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

Revision as of 15:39, 21 June 2025

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's 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 $P(z_2)+P(z_1)+P(z_3)+P(z_4)=3+4-1=\boxed{006}.$

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 $P(z_1)+P(z_2)+P(z_3)+P(z_4)=S_6-S_5-S_3-S_2-S_1=\boxed{6}$ .

~ Nafer

Solution 3

$P(x) = x^{2}Q(x)+x^{4}-x^{3}-x.$

So we just have to find: $\sum_{n=1}^{4} z^{4}_n - \sum_{n=1}^{4} z^{3}_n - \sum_{n=1}^{4} z_n$ .

And by Newton's Sums this computes to: $11-4-1 = \boxed{006}$ .

~ LuisFonseca123

Solution 4

If we scale $Q(x)$ by $x^2$ , we get $x^6-x^5-x^4-x^2$ . In order to get to $P(x)$ , we add $x^4-x^3-x$ . Therefore, our answer is $\sum_{n=1}^{4} z^4_n-z^3_n-z_n$ . However, rearranging $Q(z_n) = 0$ , makes our final answer $\sum_{n=1}^{4} z^2_n-z_n+1$ . The sum of the squares of the roots is $1^2-2(-1) = 3$ and the sum of the roots is $1$ . Adding 4 to our sum, we get $3-1+4 = \boxed{006}$ .

~ Vedoral

Solution 5

Let $S_k$ = $z_1^k+z_2^k+z_3^k+z_4^k$

By Newton's Sums,

$S_1-1=0$

$S_2-S_1-2=0$

$S_3-S_2-S_1=0$

$S_4-S_3-S_2-4=0$

$S_5-S_4-S_3-S_1=0$

$S_6-S_5-S_4-S_2=0$

Solving for $S_1,S_2,S_3,S_4,S_5,S_6$ , we get $S_1=1, S_2=3, S_3=4, S_4=11, S_5=16, S_6=30$

$P(z_1)+P(z_2)+P(z_3)+P(z_4)=S_6-S_5-S_3-S_2-S_1=\boxed{006}$

Solution 6 (Short and simple)

$P(x)$ and $Q(x)$ look very similar, so let's try subtracting $Q(x)$ multiplied with something from $P(x)$ since we know $Q(z_n) = 0$ .

Notice that $P(x) - (x^2 + 1) \cdot Q(x) = x^2 - x + 1$ . This tells us that our answer is the sum of the squares of the roots of $Q(x)$ minus the sum of the roots of $Q(x)$ plus $1 \cdot 4$ .

By Newton's Sums the sum of the squares of the roots is 3, and by Vieta's Formulas the sum of the roots is 1. So our answer is just $3 - 1 + 4 = \boxed{006}$ .

~sharmaguy

Video Solution by Sal Khan

https://www.youtube.com/watch?v=ZSESJ8TeGSI&list=PLSQl0a2vh4HCtW1EiNlfW_YoNAA38D0l4&index=14 - AMBRIGGS

[rule]

Nice!-sleepypuppy

@@ Line 73: / Line 73: @@
 <math>P(z_1)+P(z_2)+P(z_3)+P(z_4)=S_6-S_5-S_3-S_2-S_1=\boxed{006}</math>
+==Solution 6 (Short and simple)==
+<math>P(x)</math> and <math>Q(x)</math>  look very similar, so let's try subtracting <math>Q(x)</math> multiplied with something from <math>P(x)</math> since we know <math>Q(z_n) = 0</math>.
+Notice that <math>P(x) - (x^2 + 1) \cdot Q(x) = x^2 - x + 1</math>. This tells us that our answer is the sum of the squares of the roots of <math>Q(x)</math> minus the sum of the roots of <math>Q(x)</math> plus <math>1 \cdot 4</math>.
+By [[Newton's Sums]] the sum of the squares of the roots is 3, and by [[Vieta's Formulas]] the sum of the roots is 1. So our answer is just <math>3 - 1  + 4 = \boxed{006}</math>.
+~[[sharmaguy]]
 == Video Solution by Sal Khan ==

2003 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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