Achievable logoAchievable logo
AMC 10/12
Sign in
Sign up
Purchase
Textbook
Practice exams
Feedback
Community
How it works
Resources
Exam catalog
Mountain with a flag at the peak
Textbook
1. Introduction
2. Algebra
2.1 Manipulation
2.2 Functions (polynomial)
2.3 Functions (trigonometric)
2.4 Functions (exponential and logarithmic)
2.5 Functions (sequences)
2.6 Functions (other)
3. Geometry
4. Number theory
5. Probability
6. Combinatorics
7. What's next?
Achievable logoAchievable logo
2.2 Functions (polynomial)
Achievable AMC 10/12
2. Algebra

Functions (polynomial)

4 min read
Font
Discuss
Share
Feedback

Polynomial functions

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.

Definitions
Pascal’s triangle
An arrangement of numbers into a particular two-dimension pattern with a few important properties, including these:
  • It’s organized into rows
  • Each row can be calculated easily from the row above it
  • There are an infinite number of rows
  • It gives answers to many important mathematical questions, including “what are the coefficients of the expansion of (x+y)n?”, no matter how big n is
A root of a polynomial
Also known as a “zero,” this is a number that can be plugged in for x in a polynomial expression such that the value of the whole polynomial will then be zero
The Rational Root Theorem
This theorem, which you can look up by name, allows you to look at a polynomial and narrow down your search for possible roots to a very small number of possible options, at least one of which is then guaranteed to work.
Synthetic division
Sometimes also known as synthetic substitution, this is a technique for dividing one polynomial by hand, but without writing any of the unnecessary details of the process along the way, thus speeding up the process by a factor of three or so as compared to regular long division
de Moivre’s theorem
A technique for easily calculating square roots and higher-order roots of complex numbers, by first converting the numbers to geometric representations that follow certain easy-to-work-with rules

A few notes on the above definitions:

For the Rational Root Theorem, Brilliant’s explanation and example are a good place to start

For de Moivre’s Theorem, AoPS’s statement of the theorem is helpful, but not their proof; their discussion of complex numbers, especially the sections on Operations and Alternate Forms, may be necessary to understand the statement.

For synthetic division, start with Purple Math then turn to YouTube here and here.

Example problems

  • 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 x4+4x3y+6x2y2+4xy3+y4.

If you write it as 1x4+4x3y+6x2y2+4xy3+1y4 (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 16=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 ks that are multiples of 60 degrees or π/3 radians
  • …and that the imaginary ones are therefore ±21​±2i3​​

Sign up for free to take 16 quiz questions on this topic

All rights reserved ©2016 - 2025 Achievable, Inc.