Difference between revisions of "2017 AMC 10A Problems/Problem 24"

Revision as of 17:35, 8 February 2017

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

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

2017 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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All AMC 10 Problems and Solutions

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

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 -10r&=c\\
 \end{align*}</cmath>
+==See Also==
+{{AMC10 box|year=2017|ab=A|num-b=8|num-a=10}}
+{{MAA Notice}}

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Difference between revisions of "2017 AMC 10A Problems/Problem 24"

Revision as of 17:35, 8 February 2017

Problem

Solution

See Also