a factor and then factorising the resulting quadratic? :
    2 \tan ^3 \theta - 33 \tan^2 \theta - 6 \tan \theta + 11 = 0

    This cubic rings a bell. Does it by any chance emerge from a STEP question where one of the roots is suggested by the earlier part of the question?
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    I think I have it. When asked as STEP I 2005 Q4, you have to guess the root by noticing that f(0) and f(1) have different signs.

    But anyone who tackled STEP I 2001 Q4 was given a big hint in the earlier part of the question, which 2005 candidates might have remembered.
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    That would make a lot more sense. :yep:
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    (Original post by around)
    any divisor of the form (x-a) has a|11 (can you see why).
    This is somewhat incomplete: http://en.wikipedia.org/wiki/Rational_root_theorem

    I'll also note that "can you see why" is perhaps optimistic - I wouldn't say it's obvious.
 
 
 
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