Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(Created page with "==Problem== For certain real numbers <math>a</math>, <math>b</math>, and <math>c</math>, the polynomial <cmath>g(x) = x^3 + ax^2 + x + 10</cmath>has three distinct roots, and...")
 
(Start of Solution)
Line 3: Line 3:
  
 
<math>\textbf{(A)}\ -9009 \qquad\textbf{(B)}\ -8008 \qquad\textbf{(C)}\ -7007 \qquad\textbf{(D)}\ -6006 \qquad\textbf{(E)}\ -5005</math>
 
<math>\textbf{(A)}\ -9009 \qquad\textbf{(B)}\ -8008 \qquad\textbf{(C)}\ -7007 \qquad\textbf{(D)}\ -6006 \qquad\textbf{(E)}\ -5005</math>
 +
 +
==Solution==
 +
<math>f(x)</math> must have four roots, three of which are roots of <math>g(x)</math>. Using the fact that every polynomial has a unique factorization into its roots, and since the leading coefficient of <math>f(x)</math> and <math>g(x)</math> are the same, we know that
 +
 +
<cmath>f(x)=g(x)(x-r)</cmath>
 +
 +
where <math>r\in\mathbb{C}</math> is the fourth root of <math>f(x)</math>. Substituting <math>g(x)</math> and expanding, we find that
 +
 +
<cmath>f(x)=(x^3+ax^2+x+10)(x-r)=x^4+(a-r)x^3+(1-ar)x^2+(10-r)x-10r.</cmath>
 +
 +
Comparing coefficients with <math>f(x)</math>, we see that
 +
 +
<cmath>\begin{align*}
 +
a-r=1\\
 +
1-ar=b\\
 +
10-r=100\\
 +
-10r=c\\
 +
\end{align*}</cmath>

Revision as of 16:30, 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

\[f(x)=(x^3+ax^2+x+10)(x-r)=x^4+(a-r)x^3+(1-ar)x^2+(10-r)x-10r.\]

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

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_24&oldid=82895"