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anonymous

  • 5 years ago

x^3-34x-12 find the solution set and an actual root

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  1. anonymous
    • 5 years ago
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    There's a theorem called The Rational Root Theorem that says, if a polynomial is to have rational roots, then they will be of the form,\[x=\frac{p}{q}\]where p and q are co-prime (i.e., p and q have no common factors other than 1), and p is a plus/minus factor of the constant (here, -12) and q is a plus/minus factor of the coefficient of the highest powered x (here, +1). So here, the +/- factors of 12 are\[\pm \left\{ 1,2,3,4,6,12 \right\}\]and the +/- factor of 1 is, funnily enough,\[\pm \left\{ 1 \right\}\]So if your polynomial has rational roots, they will be from the set of all combinations of \[\frac{\pm 1,2,3,4,6,12}{\pm 1}\]If you try +6, you'll find your polynomial ends up being zero. This means\[(x-6)\]is a factor of your cubic.

  2. anonymous
    • 5 years ago
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    Now that you have one factor of your cubic, you can use synthetic or polynomial division to find the quadratic, Q(x), where\[x^3-34x-12=(x-6)Q(x)\]That Q(x) is found to be,\[Q(x)=\frac{x^3-34x-12}{x-6}=x^2+6x+2\]That is,\[x^3-24x-12=(x-6)(x^2+6x+2)\]

  3. anonymous
    • 5 years ago
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    You can find the solutions of the quadratic using the quadratic formula or by completing the square. Using the quadratic formula gives\[x=\frac{-6 \pm \sqrt{6^2-4.1.2}}{2.1}=\frac{-6\pm \sqrt{28}}{2}=-3 \pm \sqrt{7}\]The solution set is therefore,\[\left\{ 6, -3+\sqrt{7},-3-\sqrt{7} \right\}\]

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