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 ==
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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 00:08, 8 February 2009

Problem

The polynomial $p(x) = x^3+ax^2+bx+c$ has the property that the average of its zeros, the product of its zeros, and the sum of its coefficients are all equal. The $y$-intercept of the graph of $y=p(x)$ is 2. What is $b$?

$(\mathrm{A})\ -11 \qquad (\mathrm{B})\ -10 \qquad (\mathrm{C})\ -9 \qquad (\mathrm{D})\ 1 \qquad (\mathrm{E})\ 5$

Solution

We are given $c=2$. So the product of the roots is $-c = -2$ by Vieta's theorem. Vieta's also tells us $\frac{-a}{3}$ is the average of the zeros, so $\frac{-a}3=-2 \implies a = 6$. We are also given that the sum of the coefficients is -2, so $1+6+b+2 = -2 \implies b=-11$. So the answer is $\mathrm{A}$.

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