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,</math> which simplifies to <math>k(k^2+8k+8)=0. 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> | + | 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 <math>\boxed{\textbf{(A)}\ -8}.</math> | + | 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
