Difference between revisions of "2002 AMC 10P Problems/Problem 11"
(→Solution 1) |
(→Solution 1) |
||
| Line 16: | Line 16: | ||
== 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, | + | 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 </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> |
| − | Thus, our answer is | + | Thus, our answer is </math>\boxed{\textbf{(A)}\ -8}. |
== 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}} | ||
Revision as of 21:10, 14 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 
k(k^2+8k+8)=0. 0
0.
8^2 -4(1)(8)=32 > 0,
-8.
\boxed{\textbf{(A)}\ -8}.
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. 
