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.

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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.

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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]