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

Revision as of 12:56, 7 October 2018

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

@@ Line 3: / Line 3: @@
 == Solution ==
-<math>{{Q(z_1)=0}}</math>, so
+When we use long division to divide <math>P(x)</math> by <math>Q(x)</math>, the remainder is <math>x^2-x+1</math>.
-<math>z_1^4-z_1^3-z_1^2-1=0</math>
-therefore <cmath>-z_1^3-z_1^2=-z_1^4+1.</cmath>
+So, since <math>z_1</math> is a root, <math>P(z_1)=(z_1)^2-z_1+1</math>.
-Also   <cmath>z_1^4-z_1^3-z_1^2=1 </cmath>
-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>
-<cmath>P(z_1)=z_1^6-z_1^5-z_1^3-z_1^2-z_1</cmath> can now be
-<cmath>P(z_1)=z_1^2-z_1+1</cmath>
 Now this also follows for all roots of <math>Q(x)</math>

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 12:56, 7 October 2018

Problem

Solution

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