2010 AMC 12B Problems/Problem 21
Problem 21
Let , and let be a polynomial with integer coefficients such that
, and
.
What is the smallest possible value of ?
Solution
We observe that because , if we define a new polynomial such that , has roots when ; namely, when .
Thus since has roots when , we can factor the product out of to obtain a new polynomial such that .
Then, plugging in values of we get
Thus, the least value of must be the . Solving, we receive , so our answer is .
To complete the solution, we can let , and then try to find . We know from the above calculation that , and . Then we can let , getting . Let , then . Therefore, it is possible to choose , so the goal is accomplished. As a reference, the polynomial we get is
See also
2010 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 20 |
Followed by Problem 22 |
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