Difference between revisions of "2003 AMC 12B Problems/Problem 20"

Revision as of 17:12, 7 April 2014

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 $b = -an = -2 \Rightarrow \mathrm{(B)}$ .

2003 AMC 12B (Problems • Answer Key • Resources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 23: / Line 23: @@
 Two of the roots of <math>f(x) = 0</math> are <math>\pm 1</math>, and we let the third one be <math>n</math>. Then
 <cmath>a(x-1)(x+1)(x-n) = ax^3-anx^2-ax+an = ax^3 + bx^2 + cx + d = 0</cmath>
-Notice that <math>f(0) = d = an = 2</math>, so <math>b = -an = -2</math>.
+Notice that <math>f(0) = d = an = 2</math>, so <math>b = -an = -2 \Rightarrow \mathrm{(B)}</math>.
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