# Algebra

This is an online resource for instructors and students. While the material is designed to be taught to strong middle school students, these notes are written for instructors who are invited to guide and discuss topics with their students.

## Fundamentals

### The Integers

The first kinds of numbers discovered were the natural numbers: $0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, \dots$

There are (infinitely) many natural numbers, and there are many things that we can do with them. For example, we can add any two natural numbers to obtain another. It does not matter how big the numbers become: their sum is always still a natural number.

We may also multiply two natural numbers to obtain another. Again, it does not matter how big or small the numbers are: multiplication is something we can do with any natural numbers.

Natural numbers are great for representing “things”. In particular, they're great for representing “some number of things”. But they fall short when we want to represent less than nothing. But you can’t have less than nothing, so why would we ever want that?

The desire for a more general kind of number comes from the desire to represent change. If yesterday there were $20$ students in attendance, and today there are $24$ students in attendance, then we can say $4$ more people attended today than did yesterday. But what if tomorrow there will again be $24$ students in attendance? Then how many more students will have attended? Even worse, what if the day after tomorrow, there will be $20$ students in attendance again. Then what is the change in the number of students?

It is not that these problems cannot be solved with positive natural numbers. In fact, they certainly can. The number of students could increase by $4$, and we have a number for that. Or the number of students could decrease by $4$. We already have a number for that.

So to express a change in a quantity, we must convey two pieces of information: firstly, in what direction is the change; and secondly, by how much the quantity is changed. This seems quite complicated. It would be simpler to introduce a new kind of quantity that could represent these changes more conveniently and more compactly. This, of course, is the motivation for negative numbers.

Negative numbers do not “exist” in the real world. We cannot have a negative number of people in a class. But they provide an easy way to talk about change: if positive numbers represent increase, then negative numbers represent decrease.

The introduction of negative numbers allows us to expand our set of mathematical objects to the integers, $\dots, -3, -2, -1, 0, 1, 2, 3, \dots$. As with natural numbers, we still have these two useful properties of integers, which we will call closure properties (think of a room withh closed door; if applying these operations inside the room of the integers, you never need to worry about anything outside the room):

• The sum of any two integers is again an integer.

• The product of any two integers is again an integer.

But in addition, we have a third closure property, which enables us to describe changes:

• The difference of any two integers is again an integer.

#### Exercise 1: Arithmetic with Integers

Evaluate each of the following expressions:

1. $1 + (-1) =$

2. $-1 - (-2) =$

3. $-3 \times 4 =$

4. $(-4) \times (-10) =$

##### Solution
1. $1 + (-1) = \boxed{0}$

2. $-1 - (-2) = \boxed{1}$

3. $-3 \times 4 = \boxed{-12}$

4. $(-4) \times (-10) = \boxed{40}$

### Variables

When we first learn arithmetic, we only have to worry about natural numbers. As we continue to develop new techniques, we find new kinds of numbers — like the integers mentioned above. Let’s take a step back and examine the philosophy behind what numbers really are.

We are familiar with everyday objects: things we can see, feel, or otherwise interact with. Mathematical objects are like those, but more abstract. In the mathematical world, you can interact with mathematical objects through statements of fact. A common example of such a statement is equality: $1 + 1 = 2$ is a statement of fact that the mathematical object referred to by $1 + 1$ is in fact the same as the mathematical object referred to by $2$.

Numbers are the most common mathematical objects, but as we will see later, they are not the only kind. Typically, we study mathematical objects because of some motivation from a real-world problem. In the real world, we may talk of quantity, location, a process, or any number of other things. To use mathematics to solve these problems, we must translate these real-world concepts into mathematical objects by using a model. Inside the model, we have many mathematical objects. Some of these objects represent the information given to us by the problem. Other objects might represent information we need to figure out — they might not be known from the start!

There are three ways we can describe a mathematical object in a simple model:

• A literal: for example, just a number, like $123$.

• A variable: a letter that we understand the meaning of, but we may or may not know the value of, like $x$. Essentially, we are naming a mathematical object. It can sometimes be useful to name known objects (for example, if they are complicated), but usually we only use variables for unknown objects.

• An expression: a combination of literals, variables, and operations, like $x - 123$.

For instance, a real world problem is as follows:

#### Exercise 2: More Lumber Is Required

Yahui and Zhen want to build a wooden treehouse. $100$ planks are required. Yahui has $33$ in her shed, and Zhen has $25$ in his shed. How many additional planks must they buy to complete the treehouse?

##### Solution

We can use whole numbers to model this problem. First, let’s give names (variables) to all mathematical objects (numbers) that the problem gives us or requests that we find. Let $y$ be the number of planks that Yahui has, $z$ be the number of planks that Zhen has, $r$ be the total number of planks required, and $x$ be the number of additional planks they must buy. The equation is: $y + z + x = r$

This equation is the governing equation of our model. We are given the values of $y$, $z$, and $r$, and we’re asked to find the value of $x$. (This is not always possible!) As mentioned above, we don’t need to use names for the known values, so we can substitute the literals: $33 + 25 + x = 100$

To find the unknown value we must add to $33 + 25 = 58$ to arrive at $100$, we can use subtraction. That is, $x = 100 - 58 = \boxed{42}$

Another way to understand what we have done in the above step is that $58 + x$ and $100$ refer to the same mathematical object, $100$. Therefore, we can subtract $58$ from both of them, and they will still be different names for the same object. That is, $58 + x - 58 = 100 - 58$. By using the properties of addition that we are familiar with, the left hand side is just another name for $x$ — and so we obtain the result seen above.

In this example, all of the variables we used represent specific mathematical objects. Three of them were immediately given to us in the question. The other, $x$, still represented a specific mathematical object, but we had to figure it out.

It is not always the case that variables represent specific mathematical objects — sometimes, we can attach a quantifier to a variable, to say that a statement is true of all mathematical objects of a certain type at once.

#### Exercise 3: Properties of Whole Number Addition

Give a concrete example for each of the following properties of addition of whole numbers:

1. For all integers $a$, $a + 0 = a$.

2. For all integers $a$ and $b$, $a + b = b + a$.

3. For all integers $a$, $b$ and $c$, $(a + b) + c = a + (b + c)$.

##### Solution

Solutions may vary.

1. Take $a = 2$. Then $2 + 0 = 2$.

2. Take $a = 3$ and $b = 7$. Then $3 + 7 = 10 = 7 + 3$.

3. Take $a = 1$, $b = 2$ and $c = 3$. Then $(1 + 2) + 3 = 3 + 3 = 6 = 1 + 5 = 1 + (2 + 3)$.

Note that these concrete examples are applications of, not justifications for, the properties in question. It can be difficult to give a formal justification for true properties that involve quantifiers. However, if a statement is false, it is often much easier: we can simply give a single concrete example which does not satisfy the statement.

#### Exercise 4: Counterexamples

Give a counterexample for each of the following incorrect statements about whole numbers:

1. For all integers $a$, $a + a > a$.

2. For all integers $a$ and $b$, $a - b = b - a$.

3. For all integers $a$, $b$ and $c$, $a + b = a + c$.

##### Solution

Solutions may vary.

1. Take $a = 0$. Then $a + a = 0 + 0 = 0 \not > 0 = a$.

2. Take $a = 1$ and $b = 0$. Then $a - b = 1 - 0 = 1 \ne -1 = 0 - 1 = b - a$.

3. Take $a = 0$, $b = 0$, and $c = 1$. Then $a + b = 0 + 0 = 0 \ne 1 = 0 + 1 = a + c$.

Note that even though each of the statements is false, they all have certain cases where they do hold. In the future, we will see techniques to find out exactly which cases the statement is true in.

## Rational Numbers

The motivation behind fractions is similar to that of expanding the whole numbers to the integers. Many things in the real world are divisible into portions. For instance, we can cut a cake into slices. Natural numbers are good at counting wholes, but bad at measuring parts of wholes.

To solve this problem, we introduce the concept of splitting a single whole into $n$ equal parts, where $n$ is a positive whole number. (Later, we will allow $n$ to be negative, but it cannot be zero — it is not possible to split something into zero equal parts.) We write this as $\frac{1}{n}$, and call each of the equal parts an $n$th part. (If $n := 10$, then they are called tenths.) We may sometimes require more than one such equal part. If we want $m$ parts where $m$ is any integer, we will write it as $\frac{m}{n}$.

All numbers that can be formed from fractions of integers are called rational numbers. We can talk about rational numbers as an extension to the integers, just like how we did with integers above, as an extension to the natural numbers. Each integer is still a rational number: for every integer $n$, and $n = \frac{n}{1}$.

There are various simple properties of fraction addition and subtraction, and multiplication, by counting the number of of $n$th parts:

1. For all integers $a$, $b$, and $c$, if $c > 0$, then $\frac{a}{c} + \frac{b}{c} = \frac{a+b}{c}$.

2. For all integers $a$, $b$, and $c$, if $c > 0$, then $\frac{a}{c} - \frac{b}{c} = \frac{a-b}{c}$.

3. For all integers $a$, $b$, and $c$, if $c > 0$, then $\frac{a}{c} \times b = \frac{ab}{c}$.

With the integers, we could not in general perform exact division on any two integers. Fractions cover the case of dividing almost any two integers (almost, since the denominator may not be $0$). But do we still have closure under addition, subtraction, and multiplication? Moreover, do we almost have closure under division (if we disallow a zero divisor)? We only saw some special cases of these operations above. We need to introduce stronger techniques — a general way to do addition, subtraction, and multiplication of fractions, to show that this closure is indeed the case.

### Fraction Multiplication

We saw above how to multiply a fraction by an integer. This has the same meaning as multiplying two integers; we can think of it as adding (maybe in the negative direction) repeated copies of the fraction. But it is harder to extend this idea toward multiplying two fractions: what do we mean when we say we want $\frac{1}{2}$ copies of $\frac{1}{3}$?

Luckily, there is in fact a single reasonable meaning for this. Recall that $\frac{m}{n}$ means to split a single whole into equal $n$th parts, then take $m$ copies of the $n$th parts. We can replace the single whole with something that is itself an arbitrary rational number: let $q \times \frac{m}{n}$ mean to split $q$ into $n$th parts, and then take $m$ of those. Then, $1 \times \frac{m}{n} = \frac{m}{n}$, as we would expect.

How do we split $\frac{a}{c}$ into $n$th parts? We can split each of the $a$ copies of the $b$th parts into $n$ parts, then only take one of every $n$. Each small part is a $b$th part split further into $n$ parts. In a single whole, there would be $c\times n$ equal such parts, so these small parts are in fact $\frac{1}{c\times n}$ each.

Therefore, we can justify the following fact (or definition, or a sort) about fraction multiplication:

#### Fact 1: Fraction Multiplication

Let $a$, $b$, $c$ and $d$ be integers with $c \ne 0 \ne d$. Then $\frac{a}{c} \times \frac{b}{d} = \frac{a \times b}{c \times d}$

### Equivalence Classes

It happens to be the case with fractions that distinct ordered pairs might represent the same quantity. For instance, $\frac{3}{6}$ and $\frac{7}{14}$ are different pairs of numbers, but they represent the same fraction: one half. All the fractions that represent the same particular quantity form a so-called equivalence class. In a sense, this is not very different from $2 - 5$ and $3 - 6$ representing the same change in quantity $-3$.

How can we decide whether two fractions represent the same quantity? That is, suppose that $\frac{a}{c}$ and $\frac{b}{d}$ are rational numbers. Are they equal? In the case where the denominator is the same, this is easy to answer: just compare the numerators. Rational numbers with the same denominator are equal if and only if the numerators are equal.

#### Fact 2: Equivalence of Fractions With Equal Denominator

Let $a$, $b$, and $c$ be integers, and $c \ne 0$. Then $\frac{a}{c} = \frac{b}{c} \iff a = b$

If the denominators are not the same, we need to rewrite the two fractions to have the same denominator. We can do this by first noticing the following fact, which we can obtain from our knowledge of fraction multiplication:

#### Fact 3: Common Factor of Numerator and Denominator

Let $a$, $b$ and $c$ be integers, and $c \ne 0 \ne b$. Then $\frac{a}{c} = \frac{a}{c} \times 1 = \frac{a}{c} \times \frac{b}{b} = \frac{a \times b}{c \times b}$

This fact allows us to rewrite $\frac{a}{c}$ as $\frac{a\times d}{c\times d}$, and $\frac{b}{d}$ as $\frac{b\times c}{d\times c}$. Now the denominators are the same (remember $c\times d = d\times c$ for any integers $c$ and $d$). So we can simply compare $a\times d$ with $b\times c$!

Visually, we are multiplying the top-left with the bottom-right, and the top-right with the bottom-left. This makes a cross shape, so one way to remember this technique is that it is often called “cross-multiplication”.

#### Fact 4: Cross-multiplication

Let $a$, $b$, $c$ and $d$ be integers, and $c \ne 0 \ne d$. Then $\frac{a}{b} = \frac{c}{d} \iff a\times d = b\times c$

We have seen above the example of rewriting fractions with a common denominator in order to compare them. But another use of fractions with a common denominator is that they are easy to add and subtract. We can apply the same technique:

#### Exercise 5: Fraction Addition and Subtraction

Rewrite the fractions using a common denominator in order to calculate:

1. $\displaystyle\frac{1}{2} + \frac{1}{3} =$

2. $\displaystyle\frac{1}{2} - \frac{1}{3} =$

3. $\displaystyle\frac{1}{3} - \frac{2}{-3} =$

##### Solution
1. $\displaystyle\frac{1}{2} + \frac{1}{3} = \frac{1\times 3}{2\times 3} + \frac{1\times 2}{3\times 2} = \frac{3}{6} + \frac{2}{6} = \boxed{\frac{5}{6}}$

2. $\displaystyle\frac{1}{2} - \frac{1}{3} = \frac{1\times 3}{2\times 3} - \frac{1\times 2}{3\times 2} = \frac{3}{6} - \frac{2}{6} = \boxed{\frac{1}{6}}$

3. $\displaystyle\frac{1}{3} - \frac{2}{-3} = \frac{1}{3} - \frac{2\times(-1)}{-3\times(-1)} = \frac{1}{3} - \frac{-2}{3} = \boxed{\frac{1}{3}}$

In general, we can derive formulas for addition and subtraction of fractions, but you should not memorize them. It is more useful to understand the process of arriving at the formulas.

#### Fact 5: Fraction Addition & Subtraction

Let $a$, $b$, $c$ and $d$ be integers with $c \ne 0 \ne d$. Then $\frac{a}{c} + \frac{b}{d} = \frac{a \times d}{c \times d} + \frac{b \times c}{d \times c} = \frac{a\times d + b\times c}{c \times d}$ and $\frac{a}{c} - \frac{b}{d} = \frac{a \times d}{c \times d} - \frac{b \times c}{d \times c} = \frac{a\times d - b\times c}{c \times d}$

Notice the resemblance to cross-multiplication. This is not accidental! From the subtraction formula, we see that if two fractions are equal, their difference is zero, and vice versa.

### Simplification

Often, given some fraction, we want to find the equivalent fraction with the smallest possible positive integer denominator. This is called the simplest form and is is useful for various reasons:

• Smaller positive integer denominators are easier for people to understand.

• Two fractions that are equal will have the same simplest form, so fractions in simplest form are easy to compare.

Reducing a fraction to simplest form is a matter of finding the largest common factor of both the numerator and denominator, and then dividing both the numerator and denominator by it.

#### Exercise 6: Fraction Operations and Simplification

Compute each of the following, then reduce it to simplest form.

1. $\displaystyle \frac{3}{8} \times \frac{2}{7} =$

2. $\displaystyle \frac{5}{9} \times \frac{2}{5} =$

##### Solution
1. $\displaystyle \frac{3}{8} \times \frac{2}{7} = \frac{6}{56} = \boxed{\frac{3}{28}}$

2. $\displaystyle \frac{5}{9} \times \frac{2}{5} = \frac{10}{45} = \boxed{\frac{2}{9}}$

#### Exercise 7: A Telescoping Product

Compute and reduce to simplest form: $\frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} \times \dots \times \frac{99}{100}$

##### Solution

Each fraction in this product, except for the last one, has a numerator which is the same as the denominator of the following fraction. These will cancel out if we multiply the fractions. For instance, $\frac{1}{2} \times \frac{2}{3} = \frac{1\times 2}{2 \times 3}$, and we can divide $2$ from both the numerator and the denominator to get $\frac{1}{3}$.

In this manner, all the numbers except for the first $1$ in the numerator and the last $100$ in the denominator will get cancelled out. So we are left with $\boxed{\frac{1}{100}}$.

## Linear Equations

### Ratios and Rates

Frequently, we may know certain quantities not in absolute terms, but only in relative terms. What does this mean? Let’s say you see two weights on the ground, labeled A and B. You might notice that B is twice as hard to lift up as A. Without a scale, it is hard for you to measure the actual weight of the objects, but you might be able to estimate the ratio of their weights.

In another example, suppose you are counting cars as they pass by on a highway. You might notice that for every $5$ personal cars you count, you see about one truck. It might be hard for you to estimate how many cars are passing by each minute, since it is hard to guess how long a minute is, but you could estimate the ratio of personal cars to trucks on this highway.

We will see a couple of word problems that involve known ratios, and try to determine the absolute quantities using additional information provided to us.

#### Exercise 8: Carcross Car Count

The community of Carcross, Yukon is quite small, with a population of only 301. Caroline counts the number of cars that passed her house over an hour and noticed that:

• There were $15$ cars that passed in total.

• All cars were either blue or silver.

• Twice as many cars were blue than silver.

How many blue cars passed by? How many silver cars?

##### Solution

Let $b$ and $s$ be integers representing the number of cars that passed her house. Then our equations are: \begin{aligned} b + s &= 15 \\ b &= 2s \\ \end{aligned}

We can substitute the second equation into the first equation, since $2s$ and $b$ refer to the same mathematical object. Thus: \begin{aligned} 2s + s &= 15 \\ 3s &= 15 \\ \textcolor{blue}{\frac{1}{3}} \times 3s &= \textcolor{blue}{\frac{1}{3}} \times 15 \\ s &= 5 \\ \end{aligned} so there were $\boxed{5}$ silver cars. Then we can substitute this back into that second equation $b = 2s$. So $b = 2 \times 5$, and so $b = 10$. Therefore there were $\boxed{10}$ blue cars.

#### Exercise 9: Produce Price Sum

A supermarket stocks four kinds of produce: apples, oranges, tomatos, and potatos. Apples cost twice as much as oranges, and oranges cost twice as much as tomatos. August bought $1\,\mathrm{kg}$ of each kind of produce, and the total price was $\20$.

Can we figure out what was the price of tomatos? If so, what was it?

##### Solution

Let $a$, $o$, $t$, and $p$ be rational numbers representing the prices of apples, oranges, tomatos, and potatos, all in dollars per kilogram. Then our equations are \begin{aligned} a &= 2o \\ o &= 2t \\ a + o + t + p &= 20 \\ \end{aligned}

If we substitute the first and second equations into the third, we get $4t + 2t + t + p = 20$ and thus $7t + p = 20$. But there is a problem: there are multiple solutions to this! There are even multiple integer solutions; for instance, maybe $t = 2$ and $p = 6$, or $t = 1$ and $p = 13$. So we do not have enough information to figure out the price of tomatos.

#### Exercise 10: An Unlikely Sprint?

Miran, Gosse, and Brayan participated in a $100\,\mathrm{m}$ sprint. Miran tells you that she won and was twice as fast as Brayan. Gosse agrees that Miran won, and says he was close behind with a time only $20\%$ higher than Miran’s. Brayan says that he came in last with a time $8\,\mathrm{s}$ longer than Gosse’s time.

You know, however, that sometimes Miran, Gosse, and Brayan aren’t the most reliable. Is it mathematically possible for all of these accounts to be accurate? If so, do we have enough information to determine what were each of their times? If so, calculate the times.

##### Solution

Let $x$ denote Miran’s time, $y$ denote Gosse’s time, and $z$ denote Brayan’s time. Based on what everyone said, the equations are: \begin{aligned} x &= \frac{1}{2} z \\ y &= 1.2 x \\ z &= y + 8\,\mathrm{s} \\ \end{aligned}

Substitute the expression for $x$ given by first equation into the second equation, to get $y = 1.2 \times \textcolor{blue}{\frac{1}{2} z} = \frac{3}{5} z$

Now substitute this into the third equation, to get \begin{aligned} z &= \textcolor{blue}{\frac{3}{5} z} + 8\,\mathrm{s} \\ \textcolor{blue}{-\frac{3}{5} z} + z &= \textcolor{blue}{-\frac{3}{5} z} + \frac{3}{5} z + 8\,\mathrm{s} \\ \frac{2}{5} z &= 8\,\mathrm{s} \\ \textcolor{blue}{\frac{5}{2}} \times \frac{2}{5} z &= \textcolor{blue}{\frac{5}{2}} \times 8\,\mathrm{s} \\ z &= 20\,\mathrm{s} \\ \end{aligned}

We can now substitute this back into $y = \frac{3}{5} z$ to get $y = 12\,\mathrm{s}$, and into $x = \frac{1}{2} z$ to get $x = 10\,\mathrm{s}$ (very fast indeed, maybe suspiciously so!).

We can check that these times match all three of the equations above, so it is mathematically possible and unique. This doesn’t mean that the statements were accurate, but they are not mathematically contradictory.

#### Exercise 11: Raccoon Population Growth

The number of raccoons in the city of Raccoonville is plotted on the following chart:

If the current trend continues, by what year will there be 180 raccoons in Raccoonville?

##### Solution

In this problem, we have to figure out the rate of increase of raccoons from the chart. The trend seems to be a straight line with an increase of $10$ raccoons every year. We can assume this trend will continue as the question asks us in that hypothetical. Let $x$ denote the number of years after $2019$. Then the number of raccoons will be $130 + 10x$. We want to solve: \begin{aligned} 130 + 10x &= 180 \\ \textcolor{blue}{-130} + 130 + 10x &= \textcolor{blue}{-130} + 180 \\ 10x &= 50 \\ \frac{10x}{\textcolor{blue}{10}} &= \frac{50}{\textcolor{blue}{10}} \\ x &= 5 \\ \end{aligned}

Since this means $5$ years after $2019$, the year that there will be $180$ raccoons in Raccoonville is $2019 + 5 = \boxed{2024}$.

### A General Approach

The examples above all have the same general form, where we have a number of equations of the form $ax = b$, where $a$, $b$, and $x$ are rational numbers, and we know $a$ and $b$ (but not $x$). Equations of this form are called “linear equations”. Why are they linear? Intuitively, one reason is that if we draw a line graph of the value of $ax$ as we increase the value of $x$, we will find a straight line:

The solution to $ax=b$, if one exists, is simply where this straight line reaches a vertical height of $b$. We saw a general technique to do this if $a\ne 0$: we can multiply both sides by $\frac{1}{a}$ (or equivalently, divide both sides by $a$). Thus the solution is $x = \frac{b}{a}$.

But what if $a = 0$? In this case, we cannot divide by $a$, since division by $0$ is meaningless. Instead, we have the flat blue line in the graph. Obviously, this line will never reach any vertical height except $0$! Therefore, there is no solution if $b \ne 0$. If $b = 0$, then we still have no information about $x$: any rational number will do. In this case, there are multiple solutions.

## Real Numbers

### Exponents

Recall that a positive exponent represents repeated multiplication, much like how a positive multiplier represents repeated addition. We can express this rule recursively using the following identity: $x^{n+1} = xx^n$ which says that if you increase the exponent by $1$ it is the same as multiply one more copy of the base.

#### Exercise 12: Positive Integer Exponents

Evaluate each expression. Write your answer as an integer in literal form.

1. $2^4=$

2. $3^2=$

3. $10^6=$

##### Solution
1. $2^4=2\times 2\times 2\times 2=\boxed{16}$

2. $3^2=3\times 3=\boxed{9}$

3. $10^6=10\times 10 \times 10\times 10\times 10\times 10=\boxed{1000000}$

There are a variety of facts about positive integer exponents that we can justify using the properties of multiplication. Here are a few. You do not need to memorize these, but it is helpful to understand why they are true.

#### Fact 6: Sum of Exponents

If $a$ is a rational number, and $n$ and $m$ are positive integers, then $a^m \times a^n = a^{m+n}$. That is: $a^m \times a^n = \underbrace{a\times\dots\times a}_{m\text{ times}} \times \underbrace{a\times\dots\times a}_{n\text{ times}} = \underbrace{a\times\dots\times a}_{m+n\text{ times}} = a^{m+n}$

(Notice how when $n=1$, this is just the recursive rule we discussed above.)

#### Fact 7: Product of Exponents

If $a$ is a rational number, and $n$ and $m$ are positive integers, then ${(a^m)}^n = a^{mn}$. That is: ${(a^m)}^n = \underbrace{\left(\underbrace{a\times\dots\times a}_{m\text{ times}}\right) \times \dots \times \left(\underbrace{a\times\dots\times a}_{m\text{ times}}\right)}_{n\text{ times}} = \underbrace{a\times\dots\times a}_{mn\text{ times}} = a^{mn}$

It is frequently useful to extend the system of exponents to non-positive numbers, which can be done by applying the recursive rule in the other direction. Thus we can derive that $x^0 = 1$ and that $x^{-1} = \frac{1}{x}$ for all non-zero values of $x$. We can check that this extension retains the sum and product rules of exponents that we mentioned above, which is a useful feature.

#### Exercise 13: Negative and Zero Exponents

Evaluate each expression. Write your answer in simplest form as a fraction or as an integer literal.

1. ${-1}^{-1}=\boxed{-1}$

2. $4^{-2}=\boxed{\frac{1}{16}}$

3. ${999}^0=\boxed{1}$

4. ${\left(\frac{-17}{4}\right)}^0=\boxed{1}$

A useful application of exponents is in shrinking large numbers to an more humanly understandable format. Indeed, we have a poor conception of how large certain numbers are. In science, it's common to see numbers way too large to count or way too small to visualize. Scientists have developed notation using exponents to make comparing such numbers easier. In scientific notation, a number $x$ is written as $y\times 10^n$, where $y$ is a number with exactly one non-zero decimal digit before the decimal point, and $n$ is a (positive, negative, or zero) exponent.

#### Exercise 14: Scientific Notation

Express in scientific notation.

1. $1234 = \boxed{1.234\times 10^3}$

2. $0.000987 = \boxed{9.87\times 10^{-4}}$

A natural question to ask after having defined negative exponents is: what about rational exponents? Could those be useful? In fact, for a positive rational number base, we may sometimes define rational exponents in a way that preserves both the sum and product laws of exponents mentioned above. The only way to do this is to ensure that ${\left(x^{\frac{1}{n}}\right)}^n = x$, that is, $x^{\frac{1}{n}}$ must be the $n$th root of $x$. We can also write that as $\sqrt[n]{x}$. With this definition and the product law, we can define $x^q$ for any positive rational base $x$ and any rational exponent $q$.

#### Exercise 15: Fractions, Exponents & Radicals

Evaluate each expression. Write your answer in simplest form as a fraction, or as an integer using the place value system.

1. $4^{\frac{1}{2}}=\boxed{2}$

2. $9^{\frac{3}{2}}=\boxed{27}$

3. ${\left(\frac{2}{3}\right)}^3=\boxed{\frac{8}{27}}$

4. $\sqrt{\frac{16}{25}}=\boxed{\frac{4}{5}}$

5. $\sqrt[4]{\frac{256}{81}}=\boxed{\frac{4}{3}}$

With integer exponents of rational numbers, we are always guaranteed that the result exists and is a rational number (since we compute these exponents by multiplying and dividing rational numbers, which are closed under these operations). As we will see later, when rational exponents are concerned, the result may not exist as a rational number.

### Definition of a Real Number

In previous sections, we were careful to only discuss square roots and other rational exponents when we knew that there was in fact a rational number that worked. In general, we cannot assume this is always the case.

We can give a proof that no rational number is equal to $\sqrt{3}$. One way to see this is a so-called proof by contradiction. In this kind of argument, we assume that $\sqrt{3}$ is in fact rational. That is, if $\sqrt{3} = \frac{p}{q}$ for some integers $p$ and $q\ne 0$ in simplest form, then $\frac{p^2}{q^2} = 3$, so $p^2 = 3q^2$. We see that $3$ is a factor of the right hand side, so it must also be a factor of the left hand side. But the left hand side is a square, so $p=3m$ for some integer $n$. Then $9k^2 = 3q^2$ so $3m^2 = q^2$. Now $3$ is a factor of the left hand side, so it should also be a factor of the right hand side. But the right hand side is a square, so $q=3n$ for some integer $n$. But then $\frac{p}{q} = \frac{3m}{3n}$ is clearly not in simplest form, so we have reached an absurd state — a contradiction. But of course, this is not possible, so something has gone wrong. Our argument is correct, so it must be our assumption that was wrong. The assumption we made was that $\sqrt{3}$ is a rational number.

Many of you will probably be uneasy with how we are talking about $\sqrt{3}$ as if it must exist, when we have already shown that no rational number squares to $3$. We have already complicated things by introducing fractions to the easier world of the integers! If we make the claim that $\sqrt{3}$ should exist as a number, then we run the risk of making things even more complicated and difficult. We don’t, in general, say that everything which doesn’t exist must be a new kind of number (we are happy to say that $\frac{1}{0}$ simply does not exist).

There is in fact a good reason, however, to suggest that $\sqrt{3}$ might be a useful number to have. The reason for this is that we can already get very very close to a potential square root of $3$! An example of this is the rational number $\frac{3900231685776981}{2251799813685248} \approx 2.9999999999999997$. In fact, we can get arbitrarily close. A way to visualize this is to see a graph that maps numbers to their squares.

We can start by plotting a point on some axes for integer values. The horizontal distance represents the number $x$, which we vary to take on the integer values we want to show. The vertical distance represents the square of that number, $x^2$.

Of course, we can also take the square of rational numbers. We can think of this as increasing the precision of our graph by plotting more points, for example, every $0.1$.

If we imagine that we continue this process, getting more and more precision, we would expect this curve to become continuous. We can see that it reaches a vertical value of $3$ at some point, and we saw earlier that no such rational number point exists. But the curve suggests that we can define a new kind of number that is on the number line, and we can get close to using rational numbers, but can’t get exactly there. This concept is called a real number.

There are many ways to formally define a real number, but it is not necessary to understand such a definition to understand what a real number is. Intuitively, we expect real numbers to plug the holes in continuous lines that we can draw. We can get as close as we want to a real number using rational numbers. In fact we don’t even need to use all rational numbers. All finite decimals are rational numbers, and by adding more decimal points, we can get closer and closer to any rereal number we are interested in. $\sqrt{3} \approx 1.73$, but an even better approximation is $\sqrt{3} \approx 1.732$. We can keep getting more and more precise, but we can never reach the number itself because it is not rational (hence not a decimal).

### Operations on Real Numbers

We are able to add, subtract, multiply, and divide (except by zero) real numbers, just as we can with rational numbers. (Remember that we can get as close as we want to a real number using rational numbers. So it makes sense that real numbers behave almost identically to rational numbers!)

However, unlike rational numbers, it is not always possible to write real numbers in a canonical simplest form. Instead, we can use algebraic techniques to make expressions look simpler from a human perspective.

#### Exercise 16: A Linear Equation using Real Numbers

Solve the following equation for real number $x$: $\sqrt{2} x = 2$

##### Solution

We divide both sides by $\sqrt{2}$, yielding: $\frac{\sqrt{2}}{\sqrt{2}} x = \frac{2}{\sqrt{2}}$

Then, simplifying, $x = \frac{2}{\sqrt{2}} = 2\cdot \frac{1}{\sqrt{2}} = 2\cdot \frac{1}{2^{\frac{1}{2}}} = 2^2\cdot 2^{-\frac{1}{2}} = 2^{1 - \frac{1}{2}} = 2^{\frac{1}{2}} = \boxed{\sqrt{2}}$

At this stage, it is helpful to understand some supplementary material on sets. This material is on a seperate page because it is not strictly related to what we are studying right now about algebra, but the notational conveniences of the material will be useful.

## Polynomials

### Linear Equations, Again

Recall earlier when we solved linear equations of the form $ax = b$, where $a$ and $b$ are known rational (or real) numbers, and $x$ is an unknown rational (or real) number. The solution is to divide both sides of the equation by $a$, which is valid because $ax$ and $b$ are different names for the same rational (or real) number. This results in the solution $x = \frac{b}{a}$.

Note that many equations that may look different are really linear equations after some rearranging. For example, $ax + b = 0$ can be rearranged into the linear equation $ax = -b$. $ax + b = c$ can be rearranged into the linear equation $ax = c - b$. We will say that a linear equation that looks like $ax + b = 0$ is in “standard form”.

Let’s now look at a graphical method to solve a linear equation in standard form. What we will do is rewrite the right hand of the equation from $0$ to another variable $y$. We will then draw a graph, similar to what we did earlier when we introduced real numbers. Let us first consider the linear equation $2x - 6 = 0$.

Note that the equation $2x - 6 = y$ is more general than $2x - 6 = 0$. If we set $y := 0$, then we get back the original equation $2x - 6 = 0$. Therefore, to solve this equation we can look on the graphical plot for all the values on the line corresponding to $y = 0$. We see that this is where the line intersects with the x-axis. This is called an x-intercept, or root, of the function $y = 2x - 6$. From the plot, we see that the only root is $x = 3$, which corresponds to the only solution to this equation.

We will now use the knowledge about plots and roots to solve equations which are not linear. Let us start with a simple example.

#### Exercise 17: Roots of $x^2 - 1$

Plot $y = x^2 - 1$, determine its roots, and use this information to solve the equation $x^2 = 1$.

##### Solution

Here is our plot:

The roots are marked. They are $x = -1$ and $x = 1$, which correspond to the solutions to our equation $x^2 = 1$.

#### Exercise 18: Roots of $(x-3)(x+2)$

Without using a plot, determine the roots of $y = (x-3)(x+2)$. These are also solutions to the standard form equation $x^2 - x - 6 = 0$; explain why.

##### Solution

The roots occur when $y = 0$, so we want to solve $0 = (x-3)(x+2)$. If the product of two numbers is $0$, that means either the first number is $0$ or the second number is $0$ (or both). Therefore, the set of solutions to the equation $0 = (x-3)(x+2)$ is the union of the set of solutions to $x - 3 = 0$ and the set of solutions to $x + 2 = 0$. If $x - 3 = 0$, this is a simple linear equation, where $x = 3$ is the only solution. If $x + 2 = 0$, this is also a simple linear equation, where $x = -2$ is the only solution.

Therefore, the roots are $x \in \{-2, 3\}$.

Note that by using the distributive property, we find that $(x-3)(x+2) = (x-3)x + (x-3) \cdot 2 = x^2 - 3x + 2x - 6 = x^2 - x - 6$. So in fact, $x^2 - x - 6 = (x-3)(x+2)$ for all values of $x$. Hence the solutions of the two equations must be the same!

Exercise 18 suggests a general approach to solving equations that involve $x$ and $x^2$ might be to decompose it into the product of two components which are both linear. This process is called factoring. An expression of the form $ax^2 + bx + c$, where $a\ne 0$, $b$, and $c$ are known, is called a quadratic polynomial, just as $ax + b$ where $a\ne 0$ and $b$ are known is called a linear polynomial.

Suppose we have $(sx - u)(tx - v) = 0$, where $s$, $t$, $u$, and $v$ are known real numbers with $s \ne 0 \ne t$. Then we know the solutions are $x \in \left\{\frac{u}{s}, \frac{v}{t}\right\}$. A quadratic polynomial written this way is easy to solve! Our goal is to take a polynomial in standard form, and convert it into this factored form.

It is easier to go backwards. From factored form, we can use distributivity to expand: $(sx - u)(tx - v) = st x^2 - (sv + tu)x + uv$. But what we want to figure out is how to turn standard form into factored form.

If we can write a polynomial in standard form to look like $stx^2 - (sv + tu)x + uv$, then we have found the solutions! The standard form is $ax^2 + bx + c$, $a \ne 0$, so we need: $a = st$, $b = - (sv + tu)$, $c = uv$. This is not easy to solve (in fact, it does not have a unique solution), so we need to make some simplifications first.

First of all, we need to to fix the fact that the solutions are not unique. We can factor, for example, $2x^2 - 2 = (2x-2)(x+1)$, but we can also factor it as $2x^2 - 2 = (x-1)(2x+2)$. The roots are of course the same, because the equations $x+1 = 0$ and $2x+2 = 0$ have the same solutions. But they do not quite look the same. In order to force the factored form to be unique, we need to enforce that the linear polynomials are monic, that is, they have no leading coefficient (multiplier for the $x$ term). Instead, we will pull out those coefficients into a single multiplier for the entire quadratic polynomial.

That is, given $(sx - u)(tx - v)$, we would like to turn this into $a(x-u')(x-v')$, where $u'$ and $v'$ are new coefficients. How do we calculate $a$, $u'$, and $v'$? Let’s focus on the two linear polynomials seperately. We know that $sx - u = s(x - \frac{u}{s})$, by distributivity. Similarly, $tx - v = t(x - \frac{v}{t})$. Therefore: $(sx - u)(tx - v) = s(x - \frac{u}{s})t(x - \frac{v}{t}) = st(x - \frac{u}{s})(x - \frac{v}{t})$

Therefore, we can set $a := st$ and $u' := \frac{u}{s}$ and $v' := \frac{v}{s}$. Note that the expansion of $a(x-u')(x-v')$ into standard form is $ax^2 - a(u' + v')x + au'v'$. This is not actually any different from what we had before, but it looks somewhat closer to what we need! We also now see why we have reused $a$ for the leading coefficient in both forms — in fact, this leading coefficient will be the same going from monic factored form to standard form.

Again, we have not really made much progress — the goal is not to go from a factored form to standard form, but actually the opposite! But it turns out the new system of equations is easier to solve. We need: $b = -a(u' + v')$, $c = au'v'$. Another way to write this is $\frac{b}{a} = -(u' + v')$, and $\frac{c}{a} = u'v'$. That is: after dividing by the leading coefficient, we want the constant coefficient to be the product of the solutions, and the $x$ coefficient to be the negative sum of the solutions. We can solve this problem with trial and error.

#### Exercise 19: Factoring a Quadratic via Trial and Error

Using trial and error, factor $25 + 5x - 6x^2$.

##### Solution

In standard form, this is $-6x^2 + 5x + 25$. We can divide by the leading coefficient to obtain $6 (x^2 - \frac{5}{6}x - \frac{25}{6})$. We want two numbers whose sum is $\frac{5}{6}$, and whose product is $-\frac{25}{6}$. One of them will need to be negative! Let’s guess that $u'$ and $v'$ will be rational numbers. Write them as $\frac{m}{p}$ and $\frac{n}{q}$. We would like $mn = -25$ and $pq = 6$. There are essentially two options for $p$ and $q$, such as $1$ and $6$, or $2$ and $3$ (all other options involve negatives or just switching around $p$ and $q$). There are then six options for $m$: $-25, -5, -1, 1, 5, 25$. The value of $n$ corresponding to each of those options would be $1, 5, 25, -25, -5, -1$. Now we just need to guess and check!

• If $p=1$, $q=6$, $m=-25$, $n=1$, then the sum is $\frac{-25}{1} + \frac{1}{6} = \frac{-149}{6} \ne \frac{5}{6}$.

• If $p=1$, $q=6$, $m=-5$, $n=5$, then the sum is $\frac{-5}{1} + \frac{5}{6} = \frac{-25}{6} \ne \frac{5}{6}$.

• If $p=1$, $q=6$, $m=-1$, $n=25$, then the sum is $\frac{-1}{1} + \frac{25}{6} = \frac{19}{6} \ne \frac{5}{6}$.

• If $p=1$, $q=6$, $m=1$, $n=-25$, then the sum is $\frac{1}{1} + \frac{-25}{6} = \frac{-19}{6} \ne \frac{5}{6}$.

• If $p=1$, $q=6$, $m=5$, $n=-5$, then the sum is $\frac{5}{1} + \frac{-5}{6} = \frac{25}{6} \ne \frac{5}{6}$.

• If $p=1$, $q=6$, $m=25$, $n=-1$, then the sum is $\frac{25}{1} + \frac{-1}{6} = \frac{149}{6} \ne \frac{5}{6}$.

• If $p=2$, $q=3$, $m=-25$, $n=1$, then the sum is $\frac{-25}{2} + \frac{1}{3} = \frac{-73}{6} \ne \frac{5}{6}$.

• If $p=2$, $q=3$, $m=-5$, $n=5$, then the sum is $\frac{-5}{2} + \frac{5}{3} = \frac{-5}{6} \ne \frac{5}{6}$.

• If $p=2$, $q=3$, $m=-1$, $n=25$, then the sum is $\frac{-1}{2} + \frac{25}{3} = \frac{47}{6} \ne \frac{5}{6}$.

• If $p=2$, $q=3$, $m=1$, $n=-25$, then the sum is $\frac{1}{2} + \frac{-25}{3} = \frac{-47}{6} \ne \frac{5}{6}$.

• If $p=2$, $q=3$, $m=5$, $n=-5$, then the sum is $\frac{5}{2} + \frac{-5}{3} = \frac{5}{6}$ — we’re finally done!

So the factored form is $-6 (x - \frac{5}{2})(x + \frac{5}{3})$.

You might have noticed that this exercise was really tedious. In fact, we will find that there is a more direct way to figure out the numbers we want. Nevertheless, this primitive method can be useful sometimes when the solutions are less difficult to find.

We will introduce an additional form, besides standard form and factored form, for a quadratic polynomial. This new form will be called vertex form.

The motivation for vertex form is that the graph of any quadratic polynomial will look like a parabola, either opening upwards or downwards based on the leading coefficient. All parabolas have either a minimum or a maximum $y$-value, which is attained at exactly one point. This point is called the vertex.

The vertex form of a quadratic polynomial is: $a(x - x_0)^2 + y_0$, where $x_0$ and $y_0$ are real numbers such that $(x_0, y_0)$ are the coordinates of the vertex.

#### Exercise 20: Vertex Form to Standard Form

Rewrite $4(x + 1)^2 - 9$ in standard form.

##### Solution

We expand the square using distributivity: \begin{aligned} 4(x+1)^2 - 9 &= 4(x+1)(x+1) - 9 \\ &= 4(x^2 + 2x + 1) - 9 \\ &= 4x^2 + 8x + 4 - 9 \\ &= 4x^2 + 8x - 5 \end{aligned} which is in standard form.

#### Exercise 21: Solving a Quadratic in Vertex Form

Solve $4(x + 1)^2 - 9 = 0$.

First, add $9$ to both sides: $4(x+1)^2 = 9$

Then, divide both sides by $2$: $(x+1)^2 = \frac{9}{4}$

We know that there are two possibilities for $x+1$, i.e. $x + 1 \in \left\{\frac{-3}{2}, \frac{3}{2}\right\}$

Therefore, the two possibilities for $x$ are $x \in \left\{\frac{-5}{2}, \frac{1}{2}\right\}$

Therefore, we can solve quadratic equations if we put them in vertex form. Based on whether the parabola opens upwards or downwards, and where the vertex is relative to the $x$-axis, the equation will either have $0$, $1$, or $2$ real solutions.

Now the question becomes: given a quadratic polynomial in standard form, can we put it in vertex form? Yes, we can! We just need to use distributivity in a clever way. We know that $(x - x_0)^2 = x^2 - 2x_0 x + {x_0}^2$. Therefore, we want to get something which looks like this by manipulating the standard form expression. Let’s start with a concrete example:

\begin{aligned} 4x^2 + 8x - 5 &= 4\left(x^2 + 2x\right) - 5 && \text{so we need }x_0 = 1 \\ &= 4\left(x^2 + 2x + \textcolor{blue}{1 - 1}\right) - 5 \\ &= 4\left(x^2 + 2x + 1\right) - 4 - 5 \\ &= 4{(x + 1)}^2 - 9 \end{aligned}

So in fact, it is possible to rewrite a standard form quadratic polynomial into vertex form, and this time we did not need any trial and error. In general:

\begin{aligned} ax^2 + bx + c &= a\left(x^2 + \frac{b}{a}x\right) + c \\ &= a\left(x^2 + 2\frac{b}{2a}x + \textcolor{blue}{\frac{b^2}{4a^2} - \frac{b^2}{4a^2}}\right) + c \\ &= a\left(x^2 + 2\frac{b}{2a}x + \frac{b^2}{4a^2}\right) - \frac{b^2}{4a} + c \\ &= a{\left(x + \frac{b}{2a}\right)}^2 - \frac{b^2}{4a} + c \\ &= a{\left(x - \frac{-b}{2a}\right)}^2 - \frac{b^2 - 4ac}{4a} \end{aligned}

We can furthermore find the roots of this quadratic polynomial in vertex form, using the method above:

\begin{aligned} ax^2 + bx + c &= 0 \\ a{\left(x - \frac{-b}{2a}\right)}^2 - \frac{b^2 - 4ac}{4a} &= 0 \\ a{\left(x - \frac{-b}{2a}\right)}^2 &= \frac{b^2 - 4ac}{4a} \\ {\left(x - \frac{-b}{2a}\right)}^2 &= \frac{b^2 - 4ac}{4a^2} \\ x - \frac{-b}{2a} &= \pm \frac{\sqrt{b^2 - 4ac}}{2a} \\ x &= \frac{-b}{2a} \pm \frac{\sqrt{b^2 - 4ac}}{2a} \\ x &= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\ \end{aligned}

This formula gives us a way to find the solutions of any quadratic polynomial in standard form. We will call it the quadratic formula.

### Polynomials

Let us first define two terms that will be useful. A monomial is a whole number power of a variable $x$ multiplied by some coefficient. For example, the following are monomials:

• $x$

• $3x^2$

• $7$

• $-\frac{x^{10}}{4}$

The following are not monomials:

• $2^x$

• $x + 2$

• $x^{-1}$

A polynomial in a single variable $x$ is an expression that involves the sum of some monomials (maybe just one). All monomials are also polynomials. For example, the following are polynomials:

• $0$

• $x + 2$

• $8 + x^2 - 4x^7$

The following are not polynomials:

• $\frac{x+1}{x-1}$

• $1 + 3^x$

A polynomial equation in a single variable $x$ is an equation that has polynomials on the left and right hand sides. For example, the following are polynomial equations:

• $7x = 3$

• $1 - x - x^2 = \frac{1}{5}$

• $x^2 = x^7$

In particular, all quadratic equations and linear equations (which we saw in the previous few weeks) are also polynomial equations.

The degree of a polynomial is the highest exponent (with a non-zero coefficient). For instance, the following polynomials correspond to degrees:

• $0x^2 + x$ → 1

• $5$ → 0

• $x^{99} - x^{199}$ → 199

We will define the degree of $0$ to be $-∞$, because there is no term with a a non-zero coefficient.

• $0$ → 1

The leading coefficient is the coefficient of the highest exponent term. The constant term is the coefficient of the $x^0$ term, i.e. the monomial which does not depend on $x$ (hence, constant).

• $0x^2 + x$ → Leading coefficient: $1$, constant term: $0$

• $5$ → Leading coefficient: $5$, coconstant term: $5$

• $x^{99} - x^{199}$ → Leading coeffcient: $-1$, coconstant term: $0$

### Factoring Polynomials

We have already seen how to solve polynomial equations of degrees $1$ and $2$ (they are linear equations and quadratic equations respectively). To solve polynomial equations with higher degrees, we would ideally want to write the equation in factored form (just like what we did with quadratic equations). There are a few tools we can use to do this.

The first tool we will need to learn is called long division. The idea behind this technique is that if we know one factor of a polynomial, we can get the other factor. This is similar to the concept of division for rational numbers that we are familiar with. In fact, the technique is exactly the same, except that we use the exponents of the variables instead of place values. The idea is to consider only the leading coefficient (coefficients corresponding to the highest degree monomial) each time. Let us do some examples of long division.

#### Exercise 22: Examples of Polynomial Long Division

Use polynomial long division to compute the following quotients.

1. $\frac{x^2 - 1}{x + 1}$

2. $\frac{x^3 - 27}{x^2 + 3x + 9}$

##### Solution
1. We write \begin{aligned} x^2 - 1 &= \textcolor{blue}{x(x + 1)} + x^2 - 1 - \textcolor{red}{(x^2 + x)} \\ &= x(x + 1) - x - 1 \\ &= x(x + 1) - \textcolor{blue}{1(x + 1)} - x - 1 - \textcolor{red}{(- x - 1)} \\ &= x(x + 1) - 1(x + 1) \\ &= (x - 1)(x + 1) \\ \end{aligned} and so $\frac{x^2 - 1}{x + 1} = x - 1$.

2. We write \begin{aligned} x^3 - 27 &= \textcolor{blue}{x(x^2 + 3x + 9)} + x^3 - 27 - \textcolor{red}{(x^3 + 3x^2 + 9x)} \\ &= x(x^2 + 3x + 9) - 3x^2 - 9x - 27 \\ &= x(x^2 + 3x + 9) - \textcolor{blue}{3(x^2 + 3x + 9)} - 3x^2 - 9x - 27 - \textcolor{red}{(- 3x^2 - 9x - 27)} \\ &= x(x^2 + 3x + 9) - 3(x^2 + 3x + 9) \\ &= (x - 3)(x^2 + 3x + 9) \\ \end{aligned} and so $\frac{x^3 - 27}{x^2 + 3x + 9} = x - 3$.

We now know what to do when we have a factor of the polynomial. How do we figure out The next tool that will be useful is called the remainder theorem. In fact, the remainder theorem is more general (it tells us more) than the version we will look at, but the version we will learn is enough for our purposes.

The remainder theorem states that if $p(x)$ is a polynomial, and $a$ is some number, then $p(a) = 0$ if and only if $(x - a)$ is a factor of $p(x)$. This means that if we can guess a root of a polynomial, then we can find at least one factor of it. Let us look at a few examples.

#### Exercise 23: Applications of the Remainder Theorem

Find a single factor of each polynomial by using the remainder theorem.

1. $x^7 + 1$

2. $x^4 + x^3 + x^2 + x$

3. $10x^{10} - 9x - 1$

##### Solution
1. ${(-1)}^7 + 1 = 0$, so by the remainder theorem, $x + 1$ is a factor.

2. $0^4 + 0^3 + 0^2 + 0 = 0$, so by the remainder theorem, $x$ is a factor.

3. $10\cdot 1^{10} - 9\cdot 1 - 1 = 0$, so by the remainder theorem, $x - 1$ is a factor.

By combining the remainder theorem with the technique of long division, we can factor some polynomials for which we can easily guess a root. But we will want yet a third tool to make guessing roots easier, if they are rational. This third tool is called the rational root theorem. It allows us to use trial and error to guess all the possible rational roots; if none of them work, then we know there are no rational roots. (Recall that we did something similar to find the rational roots of a quadratic equation.)

The rational root theorem states that if we have a polynomial $p(x)$ with integer coefficients, and the leading coefficient is $a$ while the constant term is $c$, then any rational root $q = \frac{m}{n}$ (in lowest form) must satisfy the following: $m$ is a factor of $c$, and $n$ is a factor of $a$. As a special case, if $a = 1$ (so that t