2001 AMC 12 Problem 19

Below is the professionally curated solution for Problem 19 of the 2001 AMC 12, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2001 AMC 12 solutions, or check the answer key.

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Concepts:Vieta’s Formulas polynomial

Difficulty rating: 1760

19.

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

$-11$

$-10$

$-9$

$1$

$5$

Solution:

The $y$ -intercept is $P(0) = c = 2.$ By Vieta's formulas the product of the zeros is $-c = -2,$ the mean of the zeros is $-\dfrac{a}{3},$ and the sum of the coefficients is $P(1) = 1 + a + b + c.$

All three are equal to $-2.$ From $-\dfrac{a}{3} = -2$ we get $a = 6.$

Then $1 + a + b + c = -2$ becomes $1 + 6 + b + 2 = -2,$ so $b = -11.$

Thus, the correct answer is A.

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