## Friday, December 22, 2006

### Leftovers. Topic: Algebra/Polynomials. Level: AIME.

Problem: (Stanford Putnam Practice) Find the remainder when you divide $x^{81}+x^{49}+x^{25}+x^9+x$ by $x^3-x$.

Solution: Since they're both divisible by $x$, we first divide that out, and just remember to multiply the remainder by $x$ at the end. Let $xP(x)$ be the first polynomial and $xQ(x)$ be the second. we have

$P(x) = g(x)Q(x)+r(x)$

for some polynomials $g(x), r(x)$ with $\deg(r) < deg(Q) = 2$. We want to find $r(x)$. Consider the two roots of $Q(x) = x^2-1 = (x+1)(x-1)$. Plugging them into the equation, we obtain

$P(1) = r(1)$ and $P(-1) = r(-1)$.

Evaluating $P$ at those two values, we find that

$r(1) = 5$ and $r(-1) = 5$.

But since $r$ has degree less than $2$, the only possible $r$ is the constant polynomial $r(x) = 5$. Then the remainder is $xr(x) = 5x$. QED.

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

Comment: This is a super important technique when it comes to polynomial division. Using the $\deg(Q)$ roots of $Q$ and the fact that $deg(r) < deg(Q)$, we can hypothetically always determine $r$ this way, without dividing. This usually comes up when $Q$ has nice roots, so if it doesn't, look for a better way.

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

Practice Problem: (Stanford Putnam Practice) How can the quadratic equation

$\frac{(x-a)(x-b)}{(c-a)(c-b)}+\frac{(x-b)(x-c)}{(a-b)(a-c)}+\frac{(x-c)(x-a)}{(b-c)(b-a)} = 1$

have three roots $x = a, b, c$? [Reworded]

#### 1 comment:

1. It's an identity.

:O