AMC12 2003 B

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 $b = -2 \Rightarrow \mathrm{(B)}$. 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 $b = -an = -2 \Rightarrow \mathrm{(B)}$. Notice that if $g(x) = 2 - 2x^2$, then $f - g$ vanishes at $x = -1, 0, 1$ and so \[f(x) - g(x) = ax(x-1)(x+1) = ax^3 - ax\] implies by $x^2$ coefficient, $b + 2 = 0, b = -2 \Rightarrow \mathrm{(B)}$. The roots of this equation are $-1, 1, \text{ and } x$, letting $x$ be the root not shown in the graph. By Vieta, we know that $-1+1+x=x=-\frac{b}{a}$ and $-1\cdot 1\cdot x=-x=-\frac{d}{a}$. Therefore, $x=\frac{d}{a}$. Setting the two equations for $x$ equal to each other, $\frac{d}{a}=-\frac{b}{a}$. We know that the y-intercept of the polynomial is $d$, so $d=2$. Plugging in for $d$, $\frac{2}{a}=-\frac{b}{a}$. Therefore, $b=-2 \Rightarrow \boxed{B}$ From the graph, we have $f(0)=2$ so $d=2$. Also from the graph, we have $f(1)=a+b+c+2=0$. But we also have from the graph $f(-1)=-a+b-c+2=0$. Summing $f(1)+f(2)$ we get $2b+4=0$ so $b = -2 \Rightarrow \mathrm{(B)}$. Solution by franzliszt