## Monday, November 21, 2005

### Polynomial Power. Topic: Algebra/Polynomials. Level: AIME.

Problem: (2006 Mock AIME 1 - #14) Let $P(x)$ be a monic polynomial of degree $n \ge 1$ such that

$[P(x)]^3 = [P(x)]^2-P(x)+6$

for $x = 1, 2, \ldots , n$. Let $n_0$ be the smallest $n$ such that $P(0) > (P(1)+P(2)+ \cdots +P(n))^3$.
Find the remainder when $P(0)$ is divided by $1000$ given $n = n_0$.

Solution: Well the condition we're given as it is doesn't look particularly helpful in defining the polynomial, especially with the cube. So we try and factor it (almost always a safe bet) and we find

$(P(x)-2)([P(x)]^2+P(x)+3) = 0$.

But we notice that $[P(x)]^2+P(x)+3 = \left(P(x)+\frac{1}{2}\right)^2 + \frac{11}{4} > 0$. So then we must have $P(x)-2 = 0$ for $x = 1, 2, \ldots, n$.

Consider the polynomial $H(x) = P(x)-2$. It has zeros at $x = 1, 2, \ldots, n$ and is also monic with degree $n$. Hence $H(x) = (x-1)(x-2)\cdots(x-n) \Rightarrow P(x) = (x-1)(x-2)\cdots(x-n)+2$.

Then we have $P(0) = (-1)^n\cdot n!+2 > (P(1)+P(2)+\cdots+P(n))^3 = (2n)^3 = 8n^3$. Checking, we find the first $n$ such that this is true is $n = 8$. Therefore, $P(0) = (-1)^8 \cdot 8!+2 = 40322$. So our answer is $322$. QED.

--------------------

Pracitice Problem #1: Factor $x^4+64$.

Practice Problem #2: Let $P(x)$ be a monic polynomial of degree $n$ such that $P(x) = x$ for $x = 1,2,\ldots,n$. Find a closed expression for $P(n+1)$.

1. #1. x^4 + 64 = x^4 + 16x^2 - 16x^2 + 64
= (x^2 + 8 )^2 - (4x)^2
= (x^2 + 4x + 8 ) (x^2 - 4x + 8 )

Took me awhile to realize there were no monomial factors =P

#2. P(x) = (x - 1)(x - 2)(x - 3)...(x - n) + x
P(n+1) = n! + n + 1

2. Why are you making weird faces? =P Anyway, your answers are correct; I should post harder problems.

3. dang slick solution to #14 haha

4. 1. $x^4+64 = (x^2+8)^2-2x^2(8)=(x^2+8)^2-(4x)^2=(x^2+4x+8)(x^2-4x+8).$