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

Revision as of 23:35, 4 July 2015

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

${{Q(z_1)=0}}$ therefore $z_1^4-z_1^3-z_1^2-1=0$

therefore $-z_1^3-z_1^2=-z_1^4+1.$

Also $z_1^4-z_1^3-z_1^2=1$

So $z_1^6-z_1^5-z_1^4=z_1^2$

So in $P(z_1)=z_1^6-z_1^5-z_1^3-z_1^2-z_1$

Since $-z_1^3-z^2=-z_1^4+1$ and $z_1^6-z_1^5-z_1^4=z_1^2$

$P(z_1)=z_1^6-z_1^5-z_1^3-z_1^2-z_1$ can now be $P(z_1)=z_1^2-z_1+1$

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

@@ Line 10: / Line 10: @@
 Also   <cmath>z_1^4-z_1^3-z_1^2=1 </cmath>
-So     <math></math>z_1^6-z_1^5-z_1^4=z_1^2<math> </math>
+So     <cmath>z_1^6-z_1^5-z_1^4=z_1^2</cmath>
 So in <cmath>P(z_1)=z_1^6-z_1^5-z_1^3-z_1^2-z_1</cmath>
-Since <math> -z_1^3-z^2=-z_1^4+1.</math> and <math>z_1^6-z_1^5-z_1^4=z_1^2</math>
+Since <math> -z_1^3-z^2=-z_1^4+1</math> and <math>z_1^6-z_1^5-z_1^4=z_1^2</math>
 <cmath>P(z_1)=z_1^6-z_1^5-z_1^3-z_1^2-z_1</cmath> can now be

2003 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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Difference between revisions of "2003 AIME II Problems/Problem 9"

Revision as of 23:35, 4 July 2015

Problem

Solution

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