2017 AMC 10A Problems/Problem 24

Problem

For certain real numbers $a$ , $b$ , and $c$ , the polynomial $g(x) = x^3 + ax^2 + x + 10$ has three distinct roots, and each root of $g(x)$ is also a root of the polynomial $f(x) = x^4 + x^3 + bx^2 + 100x + c.$ What is $f(1)$ ?

$\textbf{(A)}\ -9009 \qquad\textbf{(B)}\ -8008 \qquad\textbf{(C)}\ -7007 \qquad\textbf{(D)}\ -6006 \qquad\textbf{(E)}\ -5005$

Solution

$f(x)$ must have four roots, three of which are roots of $g(x)$ . Using the fact that every polynomial has a unique factorization into its roots, and since the leading coefficient of $f(x)$ and $g(x)$ are the same, we know that

$f(x)=g(x)(x-r)$

where $r\in\mathbb{C}$ is the fourth root of $f(x)$ . Substituting $g(x)$ and expanding, we find that

$\begin{align*}f(x)&=(x^3+ax^2+x+10)(x-r)\\ &=x^4+(a-r)x^3+(1-ar)x^2+(10-r)x-10r.\end{align*}$

Comparing coefficients with $f(x)$ , we see that

$\begin{align*} a-r&=1\\ 1-ar&=b\\ 10-r&=100\\ -10r&=c.\\ \end{align*}$

Let's solve for $a,b,c,$ and $r$ . Since $10-r=100$ , $r=-90$ , so $c=(-10)(-90)=900$ . Since $a-r=1$ , $a=-89$ , and $b=1-ar=-8009$ . Thus, we know that

$f(x)=x^4+x^3-8009x^2+100x+900.$

Taking $f(1)$ , we find that

$\begin{align*} f(1)&=1^4+1^3-8009(1)^2+100(1)+900\\ &=1+1-8009+100+900\\ &=\boxed{\bold{(C)\text{ }}\text{-}7007}.\\ \end{align*}$

2017 AMC 10A (Problems • Answer Key • Resources)
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2017 AMC 10A Problems/Problem 24

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