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anonymous
 5 years ago
x^334x12 find the solution set and an actual root
anonymous
 5 years ago
x^334x12 find the solution set and an actual root

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0There'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 coprime (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\[(x6)\]is a factor of your cubic.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Now that you have one factor of your cubic, you can use synthetic or polynomial division to find the quadratic, Q(x), where\[x^334x12=(x6)Q(x)\]That Q(x) is found to be,\[Q(x)=\frac{x^334x12}{x6}=x^2+6x+2\]That is,\[x^324x12=(x6)(x^2+6x+2)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You can find the solutions of the quadratic using the quadratic formula or by completing the square. Using the quadratic formula gives\[x=\frac{6 \pm \sqrt{6^24.1.2}}{2.1}=\frac{6\pm \sqrt{28}}{2}=3 \pm \sqrt{7}\]The solution set is therefore,\[\left\{ 6, 3+\sqrt{7},3\sqrt{7} \right\}\]
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