## Understanding Exponents

When working with exponents, there are two new words that come up: exponent and base. As an example, in $x^2$, 2 is the exponent while $x$ is the base and in $4^{12}$, 12 is the exponent while $4$ is the base.

As long as the exponent is a positive whole number, you can think of it as telling you how many times you should multiply the base by itself.

$2^3=2\cdot2\cdot2=8$
$x^3=x\cdot x\cdot x$
$5^2=5\cdot5=25$

As you can see,  $4^{12}$ means “4 multiplied by itself 12 times” and that is a really big number. Many values that you calculate from exponents will be quite large.

There are a couple of things to note:

• Anything with an exponent of zero is defined to be 1. So,  $4^{0}=1$,  $100^{0}=1$.
• Ok right above, when I said anything, I meant “almost anything”. $0^0$ is an indeterminate form. For everything else however, the rule above holds.
• An exponent of 1 is the same as just writing the number by itself:  $4^{1}=4$.

## Negative Exponents

Negative exponents are treated a bit differently. By definition, if we have a positive number n and a nonzero b, then the following rule holds:

$b^{-n} = \dfrac{1}{b^n}$

Using this rule, if you see a value with a negative exponent, then it can be rewritten as 1 “over” that term with a positive exponent. Here are a few examples of how this works.

$3^{-2} = \dfrac{1}{3^2} = \dfrac{1}{9}$
$10^{-1} = \dfrac{1}{10^1} = \dfrac{1}{10}$
$(-2)^{-4} = \dfrac{1}{(-2)^4} = \dfrac{1}{16}$

As you can see, it is simply about following the formula when dealing with negative exponents. After this, you can simply calculate the value using the rules above.

## Fractional Exponents

In algebra and more advanced math, it is very common to see an exponent of one-half or some other fraction. This is one way to represent terms involving roots, or radical terms. By definition (if n is nonzero):

$b^{\frac{m}{n}}= \sqrt[n]{b^m}$

Working with these types of terms (radical terms) is a more advanced topic and will be the subject of a different article, but it is important that you be able to rewrite such terms using the definition as below.

$3^{\frac{1}{2}}=\sqrt{3^1}=\sqrt{3}$
$4^{\frac{2}{3}}=\sqrt[3]{4^2} = \sqrt[3]{16}$

As you study terms involving exponents, you will find that you need to combine rules quite often. For example, the rule above can be combined with the rule for negative exponents to simplify even more complex expressions.

## Using a Calculator

Finding the value of terms with exponents on scientific or graphing calculators is quite easy. The usual key you will need will use a carat symbol ^ to represent that the next number is an exponent. So, $3^4$ can be represented as 3^4 in most calculators. On the TI83 or TI84:

You can also always use the carat symbol on google (just type it into google! they now do automatic calculations for you) or wolfram alpha.

## Understanding y-Intercepts

It doesn’t matter what function you are looking at, the y intercept is the point at which the graph of that function crosses the y-axis. That means the point can always be written (0, c) for some number c.

The y-intercept is at y = -1. As a point, we would write (0, -1)

### Finding y Intercepts

Along the y-axis, the x value of any point is zero. This means that to find the y intercept of any function, we can just let x = 0 and solve for y. For example, consider the function $y=\frac{1}{3}x+4$. If we let $x = 0$, then $y=\frac{1}{3}(0)+4 = 4$. As a point, this would be the point (0, 4).

Similarly, the function in the graph at the top of the page is, $y=x^2-1$. When $x = 0$, $y = 0^2 - 1 = -1.$, meaning the y intercept is -1, the same as on the graph. If this is still confusing, in the video to the right, I go through a couple more examples.

### A Closer Look at the y Intercept

Now that we have seen how to find them, there are two interesting questions that can come up:

1. Can a function have more than one y intercept?
2. Can a function have no y intercept?

In answering these, remember that a function can only have one output (y-value) for each input (x-value). If a function has more than one y intercept, that means that there are two y’s for the input $x = 0$. This isn’t possible, so no, it is not possible for a function to have more than one y-intercept.

What about no y intercept? Well, take a look at the graph of $y=\frac{1}{x}$ below. It never crosses the y-axis, so it has no y-intercept.

This function has no y-intercept.

What happened? If you let $x=0$, you get $y = \frac{1}{0}$ which is undefined. This is just one example, but there are other similar functions, so yes, it is certainly possible for a function to not have a y-intercept.

### Using a Graphing Calculator

Graphing calculators can be used to find x and y intercepts pretty quickly. A while back, I made a short video that can show you how this works. You can find it on the mathbootcamps youtube page.

Well, you can take a look at my discussion on x intercepts right over here!

## Understanding x Intercepts

Given the graph of any function like y = x – 2, the x intercept is simply the point where the graph crosses the x-axis. There might be just one such point or many. In either case, the idea stays the same. Another way to think about this is, the x intercept is the point on the graph where y = 0. This can be used to easily find the x intercept for any graph. In this guide, we will talk about how to find x intercepts using a graph, an equation, and a graphing calculator.

### Finding the x-intercept or intercepts using a Graph

As mentioned above, functions may have one, zero, or even many x intercepts. These can be found by looking at where the graph of a function crosses the x-axis. This point (or these points) is graphed for each of the functions below.

Graph of $y = x - 2$

The x intercept is 2

Graph of $y = x^2 + 1$

The x intercepts are -1 and 1.

Graph of $y = x^4-8x^3-49x^2+260x+300$

The x intercepts are -6, -1, 5, and 10.

### How to Find x Intercepts

When the graph of a function is crossing the x-axis, what does the point look like as an ordered pair? For instance, is the x-intercept is said to be 5, what point is this? Well, since there is no “height” to that point, we would write it as (5, 0). In fact, any x-intercept would have a 0 as the y-value.

We can use this to find the x-intercept of intercepts of any function if we have the equation. The general rule is let y = 0 and solve for x. How easy this is depends entirely on the equation!

As an example, let’s use the first equation graphed above: $y = x - 2$. If we let $y = 0$, then we get the equation $0 = x - 2$ This equation can be solved by using the usual rules for solving linear equations: we just add 2 to both sides and find the x intercept is $x = 2$.

Things are a little more complicated with the second example. If $y = 0$ in the equation $y = x^2 - 1$, then we get the quadratic equation $0 = x^2 - 1$. Adding 1 to both sides, we find that $x^2 = 1$ which has solutions 1 and -1 by the square root rule.

Of course, the idea remains the same even when we get to the last (much more complicated) equation $y = x^4-8x^3-49x^2+260x+300$. Setting y to zero and solving for x would give us all of the x-intercepts we see on the graph. However, this is a bit more algebra than we need for this article so we will save that equation for another day. Instead, if you need a few more examples, I have gone through some different types of functions in this video from the mathbootcamps youtube!

### Finding x-intercepts on a Graphing Calculator

This is a bit more involved, but on this old youtube video I go through how to find both x and y intercepts using a TI 84.

## Visualizing Matrix Multiplication

Matrix multiplication is just one of those things that is not intuitive – at least not at first. You have just had so many years of multiplication meaning one thing and then you have this entirely new definition to work with! It certainly takes some getting used to. (and if you continue to study advanced math – get used to that idea of “getting used to” things)

When I first studied matrix multiplication, I had an “aha!” moment that really helped. What was this?

Rows hit columns.

The first row of the first matrix hits every column and fill up rows in the new matrix. This process is repeated until you run out of rows in the first matrix. This is a bit difficult to imagine so I have created a small animation to help (there is no audio – just the animation):

This is really a way of visualizing the dot-product definition of matrix multiplication. For instance, when the first row of matrix 1 “hits” the first column of matrix 2, we see the sum 1(5)+2(7) results. This comes from the following definition:

$\left[ \begin{array}{c} a\\ b\\ \end{array} \right] \left[ \begin{array}{cc} c & d\\ \end{array} \right] = a(c)+b(d)$

The next entry is therefore 1(6)+2(8). At this stage, you have run out of columns to “hit” and the next row in the first matrix is used.

As you continue to study matrices, you will likely find that this way of thinking about matrix multiplication works whether we are looking at multiplying two square matrices as above or more “odd” shaped matrices like those that can come up in linear algebra and similar courses.

## Sketching Quick and Accurate Graphs of Linear Equations

While there are plenty of methods to get a good hand drawn graph of a linear equation, many plot more points than necessary and (if you aren’t careful) misrepresent the location of two important points: the x and y intercepts.

Using the intercepts to plot your line takes care of both of these problems and more importantly: its fast! In this video, I will use two examples to show you exactly how it’s done.

http://www.mathbootcamps.com/media/fastgraphsoflinearequations.mp4

## Why Can’t You Cancel the x?

At this point, I haven’t even written out any equation or expression and you already know what I’m talking about. There are problems that come up in algebra where it seems like you should be able to cancel a variable out and move on. Yet, in these very situations your professor keeps telling you it can’t happen!

It may not always seem like it, but there really is a pattern and a simple rule of when you can and can’t cancel a term or a variable. Let’s focus on rational expressions specifically. Those are expressions (like the one below) where both the numerator and the denominator are polynomials.

$\dfrac{x+5}{x^2+1}$

### When can you cancel terms in a rational expression?

You can cancel any matching factors that occur in both the numerator and the denominator. Let’s use numbers to understand this a little better:

$\dfrac{4}{14}$

In the fraction above, 2 is a factor of 4 since 2 x 2 is 4. Similarly, 2 is a factor of 14 since 2 x 7 = 14. Based on my rule, I should be able to cancel the 2’s and get a fraction that is still the same thing! Let’s see:

$\dfrac{4}{14}=\dfrac{2(2)}{2(7)}=\dfrac{2}{7}$

If you check on a calculator you will find that 4 divided by 14 is the same as 2 divided by 7 – so this rule seems to be working here. What is really going on?

### Why does this work?

Every number other than zero, when divided by itself is 1. Since a fraction simply represents division (in one sense), $\dfrac{2}{2} = 1$ and we really just have $1\times\dfrac{2}{7}$.

### Trying it with Variables

Now that everything makes some sense with numbers, let’s try variables. For us, the variables represent numbers so they will behave in the same way. Take for instance the expression

$\dfrac{x^2-x}{x^3}, x\neq 0$

The numerator has a couple of factors which we can find by factoring out an x (doesn’t that phrase make a lot more sense now?!): $x^2-x = x(x-1)$. This shows us that $x$ and $x-1$ are factors of the numerator. Looking at the denominator, we see $x$ is also a factor since $x^3=x(x^2)$. Remember, when both the numerator and the denominator share a factor, you can cancel that factor since it is really just 1.

$\dfrac{x^2-x}{x^3} = \dfrac{x(x^2-1)}{x(x^2)}= \dfrac{x^2-1}{x^2}$

The only time you can’t cancel terms in the numerator and denominator is when they are both NOT factors.. That’s why you can’t cancel $x$ in $\dfrac{x-5}{x}$. The x is not a factor of the numerator; its just a term being added. Cancelling the $x$ here would be like cancelling the $5$ in $\dfrac{5+1}{5}$ and saying that is 1. If you check it in a calculator it certainly isn’t!

A quadratic equation is any equation that can be written as $ax^2+bx+c=0$, for some numbers a, b, and c (a is nonzero). There are many ways to solve equations like this, including graphing, factoring, and using technology. One other method is using the quadratic formula.

Looking at the formula, you can see that a, b, and c are the numbers straight from your equation. Applying this formula is really just about determining the values of a, b, and c and then simplifying the results. Let’s take a look at a couple of examples.

### Example

Solve: $x^2-x=6$

Before we do anything else, we need to make sure that all the terms are on one side of the equation. As you can see above, the formula is based on the idea that we have 0 on one side.

Subtract 6 from both sides to get:

$x^2-x-6=0$

Now that we have it in this form, we can see that:

$a=1$ , $b= -1$ , $c=-6$

Why are $b$ and $c$ negative? The formula is based off the form $ax^2+bx+c=0$ where all the numerical values are being added and we can rewrite $x^2-x-6=0$ as $x^2 + (-x) + (-6) = 0$.

To keep it simple, just remember to carry the sign into the formula.

Once you have the values of a, b, and c, the final step is to substitute them into the formula and simplify.

$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}=\dfrac{-(-1)\pm\sqrt{(-1)^2-4(1)(-6)}}{2(1)}$

$=\dfrac{1\pm\sqrt{1+24}}{2}$

$=\dfrac{1\pm\sqrt{25}}{2}$.

At this stage, the plus or minus symbol ($\pm$) tells you that there are actually two different solutions:

$\dfrac{1+\sqrt{25}}{2}=\dfrac{1+5}{2}=\dfrac{6}{2}=3$

and

$\dfrac{1- \sqrt{25}}{2}=\dfrac{1-5}{2}=\dfrac{-4}{2}=-2$

Therefore the final answer would be:

$x=3\;,-2$

This particular quadratic equation could have been solved using factoring instead, and so it ended up simplifying really nicely. Often, there will be a bit more work – as you can see in the next example.

### Example

Solve: $2x^2+2x-7=0$

This time we already have all the terms on the same side. So, we just need to determine the values of a, b, and c.

$a=2$ , $b=2$ , $c=-7$

Applying the formula:

$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}=\dfrac{-2\pm\sqrt{(2)^2-4(2)(-7)}}{2(2)}$

$=\dfrac{-2\pm\sqrt{4+56}}{4}$

$=\dfrac{-2\pm\sqrt{60}}{4}$

$=\dfrac{-2\pm 2\sqrt{15}}{4}$

Notice that $2$ is a FACTOR of both the numerator and denominator, so it can be cancelled.

$=\dfrac{-1\pm\sqrt{15}}{2}$

This answer can not be simplifed anymore, though you could approximate the answer with decimals. Therefore the final answer is:

$x=\dfrac{-1+\sqrt{15}}{2}$,$\dfrac{-1-\sqrt{15}}{2}$

## Complex Solutions

When using the quadratic formula, it is possible to find complex solutions – that is, solutions that are not real numbers but instead are based on the imaginary unit, i. Recall the following definition:

$i = \sqrt{-1}$

If a negative square root comes, your equation has complex solutions which can be written in terms of i.

### Example

Solve: $x^2-2x+5=0$

Just as in the previous example, we already have all the terms on one side. So, we will just determine the values of a, b, and c and then apply the formula.

$a=1$ , $b=-2$ , $c=5$

Applying the formula:

$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}=\dfrac{-(-2)\pm\sqrt{(-2)^2-4(1)(5)}}{2(1)}$

$=\dfrac{2\pm\sqrt{4-20}}{2}$

$=\dfrac{2\pm\sqrt{-16}}{2}$

A negative value under the square root means that there are no real solutions to this equation. However, there are complex solutions. Using the definition of i, we can write:

$\sqrt{-16} = 4i$

This allows us to keep simplifying.

$=\dfrac{2\pm 4i}{2}$

$=1 \pm 2i$

The solutions to this quadratic equation are:

$x=1+2i$ , $1 - 2i$

There are three cases with any quadratic equation: one real solution, two real solutions, or no real solutions (complex solutions). Each case tells us not only about the equation, but also about its graph – something we will study more in a later article!

## A Quadratic Equation Solver from Wolfram Alpha

To use this solver, your quadratic equation should be written in the form $ax^2+bx+c=0$. Simply input the coefficients (a,b,c) into the space below! While it will show you the steps, be aware that wolfram alpha (the program behind this widget) likes to complete the square for just about everything which may not be the most straightforward way to get your answer.

(I will probably add cool wolfram alpha widgets here over time but if you are looking for one for your class of website head on over this way: http://www.wolframalpha.com/widgets/

## Understanding The Laws of Exponents

The laws of exponents (sometimes called the rules of exponents) allow us to simplify expressions by using our understanding of what an exponent means in much the same way that combining like terms uses the idea of what multiplication means. In both cases, the idea is to write algebraic expressions in the most simple way possible.

As a brief reminder, an exponent tells you how many times to multiply a number, variable or expression times itself. For example:

$x^4=x \cdot x\cdot x\cdot x$

$(x+5)^3=(x+5)(x+5)(x+5)$

All of the rules require that the terms have the SAME BASE. The base is whatever is being taken to a power. Above, $x$ is the base in the first example and $x+5$ is the base in the second example.

## When multiplying two terms with the same base, add the exponents

$x^{m}x^{n}=x^{m+n}$

### Examples:

• $x^{9}x^{3}=x^{9+3}=x^{12}$
• $y^{4}y^{10}=y^{4+10}=y^{14}$

### Why does this work?

If we have $3^43^2$, this means to multiply $3^4$ by $3^2$. But, $3^4=3\cdot3\cdot3\cdot3$ while $3^2=3\cdot3$. Looking at the original example, $3^43^2 = (3\cdot3\cdot3\cdot3)(3\cdot3)$ and if you count the number of 3’s being multiplied, there are 6. So really, $3^43^2$ is a complicated way to write $3^6$.

## When dividing two terms with the same base, subtract the exponents

$\dfrac{x^m}{x^n}=x^{m-n}$

### Examples:

• $\dfrac{y^7}{y^4}=y^{7-4}=y^3$
• $\dfrac{x^2}{x^5}=x^{2-5}=x^{-3}=\dfrac{1}{x^3}$

### Why does this work?

As an example, think about $\dfrac{5^4}{5^2}$. This can be rewritten by writing out the multiplication $\dfrac{5\cdot5\cdot5\cdot5}{5\cdot5}$. At this point, we can cancel any factors that are shared by the numerator and the denominator and are left with $\dfrac{5\cdot5}{1}=5^2$. The “rule” is essentially doing this work for us by subtracting.

## When taking a “power to a power”, multiply the exponents

$(x^m)^n=x^{mn}$

### Examples:

• $(x^4)^4=x^{16}$
• $(-5x)^2=(-5)^2x^2=25x^2$

### Why does this work?

Let’s use another example: $(8^2)^3$. If an exponent tells us how many times to multiply something by itself, then this means to multiply $8^2$ by itself 3 times: $8^28^28^2$. Now, we have three terms with exponents multiplied that all have the same base. This is the first rule of exponents so we add them: $8^28^28^2 = 8^6$. This rule shortcuts this process.

## Other properties of exponents

When you begin simplifying expressions with exponents, there will more other properties you may find helpful.

• Anything to the 1 power is just the same term back again. $23^1=23$
• Anything other than zero the 0 power is 1. $(x+y+z^{12})^0=1$ (this is not true for $0^0$ which is an indeterminate form)
• Exponents distribute over multiplication and division only! $(xy)^2=x^{2}y^{2}$ but $(x+y)^2 \neq x^2 + y^2$. To find $(x+y)^2$ you must use FOIL.
• Negative exponents and fractional exponents have a specific meaning, which you can review here: http://www.mathbootcamps.com/understanding-exponents/.

## Video Examples

In the video below, you can see how to use the laws of exponents to simplify algebraic expressions.

## How to multiply two binomials: FOIL

A binomial is a polynomial with two terms that are either added or subtracted from each other like $3x+2$ or even something more complicated like $5xy-3x^2$. When multiplying any two polynomials, the ultimate goal is to make sure every term in the first polynomial has been multiplied by every term in the second polynomial. With two binomials, this process can be simplified into an easy to remember process that you have probably heard of: FOIL.

## FOIL: First, outer, inner, last

First: multiply the first terms of each binomial

Outer: multiply the “outer” terms in each binomial

Inner: multiply the “inner” terms in each binomial

Last: multiply the “last” terms in each binomial

If you think about it, this does in fact make sure every term is multiplied by every other term. Take the first term for instance. Once you have FOILed, it has been multiplied by the first term of the second binomial and the last (by the “outer” step). This is true for every term!

Once this process is complete, it is very likely you have some cleaning up to do in the sense of combining like terms, so you should make sure to simplify your answer once you are done.

### Example

Multiply $(3x+1)(2x-3)$

For this first example, we will show every step so you can see how they each work.

First: The first two terms are the $3x$ and the $2x$. Multiplying these will give $(3x)(2x) = 6x^2$.

Outer: The “outer terms” are the $3x$ and the $-3$. Multiplying these terms will give $(3x)(-3) = -9x$.

Inner: The inner terms are the $1$ and the $2x$. Multiplying these terms will give $(1)(2x) = 2x$.

Last: The last terms are the $1$ and the $-3$. Multiplying these terms will give $(1)(-3) = -3$.

The final answer would be the terms we got in each step above added together. The steps all together would look like this:

$(3x+1)(2x-3)$

$= (3x)(2x)+(3x)(-3)+(1)(2x)+(1)(-3)$ (apply FOIL using the steps listed above)

$= 6x^2-9x+2x-3$ (multiply)

$= 6x^2-7x-3$ (simplify)

Let’s try another example, but this time one that looks a little bit different.

### Example

Multiply $(x+y)^2$

If you think about what an exponent means, you can rewrite this as $(x+y)(x+y)$. That means that we can apply FOIL. Note that this is definitely not equal to $x^2 + y^2$!

$(x+y)^2$

$=(x+y)(x+y)$

$=x^2 + xy + xy + y^2$

$=x^2 + 2xy + y^2$

Now you can see where the general rule for squaring binomials comes from. But, there is no reason to memorize that unless you want to. You can always quickly apply FOIL to find the correct answer.

A last but important point: This method will NOT work if one of the parentheses contains more than two terms (in other words if either is not a binomial). For instance, to find $(2x-1)(x^2+5x-1)$, you would need a different method.

Also, if you have more than one binomial, FOIL may not work by itself – you may need to use FOIL on the first two binomials and then some other method depending on the results of that. This is a very nice method but make sure you apply it only when it makes sense!