2001 AMC 12 Problems/Problem 19

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 $\fbox{A}$ .

Solution 2 (inefficient)

If you don't know Vieta's formulas, you can go through the desperate route. Let $p(x) = (x-r_1)(x-r_2)(x-r_3)$ . After distributing, we find that $p(x) = x^3 - (r_1+r_2+r_3)x^2 + (r_1r_2+r_2r_3+r_1r_3)x - r_1r_2r_3$ . We know that the last term ( $c$ ) is $2$ , so $r_1r_2r_3$ must be $-2$ . We know from our equation that $b = r_1r_2+r_2r_3+r_1r_3$ . The problem gives that the average of the roots is equal to their product (their product is $-2$ ), so the sum of the roots is hence $-2\cdot3 = -6$ (giving us $a = 6$ , since a is the sum of the roots times $-1$ ). Finally, we know that the sum of the coefficients is the same as the product of the roots, so the sum of the coefficients is $-2$ . Thus $1 + a + b + c = -2$ , so $1 + 6 + b + 2 = -2$ , so $b + 9 = -2$ and $b = -11$ . The answer is then $\fbox{A}$ .

Video Solution by OmegaLearn

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

~ pi_is_3.14

Video Solution

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

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

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