Working with Polynomials and Monomials, and How to FOIL

Following up on my post explaining Monomials and Polynomials, here I would like to provide a bit more of an explanation about how to actually work with these terms. In particular, first I am going to show you how to multiply a monomial by a polynomial. Then I will similarly show you how to divide a polynomial by a monomial. And then finally, I will explain the very important concept of FOIL for multiplying to binomials. Nothing too difficult, but all key concepts of algebra. Let’s get to it!

How to multiply a Polynomial and a Monomial. It’s actually very easy, probably way easier than it sounds! Let’s step back a bit first, though, and think about how we multiply regular numbers and expressions. Consider the expression (x + y), and say that we want to multiply that by 3. How do we do that? By now, I think this is very straightforward. We first write it out, like this:

(3)(x + y), or 3(x + y)

Next, we multiply each part of the expression (x + y) by 3, and then we can drop the brackets at the end. This is called the Distributive Property, which refers to how we can remove the brackets if we are multiplying (distributing) the one term with all of the terms within the brackets. I don’t think I need to go into more detail than this for this basic operation, but leave me a comment below if you’d like a bit of help with it. Maybe I’ll do a separate post on the Distributive Property. When we’re done, we have this:

3x + 3y

Now let’s extend this method to some more complicated looking terms. Let’s consider multiplying the monomial 5x with the polynomial (3x + 1). I’ll walk through the steps below, and it hopefully will be easy to follow. All we do is abide by the Distributive Property and simply multiply the single term into the brackets with all the other terms.

5x(3x + 1)
5x(3x) + 5x(1)

Now that we have properly distributed the terms, we just have to simplify it and we’re done. The first term becomes 15x2 while the second term becomes simply 5x:

15x2 + 5x

And that is all there is to multiplying a monomial with a polynomial. The key is to follow the Distributive Property and multiply things through where appropriate, and then combine like terms if necessary and simplify!

OK, so now hopefully you followed all that. But what about if we want to go the other way? What if we want to start with a polynomial and then divide by the monomial instead? Again, it’s very similar to operations you are already familiar with. In this case, you might compare it to cancelling and reducing terms. Let’s look at an example and you’ll see what I mean.

Let’s say that we want to divide (8x + 4) by 2. Probably pretty simple, right? We just divide the 2 into all of the terms to reduce them, leaving us with the answer (4x + 1). We can think of this as a big fraction, with the (8x + 4) on the top as the numerator, and the 2 on the bottom as the denominator. Because we are adding the 8x and 4, by the laws of fractions, we can separate everything into two separate fractions with the same denominator: 8x/2 + 4/2. Now, it might be even easier to see how to reduce these to get our final answer.

The same logic applies when we want to divide a polynomial by a monomial. Consider the example of dividing (25x3 + 15x2 – 5x) by 5x. I’ll write it out as a big fraction first (typing these out is not the most obvious way of demonstrating!).

Now, just like with the example above, we can cancel and reduce things. For cancelling, remember that you can only cancel if you are performing the same operation to each term. Or alternately, if you want to split this into the multiple smaller fractions first before simplifying, if that is clearer to you, then that works as well. Either way, you will find that you arrive at the final answer of:

5x2 + 3x – 1

Note that because there is an x term in the denominator, you cancel this out of the numerator for each term by reducing each by x. That is why the powers all reduce by one. The same theory applies to factoring the 5.

Hopefully that explains the two rudimentary concepts of multiplying and dividing monomials and polynomials. Let me know in the comments below if you need any more help!

The last thing I’d like to talk about in this post is FOIL, or how to multiply two binomials. Recall that a binomial is an expression that contains two terms. This is a specific instance of multiplying two polynomials together, though clearly, since polynomials can have many terms, it is best to start simple – so we start here by multiplying two binomials! Larger expressions are not necessarily any harder, though it does become more of a challenge to not mix things up. The key to multiplying any two polynomials is to add up the products of multiplying each term in the first polynomial by each term in the second polynomial.

You have probably heard the expression FOIL when talking about multiplying polynomials, and you will probably see FOIL many times in your studies, so it is best to understand well to what it is referring! FOIL is an acronym that stands for the order of operations that you apply when multiplying two binomials. It only works for binomials, though it still abides by the concept I just outlined for multiplying larger polynomials together. It means FIRST OUTSIDE INSIDE LAST. That probably doesn’t mean much until you see it. It refers to all the pairs of products that you add together when multiplying two binomials.

Here is an example of this:

(2x + 3)(x + 5) = ?

  • First: 2x * x
  • Outside: 2x * 5
  • Inside: 3 * x
  • Last: 3 * 5

Adding these up gives:
2x2 + 10x + 3x + 15

Furthermore, because you are adding polynomials with like terms, you can simplify this expression further:
2x2 + 13x + 15

And that’s the final, reduced answer. You can apply the FOIL principle to any two binomials to arrive at their product. When you have more complicated polynomials, such as those composed of 3, 4, 5, or more monomials, you do the same type of thing. What I find easiest is to take the first term of the first polynomial, and multiply it with every term of the second polynomial. Then do the same for the second term in the first one, multiplying with every term in the second one, and so on.

Like this:

(x2 + x + 5)(x3 + x2 + 1)

First group (multiply x2 with all in the second polynomial)

  • x2 * x3
  • x2 * x2
  • x2 * 1

Second group (x)

  • x * x3
  • x * x2
  • x * 1

Third group (5)

  • 5 * x3
  • 5 * x2
  • 5 * 1

Now, you just add up all these terms, and simplify where you can:

(x2 * x3) + (x2 * x2) + (x2 * 1) + (x * x3) + (x * x2) + (x * 1) + (5 * x3) + (5 * x2) + (5 * 1)

(x5) + (x4) + (x2) + (x4) + (x3) + (x) + (5x3) + (5x2) + (5)

(x5) + 2(x4) + 6(x3) + 6(x2) + x + 5

And that’s it. A little more complicated, but as long as you keep track of what you’re doing and work your way through it, you will arrive at the answer!

Monomials and Polynomials Explained

When studying algebra, and learning how to perform more complicated rearrangements and calculations, you will frequently see math terms such as “monomials” and “polynomials.” They may sound like some kind of advanced mathematical concepts, but in reality, they are not. In fact, their definitions are really quite simple, and understanding these will help you when working with problems that include them. Understanding the words is important here, because their definitions essentially explain the concepts.

Consider first the term “monomial.” As you probably know already, the mono- part of the word refers to one (or single), just like the word monologue refers to one person speaking, or monotone refers to a sound that has a single pitch. The -nomial part of the word come from the Latin word for name, and refers to terms or expressions (think of them as math names). So, if you put the two parts together, you can see that a monomial is an expression that contains only a single term. Here are a few examples of monomials, and yes they really can be as simple as they look:

  • 3
  • x
  • 58x
  • 178293xyz
  • 51993x2y3

All of these are single terms, therefore these are all monomials. Any standalone whole number is a monomial, as are variables. You can also combine these to have many parts multiplied or divided, as seen in the examples, but you cannot add or subtract within a term. You also can’t have negative or fractional exponents. Think about a monomial as being anything where all the numbers and letters can be crammed together without having to break it up with an operation. A few things to keep in mind when you work with monomials: a monomial times a monomial is a monomial (eg. x times y is xy), and a constant time a monomial is a monomial (eg. 3 times x is 3x).

So, now that I have defined monomials as being math expressions consisting of only a single term, do you think we have a way of describing expressions that have more than one term? Ha ha! Of course!

Polynomial is the word used to identify a mathematical expression that is composed of multiple terms. Any expression with one term is a monomial, whereas any expression with more than one term is a polynomial. You can break this word down into its components as well, and realize that poly- refers to multiple or many, so now you can easily remember what this word means. Here are some examples of polynomials, and yes these can get very complicated:

  • x + 3
  • 2x – 10
  • x2 + 5x – 100
  • 81x6 – 43x5 + 23x4 – 12x3 + 2x2 + 6x –1000

All of these expressions are polynomials, and they are all made up of several monomials. Each individual term in these polynomial expressions are monomials on their own. So, as I said to consider monomials as being terms where all the parts are multiplied and crammed together into a single term, as soon as you introduce an operation such as addition, you are creating a polynomial. As you can see on that last example, I ordered all of terms from highest x exponent to lowest x exponent. The highest exponent refers to the degree of the polynomial, which I will discuss in a future post.

While polynomial in a general terms that encompasses all expressions having more than a single term, there are a few other terms that you may see used to describe some specific cases. Binomial refers to polynomials that have exactly two terms, sincebi- refers to two. Similarly, using the prefix tri- gives you the word trinomial for dealing with expressions that have exactly three terms.

So, now that you are familiar with these common math definitions, you can see that they aren’t introducing any additional advanced techniques to your math problems. They are merely words used to describe what you are already used to working with!

In my next post, I will go into a bit more detail on performing calculations with polynomials. For example, how do you divide a polynomial by a monomial, or how do you FOIL two binomials? Follow along, and you’ll learn how to do polynomials problems with ease!

What are Significant Figures?

Counting Significant Figures is a concept usually introduced to students in beginner science courses, right near the beginning. However, it is a concept that continually confuses people even beyond their school years. This is because, at first glance, significant figures (commonly referred to as “sig figs”) appears to be very similar to rounding, but that is where the difficulty arises. It is more than rounding, and so there are many more rules to follow. It actually describes the amount of uncertainty there is in a value, and is tied to that value’s accuracy and precision. Sig figs are the digits in a value that are considered to be important (significant) as a result of the precision in their measurement. Counting significant figures is a good way to learn about them.

Here are a few basic examples:

A thermometer reads 26.9 degrees Celsius. This is 3 sig figs.
A scale displays a weight of 31.22 g. This is 4 sig figs.
A stopwatch shows 58.778 seconds. This is 5 sig figs.

Did you follow that? The basic rule for counting significant figures that I followed there was that all non-zero numbers are significant. These are simple examples, but there are more rules to consider. The rules that help us identify a sig fig are not difficult, and there are not many, but they are very important to understand. As always, understanding the basics is absolutely essential to being able to move forward. So, let’s learn how to count sig figs!

As I stated above, the first and simplest rule is to consider all non-zero numbers as significant.

123 has 3 sig figs
0.007654 has 4 sig figs

The remaining rules instruct you how to deal with numbers that contain zeroes.

If a number starts with one or more zeroes on the left side, you do not count these as sig figs. You begin counting significant figures from the leftmost non-zero digit.

0.1 has 1 sig fig
0.0045 has 2 sig figs

A zero that is found between other non-zero digits always counts as significant.

101 has 3 sig figs
0.3056 has 4 sig figs

A zero found at the end of a number containing a decimal point does count as significant. However, if the zero is at the end of a non-decimal number, then it may or may not count as significant. Generally, it is safe to count the number of significant figures to the left of the ending zero(es), and then state that there may be up to how many digits you see, depending on the uncertainty of the measurement.

100.0 has 4 sig figs
100. has 3 sig figs
100 has at least 1 sig fig, but could possibly have up to 3 sig figs.
50.30 has 4 sig figs
2.03000 has 6 sig figs
0.0098700 has 5 sig figs
9,885,000 has at least 4 sig figs, but could possibly have up to 7 sig figs.

As I mentioned, the number of significant figures is often a result of the degree of precision in a measurement. However, sometimes the value must be taken in context to find the correct number of sig figs. For example, look at the number 100 above. I said it has at least 1 sig fig but possibly more. If we are talking about a count of something… say, the number of houses on a street, then that value is precise to 3 digits, and it would be fair to say that, in this case, it would have 3 sig figs. Sometimes, this can be noted with a decimal point at the end, without any extra zeros (100.).

And with that, you have now learned about all the rules for counting Significant Figures. They’re not difficult once you practice for a while and recognize where the various rules apply.

Now that we have figured out just what counts as a sig fig, the next step is being able to do math with them – add them, multiply them, etc – and to do that, of course, we have an entirely different set of rules. This is the part that ALWAYS confuses people, so I will do my best to lay it out for you so that it makes sense.

My Guest Post on the Help Teaching Blog!

I just wanted to post a quick announcement here that my guest post has been published over at the Help Teaching blog! Thanks to Lilia and her staff for giving me the opportunity to publish something on their site, which I hope will be as valuable as the rest of their content that they have to offer. In my post, I put together a list of (what I believe to be) three incredibly useful tools for math educators. I personally use two of the three tools, though that is only because one is PC-based whereas I work on a Mac! I’d love to say more about these tools here, but instead, I will direct you over to the actual post! Please head on over and give it a read, and let me know what you think. Do you use any of these tools as well? Are there any I missed that you think are even better? Let me know in the comments below!

How to Calculate the Area of a Triangle

Triangle problems are some of the most common types of mathematics problems you will find when first studying geometry and trigonometry. Some might argue that one triangle concept in particular, the Pythagorean Theorem, is one of the most famous and well-known concepts of all! I have gone into detail on this blog about several trigonometry functions that are useful to help solve triangles, but in this post, I am going to talk about the geometric side. In particular, I’m going to show you how to calculate the area of a triangle. There are actually several ways of doing this. This should probably come as no surprise to you, though. There always seems to be more than one way to do things in math!

Probably the easiest way to do this requires that you know only two things about your triangle: the length of its base, and its height. With those numbers, you simply insert them into this equation and solve for the area:
 Pretty easy, right? Do you understand why this equation works, though? Take a look at the equation again. You might recognize that it looks familiar, almost like the equation to solve the area of a rectangle – only this time, multiply by a half. So the area of a triangle is equal to half of the area of a rectangle. You can very easily see this simply by drawing any old rectangle, label the sides as “base” and “height”, and then go ahead and draw in a diagonal line that connects two of the corners. You’ve just created two equal triangles, and each triangle is half of the original rectangle! Well, that’s great if you have a triangle that has one side perfectly horizontal, and the other 90 degrees up perfectly vertical. That is literally half of a rectangle. But what about triangles that have more irregular angles?
 Like a triangle with 31, 82, and 67 degrees? No right angles there, and it certainly doesn’t look like half of a rectangle. Check this out though. It still works! Orient your triangle so that it’s long side is on the bottom. Then, draw in a line from its highest point straight down to intersect the bottom.
 So far, we have just identified its base and its height. Now, for the fun part! Separate your triangle on the vertical line you just drew, and then take your two new smaller triangles and switch sides – put the one on the left over on the right now, but right next to each other, touching at their corner. Now we have a weird structure that has the same base length, and has two vertical walls that have the same height. Now if you just draw in another line to connect the two tops, you can create a rectangle. And, if you were to measure the area of the new triangle created when you connected the tops, you would find that its area is exactly equal to the area of your first triangle, and obviously together they form the complete area of the rectangle.
 So, there you have it. Visual proof of the triangle area formula! That’s a fantastic method to use to find the area of the triangle if you know its base and height (or “altitude” as some may call it). But what about if you don’t know its height? The height of right angle triangles is easy because it is equal to one of the sides. But for other triangles, if you know the side lengths, that doesn’t always mean you can immediately determine the height. In that case, there is another formula that you can use, called Heron’s (or Hero’s) formula. You can use Heron’s formula to calculate the area of a triangle where you know the lengths of all of the sides.

Heron’s formula is sometimes referred to as the “irregular triangle calculation method,” and is named after Heron (or Hero) of Alexandria. Heron was an engineer and mathematician in Ancient Greece, who is credited with inventing many thing, including an early steam engine , a windwheel, and a vending machine. (Check out this Wikipedia page for more information on some of his inventions and accomplishments!) His formula for calculating the area of a triangle contains a few steps, but thankfully, they are not all that difficult. The first thing you have to do is to calculate the half-perimeter (or the semi-perimeter, s). You do this by simply summing the three side lengths (a, b, and c) and dividing in half. Once you’ve done that, then you just plug numbers into the following equation:
 It might be easier to remember this equation by breaking it down and looking at what each part is. If you consider that the three bracketed terms are really just the difference in side length from the half-perimeter, then all you need to remember is to multiply those three values by the half-perimeter itself, and then square root the whole thing. I know that sounds like a mouthful, but having some degree of understanding about what more complex formulas are based on is always helpful.

Somewhat related to this method is a much simplified version, which is specifically applicable to equilateral triangles. (Of course, we all know that equilateral triangles are special triangles whose sides are equal and all angles are 60 degrees.) It’s a much simpler equation to remember, without needing to make a first calculation like with Heron’s formula. Here is the formula for the area of an equilateral triangle:
 The final method that I will describe here uses trigonometry to find the area. You can use this formula on any triangle, provided that you know the length of two of the sides and the angle between them. (Note: there are variations of this formula, depending on what sides and angle are known). It’s fairly easy to remember, as long as you just think about ABC! Here’s the formula:
This goes back to my diagram above, with the inscribed “height” line in the triangle. Using a basic trig identity for sine (SOH!), you can calculate the height using the hypotenuse of one of the smaller triangles and a known angle. Then, you simply take this trig formula and substitute it in for the height term in our original triangle area formula (1/2 base x height). With some simple rearranging, you can come up with our final half absinc formula here!

That’s all the methods that I’m going to discuss here, though you now know four ways of calculating the area of a triangle! Just pick the right methods for your specific problem, and you will only be a few steps away from solving the area!

The Theorem of Pythagoras Explained

The Theorem of Pythagoras is a specific case of the Cosine Law that applies specifically to right angle triangles. With it, and given any two sides of a right angle triangle, you can find the third side. Then having solved all the sides of the triangle, you can use the standard trig identities (Sine, Cosine, Tangent) to evaluate the sizes of all the angles. Therefore, you can see how incredibly useful this very well-known mathematical theorem is! Let’s look at it in a little bit more detail.

The definition of the Theorem of Pythagoras states that the square of the hypotenuse is the sum of the squares of the other two sides. That’s the wordy version, but it’s all you need to remember how to use it. Personally, I can almost recall this phrase even more easily than the mathematical expression that defines it, and that is a very simple equation to remember! This is how quick and painless the formula is:

Theorem of Pythagoras

That is all there is to it! This is known as the Pythagorean Equation. It’s amazing that such a small equation can have such a wide-reaching application. This expression says exactly what I defined above, as it provides a simple relationship connecting the three sides of a triangle. The only condition required is that we have a right angle triangle, with c in the expression being the length of the hypotenuse. So then we have the square of the hypotenuse (c) equal to the sum of the squares of the other two sides. If you think about this relationship, you can also deduce that the hypotenuse will always be the longest side, and always less than the sum of the other two.

Now you might be wondering just WHY this equation is true. Perhaps you want proof. OK then, here is a diagram to prove this to you. What I have done is created a right triangle with sides a = 3, b = 4, and c = 5, and then built a square off of each of the triangle’s sides. (Get it? I squared each side!) If you do the math, you will find that the sum of the area of the side square and bottom square (a and b) are equal to the area of the square built off the hypotenuse. Visually, this is all that the math means.

Theorem of Pythagoras Explained

Hopefully, the diagram above is sufficient enough to demonstrate this concept. However, maybe you would prefer to see video proof that a2 + b2 = c2? Here is a fantastic video that I found that perfectly demonstrates this theorem. It is a wonderful, watertight demonstration of the theorem of Pythagoras that shouldn’t leave you with any doubt.

Here’s another page with a thought exercise on it to prove the theorem. This Pythagorean card table problem is another good demonstration of putting these mathematics concepts to good use!

This is one of the most basic trig concepts, and is probably one of the first concepts that were taught in trigonometry. In fact, there are several different variations of proofs that have been developed over the years, based on concepts of proportionality (similar triangles), geometry (Euclid’s proof, or rearrangement), or algebra and calculus. It is very important and very useful, as you will undoubtedly be able to make use of it in a wide range of solutions to math questions for years.

Thanks for visiting and reading my post about the Theorem of Pythagoras! Please click the Facebook Like button below if this was useful for you!

Trigonometry – Cosine Law

The Cosine Law works similarly to the Sine Law that I have already discussed. Actually, it may seem somewhat familiar to you. While the Cosine Law can be used on any triangle, the Pythagorean Theorem is a specific case of the Cosine Law which strictly applies to right angle triangles. It’s a bit more of an equation to remember than the Sine Law unfortunately, but it is extremely useful. Here is the equation:



Upon a quick analysis, you can see that this law requires two of the three sides in a triangle, as well as the angle opposite the unknown side (or the angle contained by both of the known sides) to provide you with the unknown third side. Here is another trick to think of it: use your thumb and index finger to form a V shape that will represent your triangle, and if you know the lengths of the two sides (fingers) and the angle between them, you can find the remaining side by using the Cosine Law. This is useful for finding the third side of any triangle when two sides and the angle between them are known.

Let’s try an example, where we will solve for the unknown sides and angles.

First, assess the information that you are given and come up with a strategy to solve this triangle. (The phrase “solve the triangle” is often used to instruct you to determine all of the missing angles and sides). In this case, there are two sides and the angle between them known – the perfect case for the Cosine Law to deduce the unknown side! Once we’ve gone that far, we will have then obtained a complete ratio (side and angle) to use in the Sine Law to solve for the remaining two angles. So, now let’s put our strategy to work!

So by the Cosine Law:

Now that we have done that, we have obtained the final side length, and therefore a complete side/angle ratio to use with the Sine Law! And now, you can find the rest of the angles to fully describe the triangle!


I’ll leave those for you to solve. But that’s it! Using a combination of the Cosine Law and Sine Law, you can completely solve any triangle that you are given. These are extremely powerful and useful equations!

Also, on a side note… as I mentioned at the start of this post, the Cosine Law is a generalization of the Pythagorean Theorem, which specifically applies to right angle triangles. You can see that if you are working with a right angle triangle and substitute in 90 degrees to the Cosine Law, it reduces down to the Pythagorean theorem:



reduces to 0, and so

…The Pythagorean Theroem! :)

Trigonometry – Sine Law

The trig functions that I’ve discussed so far (Sine, Cosine, and Tangent) will be incredibly useful to you when working specifically with right angle triangles. However, of course, not all triangles have a 90 degree angle in them. So can you still use these functions? Well, yes, but in a different way. One way is through application of the Law of Sines.

Let’s consider a triangle that has three different angles, none of which are right angles. The standard naming scheme still applies, although now there is no hypotenuse (remember, the hypotenuse is opposite to the 90 degree angle).

Triangle ABC So, now we have our triangle, what is sine law?  This trigonometry sine law says that the ratio between the sine of an angle and the side opposite to it will be equal for all three angles. In other words:
a/SinA = b/SinB = c/SinC
Since you only really work with 2 ratios at a time, with a little rearranging, you can see that:aSinB = bSinA

So let’s try some sine law examples. Say we have a triangle with 1 known angle of 40 degrees, sides 4 and 6 units long. Find angle B:

Triangle ABC values

The Sine Law says that the ratios of angle to opposite sides will be equal. With this, we know how to find an angle using sine law.  So we have:
6 x (Sin 40) = 4 x (SinB)SinB = 0.964B = 74.62 degrees

Now, if the question had asked to find all the angles in this triangle, an easy trick you can apply at this point is to simply say that since the 3 angles in the triangle will add up to 180, you can just subtract your known angles from 180 to get the third angle:

180 = 40 + 74.62 + C

C = 65.38

And now that you have angle C, you can use the Law of Sines with that angle to solve for the final unknown side:

cSinA = aSinC

c x (Sin 40) = 4 x (Sin 65.38)

c = 5.66

Whenever you are given a triangle that does not include a right angle, but you are provided with 1) an angle, 2) the length of the side opposite to that angle, and 3) any one of either the other sides or angles, then you can use the Sine Law.  Now you know how to find an angle using sine law! For another good resource on this topic, which includes a link to a video proof,  take a look at the Cool Math Stuff blog!

Trigonometry – Secant, Cosecant, Cotangent

In addition to the three basic trig functions we’ve already looked at (Sine, Cosine, Tangent), there are three other related functions. These are Secant, Cosecant, and Cotangent. These functions have similar meanings as the first three, in that they represent the ratios of various side lengths of a right angle triangle, and can be used to find angles or unknown side lengths. I will not go into extensive detail on these functions, as they are less commonly required, but I will show you what they mean.  Please remember to click on the Like button if this is helpful, and be sure to follow me with the buttons on the right side!

So far, with the help of SOHCAHTOA, we have seen that:

Sine = opposite / hypotenuse

Cosine = adjacent / hypotenuse

Tangent = opposite / adjacent

These new functions are related to the originals because they represent the inverse ratios. (Of course, inverse means you swap the top and the bottom…)

Cosecant = hypotenuse / opposite… (compare to Sine)

Secant = hypotenuse / adjacent……. (compare to Cosine)

Cotangent = adjacent / opposite…… (compare to Tangent)

Also, these functions can be abbreviated: Cosecant = Csc, Secant = Sec, Cotangent = Cot.

At the middle school and high school math level, you will rarely have a need to use these functions, but it is good for you to know what they are. However, most trig problems at this stage can easily be solved with the original three functions.  Just in case, though, it’s always good to know all the trig functions: sine, cosine, tangent, secant, cosecant, cotangent. If you’d be interested in additional material, a good place to start would be these links that explain secant, cosecant, and cotangent. You can also search that site for many other math concept explanations (just remember to come back here!)

Thanks for reading this quick “cheat sheet” version of these additional trig functions. Please remember to Like this post if you found it helpful, and bookmark my site to come back again!

Graphing: Standard Form of the Equation

Just a short explanation for what is meant by “standard form” of the equation of the line. We have been looking at line equations in the form of y=mx+b. However, you may be asked to express this in standard form, or as a standard form equation.  Graphing standard form equations will give you the exact same line as graphing something expressed as y=mx+b… standard form is just a different way of displaying the equation.

The general notation for a standard form equation is Ax + By = C, where A, B, and C are coefficients, and the x and y are the same variables we’ve been looking at but in a different position from what we recognize.

To express in standard form, you simply just rearrange the y = mx + b form such that you have x and y on the same side, equal to a number. Let’s look at some examples:

Given that y = 3 x+ 5, standard form of this is 3x – y = (-5).

Given y = (1/2)x -15, standard form of this is (1/2)x – y = 15… also, if you don’t want to have any fractions in your answer, you can multiply everything by the number in the denominator, such that we now get x – 2y = 30. Both expressions mean the same thing and will produce the same line. (In fact, convince yourself that no matter what you do to the equation, so long as you do it to both sides, the line is the same. eg. Multiply it all by 100, you get 100x-200y=30000… looks different, but it’s not! Reduce it down and see for yourself!)

For graphing standard form equations, you still might want to go from standard form to the mx+b form, for which you may need to do a bit more math, but it’s still quite straight forward.

Given 5x – 15y = 10, you just have to rearrange things to get y by itself on one side:

(-15y) = (-5x) + 10

y = (1/3)x – (2/3)

and then you can see it is a line with slope 1/3 and y-intercept (-2/3).

Both types of equations mean the same thing. They are just expressed differently, and y=mx+b gives immediate information about the line without having to do a lot of work. However, you should be able to use both forms interchangeably.  Convince yourself that graphing standard form equations will give you the same line as graphing y=mx+b equations.  They just look different because the numbers are rearranged.  This should be obvious because if you start with a standard form equation, and convert it to y=mx+b and graph it, you have only rearranged things not added or removed anything.  You do not have a new line.

Also, from these equations, you should be able to tell that whenever you have an equation with 2 variables (x and y), and there aren’t any exponents on either term, then you are dealing with a straight line. So while an equation in standard form may not immediately look like a straight line equation to you until it looks like y = mx + b, because it has an x and a y in it (without an exponent… exponents make the graph do cool things later), it is automatically a straight line!