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Difference between revisions of "2003 AIME II Problems/Problem 9"

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(→‎Solution 2)
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Let <math>S_k=z_1^k+z_2^k+z_3^k+z_4^k</math> then by [[Vieta's Formula]] and [[Newton's Sums]] we have
 
Let <math>S_k=z_1^k+z_2^k+z_3^k+z_4^k</math> then by [[Vieta's Formula]] and [[Newton's Sums]] we have
 
<cmath>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</cmath>
 
<cmath>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</cmath>
−
<cmath>S_0=4</cmath>
+
<cmath>S_0==4</cmath>
−
<cmath>S_1=z_1+z_2+z_3+z_4=1</cmath>
+
<cmath>S_1=1</cmath>
−
<cmath>S_2=</cmath>
+
<cmath>S_2=3</cmath>
  
 
== See also ==
 
== See also ==

Revision as of 13:18, 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 and Newton's Sums we have \[S_{-1}=\frac{z_1z_2z_3+z_1z_3z_4+z_1z_2z_4+z_1z_2z_3}{z_1z_2z_3z_4}=0\] \[S_0==4\] \[S_1=1\] \[S_2=3\]

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

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