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Difference between revisions of "2001 AMC 12 Problems/Problem 19"

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== Solution ==
 
== Solution ==
 
We are given <math>c=2</math>. So the product of the roots is <math>-c = -2</math> by [[Vieta's formulas]]. These also tell us that <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 <math>-2</math>, so <math>1+6+b+2 = -2 \implies b=-11</math>. So the answer is <math>\fbox{A}</math>.
 
We are given <math>c=2</math>. So the product of the roots is <math>-c = -2</math> by [[Vieta's formulas]]. These also tell us that <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 <math>-2</math>, so <math>1+6+b+2 = -2 \implies b=-11</math>. So the answer is <math>\fbox{A}</math>.
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== Video Solution by OmegaLearn ==
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https://youtu.be/czk6OsfrbxQ?t=678
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 +
~ pi_is_3.14
  
 
== Video Solution ==
 
== Video Solution ==

Revision as of 06:37, 4 November 2022

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 formulas. These also tell us that $\frac{-a}{3}$ 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 $\fbox{A}$.


Video Solution by OmegaLearn

https://youtu.be/czk6OsfrbxQ?t=678

~ pi_is_3.14

Video Solution

https://www.youtube.com/watch?v=qtlXAxj4y2Y

See Also

2001 AMC 12 (Problems • Answer Key • Resources)
Preceded by
Problem 18 		Followed by
Problem 20
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 12 Problems and Solutions

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