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.

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:

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.

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.

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:

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.

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:

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”.

Addition and Subtraction

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:

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.

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.

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.

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.

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.

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.

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.

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 $q$.

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.

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.

Two Simple Quadratics

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 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.

Factoring Quadratics

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.

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.

Quadratic Formula

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.

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.

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.

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 the polynomial is monic) then the rational root theorem is called the integral root theorem, and we have that any rational root is an integer and is a factor of $c$. (You might remember the integral root theorem from a homework question a few weeks ago.)

We can now combine our knowledge of the three tools we have learned to factor polynomials:

1. If the polynomial is linear or quadratic, we already have tools to find the roots, and thus factor them (if possible) using the remainder theorem.

2. If the polynomial has integer coefficients, we can apply the rational root theorem to find all rational roots. This gives us, by the remainder theorem, some linear factors. We then can use long division to find what is left, and then repeat.

3. Whatever polynomial we have left has no rational roots. This means it is already fully factored into rational polynomials, but there might still be real roots which are not rational. If the polynomial is quadratic, then the quadratic formula can tell us what these roots are, and thus we can fully factor into real polynomials.

Linear Algebra

Previously, we have been looking at polynomial equations with one unknown variable. Recall the simplest such equations are linear equations of the form $ax = b$, where $a$ and $b$ are known real (or rational) numbers.

We will now extend our attention to polynomial equations when we have multiple unknown variables. For example, suppose that neither $x$ nor $y$ are known, and we have a linear equation $ax + by = c$. What are the solutions to this linear equation?

One way to solve this problem is to choose any value for $x$, and then we see that $ax$ becomes a known value. For instance, suppose $x = 1$. Thus we can now solve the linear equation $by = c - a$ where the right hand side is known. We can see that in this case, if $b \ne 0$, then $y = \frac{c - a}{b}$. If $b = 0$, then there are either no solutions or infinitely many solutions for $y$, depending on whether $c - a = 0$ also.

However, the number of possible values to choose for $x$ is very high — in fact, usually there are infinitely many possible values for $x$ that work! This implies that we are not looking for only a single solution, but instead we want to describe the shape of all possible solutions. We have figured out one of the solutions, so if we can describe how the solution changes when we change the assumed value of $x$, then this would be enough. The result will be some geometric shape that passes through our particular solution.

In order to visualize these shapes geometrically, we shall introduce the concept of a vector, which previously we have touched on but not discussed in detail.

Vectors

An ordered pair is two things written in an order. For example, $(3, 4)$ is an ordered pair of numbers. Ordered pairs frequently represent a single concept that is made of two components — though keep in mind that these components are not always written as in this example.

A simple ordered pair like $(3, 4)$ does not itself have much meaning, aside from being a collection of two numbers. However, we may assign an interpretation to particular ordered pairs to give them a meaning.

We saw earlier that fractions, like $\frac{a}{b}$, are made up of two components (which we called the numerator and denominator). Indeed, fractions are a kind of ordered pair $(a, b)$, with the first element of this ordered pair representing the number of fractional pieces, and the second element representing the size of a whole relative to a single fractional piece.

Another interpretation of ordered pairs is as vectors in a two-dimensional plane. The components of the vector $[a, b]$ represent the displacement in two directions. For example, the first component might represent displacement to the right, and the second component displacement toward top of the page. (Such an interpretation is called a vector space, and the choices of directions are collectively called a basis.)

We have notation for the set of two-dimensional vectors where both components are real numbers: $\mathbf{R}^2$. The superscript $^2$ denotes that the vector space is two dimensional, i.e. has two components.

It will become more clear later why we have defined this concept, which currently seems to be just a fancy way of writing down two numbers at once. First, we shall define some operations on vectors, which will have geometric meaning:

• Let $[a, b] + [c, d] = [a + c, b + d]$ be the sum of two vectors. The geometric meaning is that we displace ourselves first by $a$ in the first basis direction and by $b$ in the second basis direction. Then, we displace ourselves from the new location by $c$ in the first basis direction and by $d$ in the second basis direction.

• Let $a[b, c] = [ab, ac]$ be a scalar multiplication of a vector by real (rational) number $a$, which is called a scalar because it is not a vector. The geometric meaning is that we displace ourselves in the same direction (if $a > 0$) or in the opposite direction (if $a < 0$) as $[b, c]$, but with a step size of $a$ times.

Homogeneous Linear Equations (Two Unknowns)

Let us now return to considering the linear equations with two unknown variables. A homogeneous linear equation is an equation of the form $ax + by = 0$. Note that the only change from the more general linear equations that we have looked at is that we are forcing the $c$ in $ax + by = c$ to be $0$. For simplicity, let us assume $a \ne 0 \ne b$ (this assumption is not really required, but if $a = 0$ or $b = 0$, then this case is degenerate because we know that one of the unknown variables is not important, so we can just solve a linear equation with the other unknown). What do the solutions to this kind of equation look like? One solution that comes to mind is $[x, y] = [0, 0]$, but this solution does not tell us where to look for other solutions. Therefore, it would be useful to find solutions that are not $[0, 0]$. We shall investigate first a concrete example.

This exercise generalizes to an important property: all solutions to a linear homogeneous equation in two variables (where the coefficients are not both zero) are scalar multiples of a particular non-zero solution. We can plot these as a straight line going through the origin. The particular non-zero solution we found is any arbitrary direction on this line.

Vector Spaces

Recall from before that we saw that the solutions to a homogeneous linear equation in two variables are all scalar multiples of a particular non-zero solution. We will now increase the number of variables and seek to find a similar result. First, a useful concept is that of a vector space.

A rational (or real) vector space is a set of vectors $V$, possibly infinite, such that:

1. A zero vector is in the set, i.e. $\mathbf{0} \in V$

2. For any two vectors $u$, $v$ in the set, their sum $u+v$ is in the set

3. For any vector $v$ in the set, all rational (or real, if we are talking about a real vector space) scalar multiples $qv$ are in the set

In other words, within a vector space we can always add vectors, or multiply vectors by rational (or real) numbers, without leaving the vector space. For instance, all two-dimensional rational vectors $V = \mathbf{Q}^2 = \{[a, b]: a\in \mathbf{Q}, b\in\mathbf{Q}\}$ form a vector space (we can add and subtract these without leaving $V$). Another rational vector space is the singleton set $V = \{[0, 0]\}$. There is only one vector in here (the zero vector), but any combination of sums and scalar multiples with the zero vector will result in a zero vector, so this is still a vector space.

However, the set $U = \{[0, 0], [1, 0]\}$ is not a vector space. This is because, for example, the sum $[1, 0] + [1, 0] = [2, 0]$ is not in $U$.

What Exercise 27 tells us is that when homogeneous linear equations in two variables have solutions, those solutions form a vector space. In fact, we do not have to limit ourselves to two variables. It is an important fact that the set of solutions to homogeneous linear equations in any number of unknown variables always forms a vector space: we can always add solutions, or multiply them by rational numbers, and get another solution.

We can often describe vector spaces by giving just a few vectors, and saying that all the other vectors are going to be linear combinations: i.e. sums and scalar multiples, and sums of scalar multiples. If $u \in V$ and $v \in V$, we can see that all vectors that look like $au + bv \in V$ also (this is just by applying the 2nd and 3rd rules together). In these cases, we will write that $V = \operatorname{span}(u, v)$, to mean that all vectors in $V$ are in fact linear combinations of $u$ and $v$. Hence, we now have a brief notation to describe the solution to homogeneous linear equations.

Inhomogeneous Linear Equations

We have spent a lot of time talking about homogeneous linear equations, where the right hand side is $0$. The solution to homogeneous linear equations is always a vector space. Now we will shift our attention to linear equations in general, including those where the right hand side is not $0$. How do we solve such systems? Let us start with an example, $x + y = 2$. There is an obvious solution $[x, y] = [1, 1]$, but of course there are infinitely many other solutions. How can we describe them?

The most important observation is that this inhomogeneous equation is related to the homogeneous linear equation $x + y = 0$. Let the vector space of solutions to this equation be $V$. Any homogeneous solution $[a, b] \in V$ can be added to $[1, 1]$ to get another solution to $x + y = 2$! That is, $a + 1 + b + 1 = (a + b) + (1 + 1) = 0 + 2 = 2$. Here we see why homogeneous linear equations are important: we can always translate by a homogeneous solution to $x + y = 0$ without changing the value of $x + y$.

Therefore, we can describe all the solutions to $x + y = 2$ by simply saying $[x, y] = [1, 1] + V$. What does this mean? We mean that we start at $[1, 1]$ (some particular solution), and then translate by any solution in $V$. Since the solution to the homogeneous equation is $[x, y] \in V = \operatorname{span}([1, -1])$, another way to write it is: $x + y = 2 \iff [x, y] = [1, 1] + q[1, -1] \text{ for some }q \in\mathbf{Q}$

Linear Systems

We are now ready to combine our knowledge to solve systems of equations. A system of equations consists of multiple equations which may share unknowns, and all the equations are simultaneously true. For example: $\begin{aligned} 2x + 3y &= 7 \\ 3x + 2y &= 8 \\ \end{aligned}$ is a system of equations with two unknowns ($x$ and $y$) and two equations.

When each equation is a linear equation, we call the system a linear system. In general, the solutions to linear systems are like the solutions to linear equations: there can be no solutions, or the solutions can be described as a particular solution plus a vector space. Therefore, we should expect to be able to solve such systems in a very similar way to how we solved individual linear equations.

First, we want to rewrite the system as a homogeneous linear system, which means to set the right hand sides to $0$: $\begin{aligned} 2x + 3y &= 0 \\ 3x + 2y &= 0 \\ \end{aligned}$

To solve this homogeneous linear system, we can take the intersection of the two vector spaces determined by each equation (the intersection of two vector spaces which share a $0$ is also a vector space). In this case, the first equation is solved by the vector space $\operatorname{span}([3, -2])$, and the second equation is solved by the vector space $\operatorname{span}([2, -3])$. The intersection of these vector spaces is the zero-dimensional vector space $\{[0, 0]\}$.

We now need to find a particular solution to the system. One way to do this is by substituting each equation into the next, and creating new equations if it is necessary. In this case, we can rearrange $2x + 3y = 7$ into $y = -\frac{2}{3}x + \frac{7}{3}$. Substituting this into the second equation gives $3x - \frac{4}{3}x + \frac{14}{3} = 8$, and simplifying this gives us $\frac{5}{3}x = \frac{10}{3}$, which is a linear equation easily solved for $x$ — $x = 2$.

Now we can substitute this back into the equation $y = -\frac{2}{3}x + \frac{7}{3}$ to find that $y = \frac{3}{3} = 1$. Thus, a particular solution is $[1, 2]$. Since the vector space that solves the homogeneous linear system consists of just the zero vector, that means this particular solution is the unique solution; i.e. $x = 2$ and $y = 1$ is the only rational solution to the system.