Difference between revisions of "2002 AMC 10P Problems/Problem 11"
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== Solution 1== | == Solution 1== | ||
− | By the factor theorem, <math>x-2</math> is a factor of <math>P(x)</math> if and only if <math>P(2)=0.</math> Therefore, <math>x</math> must equal <math>2.</math> <math>P(2)=0=2^3k+2(2^2)k^2+k^3, which simplifies to < | + | By the [[Factor Theorem | factor theorem]], <math>x-2</math> is a factor of <math>P(x)</math> if and only if <math>P(2)=0.</math> Therefore, <math>x</math> must equal <math>2.</math> <math>P(2)=0=2^3k+2(2^2)k^2+k^3,</math> which simplifies to <math>k(k^2+8k+8)=0.</math> <math>k=0</math> is a trivial real <math>0.</math> Since <math>8^2 -4(1)(8)=32 > 0,</math> this polynomial does indeed have two real zeros, meaning we can use Vieta’s to conclude that sum of the other two roots are <math>-8.</math> |
− | Thus, our answer is < | + | Thus, our answer is <math>0-8=\boxed{\textbf{(A)}\ -8}.</math> |
== See also == | == See also == | ||
{{AMC10 box|year=2002|ab=P|num-b=10|num-a=12}} | {{AMC10 box|year=2002|ab=P|num-b=10|num-a=12}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 06:42, 15 July 2024
Problem
Let Find the sum of all real numbers for which is a factor of
Solution 1
By the factor theorem, is a factor of if and only if Therefore, must equal which simplifies to is a trivial real Since this polynomial does indeed have two real zeros, meaning we can use Vieta’s to conclude that sum of the other two roots are
Thus, our answer is
See also
2002 AMC 10P (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
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