Difference between revisions of "2001 AMC 12 Problems/Problem 19"
(New page: == Problem == The polynomial <math>p(x) = x^3+ax^2+bx+c</math> has the property that the average of its zeros, the product of its zeros, and the sum of its coefficients are all equal. The ...) |
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<math>(\mathrm{A})\ -11 \qquad (\mathrm{B})\ -10 \qquad (\mathrm{C})\ -9 \qquad (\mathrm{D})\ 1 \qquad (\mathrm{E})\ 5</math> | <math>(\mathrm{A})\ -11 \qquad (\mathrm{B})\ -10 \qquad (\mathrm{C})\ -9 \qquad (\mathrm{D})\ 1 \qquad (\mathrm{E})\ 5</math> | ||
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| + | == Solution == | ||
| + | We are given <math>c=2</math>. So the product of the roots is <math>-c = -2</math> by Vieta's theorem. Vieta's also tells us <math>\frac{-a}{3}</math> is the average of the zeros, so <math>\frac{-a}3=-2 \implies a = 6</math>. We are also given that the sum of the coefficients is -2, so <math>1+6+b+2 = -2 \implies b=-11</math>. So the answer is <math>\mathrm{A}</math>. | ||
Revision as of 23:08, 7 February 2009
Problem
The polynomial 
has the property that the average of its zeros, the product of its zeros, and the sum of its coefficients are all equal. The 
-intercept of the graph of 
is 2. What is 
?
Solution
We are given 
. So the product of the roots is 
by Vieta's theorem. Vieta's also tells us 
is the average of the zeros, so 
. We are also given that the sum of the coefficients is -2, so 
. So the answer is 
.
