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Math Help - How to solve inequalities: (cubic, polynomial)?

  1. #1
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    How to solve inequalities: (cubic, polynomial)?

    h3 - 40h2 + 375h - 1000 > 0 (h3 is h to the power of 3) - (h2 is h to the power of 2)

    I know the general method - finding factors of 1000 and substituting them instead of h (factor theorem) - but the numbers are way too high. Is there any other method or do I just have to keep on substituting numbers?

    (x2 + x - 12) / (x2+5x+6) < 0

    (x+1)(x-2)(2x) >= 0 (2x is 2 to the power of x
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  2. #2
    Super Member TheChaz's Avatar
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    Re: How to solve inequalities: (cubic, polynomial)?

    Quote Originally Posted by Sorena View Post
    h^3 - 40h^2 + 375h - 1000 ...

    I know the general method - finding factors of 1000 and substituting them instead of h (factor theorem) - but the numbers are way too high. Is there any other method or do I just have to keep on substituting numbers?...
    First, this isn't an inequality...!
    But to factor this, we can use synthetic division.
    Have you learned that?
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  3. #3
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    Re: How to solve inequalities: (cubic, polynomial)?

    yea my bad it's > 0. And how do u solve this usiong synthenic division?
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    Re: How to solve inequalities: (cubic, polynomial)?

    Quote Originally Posted by Sorena View Post

    I know the general method - finding factors of 1000 and substituting them instead of h (factor theorem) - but the numbers are way too high. Is there any other method or do I just have to keep on substituting numbers?
    There are other numeric methods like newton's method. I would suggest using technology (spreadsheet or graphical calculator) to find one linear factor and then use polynomial long division to find the others.


    Quote Originally Posted by Sorena View Post


    (x2 + x - 12) / (x2+5x+6) < 0

    (x+1)(x-2)(2x) >= 0 (2x is 2 to the power of x
    2^x assumes you mean 2^x if you can't use latex.

    Are you trying to solve for x here? What have you tried? What do you know about inequalities?
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  5. #5
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    Re: How to solve inequalities: (cubic, polynomial)?

    Quote Originally Posted by pickslides View Post
    There are other numeric methods like newton's method. I would suggest using technology (spreadsheet or graphical calculator) to find one linear factor and then use polynomial long division to find the others.




    2^x assumes you mean 2^x if you can't use latex.

    Are you trying to solve for x here? What have you tried? What do you know about inequalities?
    The question is: "Solve the inequality (x+1)(x-2)(2^x) >= 0 Algebraically."
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  6. #6
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    Re: How to solve inequalities: (cubic, polynomial)?

    Quote Originally Posted by Sorena View Post
    The question is: "Solve the inequality (x+1)(x-2)(2^x) >= 0 Algebraically."
    First of all, \displaystyle 2^x > 0 for all \displaystyle x, so dividing both sides by it won't change the inequality sign.

    \displaystyle \begin{align*} (x + 1)(x - 2)2^x &\geq 0 \\ (x + 1)(x - 2) &\geq 0 \\ x^2 - x - 2 &\geq 0 \\ x^2 - x + \left(-\frac{1}{2}\right)^2 - \left(-\frac{1}{2}\right)^2 - 2 &\geq 0 \\ \left(x - \frac{1}{2}\right)^2 - \frac{1}{4} - \frac{8}{4} &\geq 0 \\ \left(x - \frac{1}{2}\right)^2 &\geq \frac{9}{4} \\ \left|x - \frac{1}{2}\right| &\geq \frac{3}{2} \\ x - \frac{1}{2} \leq -\frac{3}{2} \textrm{ or }x - \frac{1}{2} &\geq \frac{3}{2} \\ x \leq -1 \textrm{ or }x &\geq 2  \end{align*}
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