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 "2003 AIME II Problems/Problem 9"

(→‎Solution 2)
(→‎Solution 2)
Line 32: Line 32:
 
<cmath>S_1=1</cmath>
 
<cmath>S_1=1</cmath>
 
<cmath>S_2=3</cmath>
 
<cmath>S_2=3</cmath>
−
Applying [[Newton's Sums]] we have <math>a_4S_k+a_3S_{k-1}+a_2S_{k-2}+a_1S_{k-1}+a_0S_{k-4}=0</math>.
+
By [[Newton's Sums]] we have  
 +
<cmath>a_4S_k+a_3S_{k-1}+a_2S_{k-2}+a_1S_{k-1}+a_0S_{k-4}=0</cmath>
  
−
Thus <math>P(z_1)+P(z_2)+P(z_3)+P(z_4)=S_6-S_5-S_3-S_2-S_1-1</math>
+
Applying the formula couples of times yields <math>P(z_1)+P(z_2)+P(z_3)+P(z_4)=S_6-S_5-S_3-S_2-S_1=\boxed{6}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 13:24, 22 December 2019

Problem

Consider the polynomials $P(x) = x^{6} - x^{5} - x^{3} - x^{2} - x$ and $Q(x) = x^{4} - x^{3} - x^{2} - 1.$ Given that $z_{1},z_{2},z_{3},$ and $z_{4}$ are the roots of $Q(x) = 0,$ find $P(z_{1}) + P(z_{2}) + P(z_{3}) + P(z_{4}).$

Solution

When we use long division to divide $P(x)$ by $Q(x)$, the remainder is $x^2-x+1$.

So, since $z_1$ is a root, $P(z_1)=(z_1)^2-z_1+1$.

Now this also follows for all roots of $Q(x)$ Now \[P(z_2)+P(z_1)+P(z_3)+P(z_4)=z_1^2-z_1+1+z_2^2-z_2+1+z_3^2-z_3+1+z_4^2-z_4+1\]

Now by Vieta's we know that $-z_4-z_3-z_2-z_1=-1$, so by Newton Sums we can find $z_1^2+z_2^2+z_3^2+z_4^2$

$a_ns_2+a_{n-1}s_1+2a_{n-2}=0$

$(1)(s_2)+(-1)(1)+2(-1)=0$

$s_2-1-2=0$

$s_2=3$

So finally $P(z_2)+P(z_1)+P(z_3)+P(z_4)=3+4-1=\boxed{6}$


Solution 2

Let $S_k=z_1^k+z_2^k+z_3^k+z_4^k$ then by Vieta's Formula we have \[S_{-1}=\frac{z_1z_2z_3+z_1z_3z_4+z_1z_2z_4+z_1z_2z_3}{z_1z_2z_3z_4}=0\] \[S_0==4\] \[S_1=1\] \[S_2=3\] By Newton's Sums we have \[a_4S_k+a_3S_{k-1}+a_2S_{k-2}+a_1S_{k-1}+a_0S_{k-4}=0\]

Applying the formula couples of times yields $P(z_1)+P(z_2)+P(z_3)+P(z_4)=S_6-S_5-S_3-S_2-S_1=\boxed{6}$.

See also

2003 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2003_AIME_II_Problems/Problem_9&oldid=113171"