# Mathematical Food for Thought

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Serves a Daily Special and an All-You-Can-Eat Course in Problem Solving. Courtesy of me, Jeffrey Wang.

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Leftovers. Topic: Algebra/Polynomials. Level: AIME. December 22nd, 2006

Problem: (Stanford Putnam Practice) Find the remainder when you divide by .

Solution: Since they’re both divisible by , we first divide that out, and just remember to multiply the remainder by at the end. Let be the first polynomial and be the second. we have

for some polynomials with . We want to find . Consider the two roots of . Plugging them into the equation, we obtain

and .

Evaluating at those two values, we find that

and .

But since has degree less than , the only possible is the constant polynomial . Then the remainder is . QED.

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Comment: This is a super important technique when it comes to polynomial division. Using the roots of and the fact that , we can hypothetically always determine this way, without dividing. This usually comes up when 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

have three roots ? [Reworded]

### One Response to “Leftovers. Topic: Algebra/Polynomials. Level: AIME.”

1. t0rajir0u Says:

It’s an identity.

:O

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