Difference between revisions of "2007 Alabama ARML TST Problems/Problem 6"

Latest revision as of 00:12, 17 June 2008

If $r$ is a root of $x^2+x+6$ , then compute the value of

$r^3+2r^2+7r+17.$

$r$ satisfies $r^2+r+6=0$ . Thus, $r^3+2r^2+7r+17 = (r^3+r^2+6r)+(r^2+r+6)+11 =$ $(r+1)(r^2+r+6)+11 = (r+1)(0)+11 = 11$ .

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 ==Solution==
-Note that <math>r^2+r+6=0</math>. Thus, <math>r^3+2r^2+7r+17 = (r^3+r^2+6r)+(r^2+r+6)+11 =</math> <math>(r+1)(r^2+r+6)+11 = (r+1)(0)+11 = 11</math>.
+<math>r</math> satisfies <math>r^2+r+6=0</math>. Thus, <math>r^3+2r^2+7r+17 = (r^3+r^2+6r)+(r^2+r+6)+11 =</math> <math>(r+1)(r^2+r+6)+11 = (r+1)(0)+11 = 11</math>.
 ==See also==
 {{ARML box|year=2007|state=Alabama|num-b=5|num-a=7}}