AIME 2003 II · 第 9 题

Solution 1

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 (Complete)

By using long division to divide $P(x)$ by $Q(x)$, we find that $P(x) = Q(x)(x^2 + 1) + (x^2 - x + 1)$

For Polynomial Remainder Theorem,

$$P(z_2)+P(z_1)+P(z_3)+P(z_4)=z_1^2-z_1+1+z_2^2-z_2+1+z_3^2-z_3+1+z_4^2-z_4+1$$ $$\Rightarrow P(z_2)+P(z_1)+P(z_3)+P(z_4)=(z_1^2+z_2^2+z_3^2+z_4^2)-(z_1+z_2+z_3+z_4)+4$$ We know that for Vieta's Formula

$$z_1+z_2+z_3+z_4=\frac{-b}{a}=\frac{-(-1)}{1}=1$$ $$x_1x_2 + x_1x_3 + x_1x_4 + x_2x_3 + x_2x_4 + x_3x_4= \frac{c}{a} = \frac{-1}{1} = -1$$ And, by Newton's Sums

$$S=(z_1^2+z_2^2+z_3^2+z_4^2)= \sum_{i=1}^{n=4} z_i^2 = \left( \sum_{i=1}^{n=4} z_i \right)^2 - 2 \sum_{i=1}^{n=4}z_i z_j \text{ with } i<j$$ So,

$$S= \sum_{i=1}^{n=4} z_i^2 = \left( \sum_{i=1}^{n=4} z_i \right)^2 - 2 \sum_{i=1}^{n=4}z_i z_j$$ $$\Rightarrow S = (z_1+z_2+z_3+z_4)-2(x_1x_2 + x_1x_3 + x_1x_4 + x_2x_3 + x_2x_4 + x_3x_4)$$ $$\Rightarrow S= 1 -2\times (-1) \Rightarrow S = 3$$ Therefore

$$\Rightarrow P(z_2)+P(z_1)+P(z_3)+P(z_4)=(z_1^2+z_2^2+z_3^2+z_4^2)-(z_1+z_2+z_3+z_4)+4$$ $$\Rightarrow P(z_2)+P(z_1)+P(z_3)+P(z_4) = (3)- 1 + 4 \therefore \boxed{ P(z_2)+P(z_1)+P(z_3)+P(z_4)=6}$$ ~fmeiramath

Solution 3

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 4

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

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 5

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 6

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 7 (if you don't know how to divide polynomials directly)

$P(x)$ and $Q(x)$ look very similar, so let's try subtracting $Q(x)$ multiplied with something from $P(x)$ since we know $Q(z_n) = 0$. To make the degrees the same, let's first multiply $Q(x)$ with $x^2$:

$Q(x) \cdot x^2 = x^6 - x^5 - x^4 - x^2$. Subtracting this from $P(x)$ gives $x^4 - x^3 - x$. The degrees of this polynomial and $Q(x)$ are the same, so let's subtract $Q(x)$: $x^4 - x^3 - x - Q(x)$ = $x^4 - x^3 - x - x^4 + x^3 + x^2 + 1 = x^2 - x + 1.$

So $P(x) - (x^2 + 1) \cdot Q(x) = x^2 - x + 1$ (this is also the remainder when $P(x)$ is divided by $Q(x)$). 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 $3 - 1 + 4 = \boxed{006}$.

~grogg007

Video Solution by Sal Khan

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

[rule]

Nice!-sleepypuppy