2003 AMC 12B Problems/Problem 20

Revision as of 18:12, 7 April 2014 by ZeBirdeh (talk | contribs) (Solution 2)

Problem

Part of the graph of $f(x) = ax^3 + bx^2 + cx + d$ is shown. What is $b$?

2003 12B AMC-20.png

$\mathrm{(A)}\ -4 \qquad\mathrm{(B)}\ -2 \qquad\mathrm{(C)}\ 0 \qquad\mathrm{(D)}\ 2 \qquad\mathrm{(E)}\ 4$

Solution

Solution 1

Since \begin{align*} -f(-1) = a - b + c - d = 0 = f(1) = a + b + c + d \end{align*}

It follows that $b + d = 0$. Also, $d = f(0) = 2$, so $b = -2 \Rightarrow \mathrm{(B)}$.

Solution 2

Two of the roots of $f(x) = 0$ are $\pm 1$, and we let the third one be $n$. Then \[a(x-1)(x+1)(x-n) = ax^3-anx^2-ax+an = ax^3 + bx^2 + cx + d = 0\] Notice that $f(0) = d = an = 2$, so $b = -an = -2 \Rightarrow \mathrm{(B)}$.

See also

2003 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 19
Followed by
Problem 21
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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