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Math Help - Solve for all possible values of x

  1. #1
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    Solve for all possible values of x

    Solve for all possible values of x Question1. x^3+3x^2-13x-15=0 Question 2. x^3-3x^2+4x-12=0
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  2. #2
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    Quote Originally Posted by oldschool999 View Post
    Solve for all possible values of x Question1. x^3+3x^2-13x-15=0 Question 2. x^3-3x^2+4x-12=0
    Just use the rational root theorem here, I get, x=-5,-1,3 for the first one.
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  3. #3
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    For the second just factor.
    x^2 \left( {x - 3} \right) + 4\left( {x - 3} \right) = \left( {x^2  + 4} \right)\left( {x - 3} \right) = 0.
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  4. #4
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    Quote Originally Posted by Plato View Post
    For the second just factor.
    x^2 \left( {x - 3} \right) + 4\left( {x - 3} \right) = \left( {x^2  + 4} \right)\left( {x - 3} \right) = 0.
    Is that the full answer?
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  5. #5
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    Quote Originally Posted by oldschool999 View Post
    Is that the full answer?
    No! You have to finish the solution.
    I, unlike some here, do not give out ready to turn in answers.
    Sorry, but I think you should do some work for yourself.
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  6. #6
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    Quote Originally Posted by Plato View Post
    No! You have to finish the solution.
    I, unlike some here, do not give out ready to turn in answers.
    Sorry, but I think you should do some work for yourself.
    Jeez, I was just asking.
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