Are you good enough with polynomial functions to do great on the AMC?

In order to answer “yes,” you should be able to do any of the following when needed, without any hints or prompting:

create Pascal’s triangle from memory

explain how it relates to binomials

explain what a root of a polynomial is

state and explain the Rational Root Theorem

divide polynomials (e.g. using synthetic division)

write and explain de Moivre’s theorem

A bit of background info

There isn’t enough time and space here to explain all of the above topics, but there is enough time and space to at least introduce them, so that’s what I’ll do here.

Given that the fourth row of Pascal’s Triangle is $14641$, write out the expansion of $(x+y)_{4}$, without doing any multiplying. Finally, briefly explain the connection here.

(spoiler)

The expansion of $(x+y)_{4}$ is $x_{4}+4x_{3}y+6x_{2}y_{2}+4xy_{3}+y_{4}$.

If you write it as $1x_{4}+4x_{3}y+6x_{2}y_{2}+4xy_{3}+1y_{4}$ (i.e. writing out the “$1$” coefficients in the first and last terms), then you can see that the fourth row of the Triangle gives the coefficients for the terms in the expansion of $(x+y)_{4}$.

That’s because each row of the triangle is made from the row above it, in more or less the same way that each $(x+y)_{n}$ is made from the $(x+y)_{n−1}$ “above” it.

What are the six (complex) sixth roots of $1$?

(spoiler)

One of them is $1$, since $1_{6}=1$. Another is $−1$, since $(−1)_{6}=1$. That’s good, but not good enough, I’m afraid.

For full credit here, you also need to know:

…that all six roots have magnitude $1$

…and that they are spaced evenly around the pole/origin

…and that they are therefore all equal to $1cis(k)$ for all $k$s that are multiples of $60$ degrees or $π/3$ radians

…and that the imaginary ones are therefore $±21 ±2i3 $

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