Difference between revisions of "AoPS Wiki talk:Problem of the Day/July 5, 2011"

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{{:AoPSWiki:Problem of the Day/July 5, 2011}}
 
{{:AoPSWiki:Problem of the Day/July 5, 2011}}
 
==Solution==
 
==Solution==
{{potd_solution}}
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Try the only possible integer roots which are <math>\pm</math> {1,2,3,6,7,14,21,42}
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3 works so divide out (x-3) to get <math>x^2 + 3x + 14</math>
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Use the quadratic formula to find the remaining roots to be <math>\frac{-3 \pm i\sqrt{47}}{2}</math>
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So, the answers are: <math> \boxed {3; \frac{-3 \pm i\sqrt{47}}{2}} </math>.

Latest revision as of 15:48, 5 July 2011

Problem

AoPSWiki:Problem of the Day/July 5, 2011

Solution

Try the only possible integer roots which are $\pm$ {1,2,3,6,7,14,21,42}

3 works so divide out (x-3) to get $x^2 + 3x + 14$

Use the quadratic formula to find the remaining roots to be $\frac{-3 \pm i\sqrt{47}}{2}$

So, the answers are: $\boxed {3; \frac{-3 \pm i\sqrt{47}}{2}}$.