2003 AMC 12B Problems/Problem 20

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Problem

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

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

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 $b = -an = -2$ .

2003 AMC 12B (Problems)
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

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