# Lesson Video: Multiplying an Algebraic Expression by a Monomial Mathematics

In this video, we will learn how to multiply an algebraic expression by a monomial.

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### Video Transcript

In this video, we will learn how to multiply an algebraic expression by a monomial. Let’s first think about what an algebraic expression is. An algebraic expression is a mathematical expression that consists of variables, numbers, and operations. 𝑥 plus three is an example of an algebraic expression. It has a variable, an operation, and a number. 𝑥 plus 𝑦 would also be an algebraic expression.

Remember, we want to be multiplying algebraic expressions by monomials. So we also need to remember that definition. A monomial is an expression that contains only one term. Three 𝑥𝑦 is a monomial. As monomials can include numbers, whole numbers, and variables that are being multiplied together, numbers by themselves are also monomials, so are variables by themselves, or even a variable that’s taken to a certain power. However, we should note that this power must be a positive integer or zero. A monomial will never have a negative power. Moving on, let’s look at some of the skills we’ll need to be able to multiply monomials by algebraic expressions.

The primary thing we need to remember is the distributive property. Here’s the distributive property represented with variables. It says that 𝑎 times 𝑏 plus 𝑐 is equal to 𝑎 times 𝑏 plus 𝑎 times 𝑐. Often, when we’re working out problems, we illustrate the distributive property using arrows. This is telling us that multiplying a number or a variable by a group of numbers added together is the same thing as doing each multiplication separately. In fact, we can already see here 𝑏 plus 𝑐 is an algebraic expression, and 𝑎 is a monomial. 𝑎 times 𝑏 plus 𝑐 is multiplying a monomial by an algebraic expression.

When working with problems like these, there’s one big thing we need to watch out for. And that would be sign errors or sign mistakes. To avoid sign mistakes, if we have something like negative 𝑎 times 𝑏 minus 𝑐, we remember that we’re multiplying the variable plus its sign by both of these terms. That means it should be negative 𝑎 times 𝑏. And then for our second term, we see that 𝑐 is also negative. We could think of this subtracting 𝑐 as adding negative 𝑐 so that when we do our distribution, we’re adding negative 𝑎 times negative 𝑐. Negative 𝑎 times 𝑏 is negative 𝑎𝑏, but negative 𝑎 times negative 𝑐 will equal positive 𝑎𝑐.

This process is no different than the regular distributive property. We just pay close attention to the sign of each value. Now that we have that out of the way, we’re ready to look at some examples.

Calculate three times 𝑥 minus 𝑥 cubed.

First, we can copy down what we’ve been given. We have the expression 𝑥 minus 𝑥 cubed and then the whole number or the monomial three. We’re trying to multiply these values together. And to do that, we should use the distributive property. It tells us that 𝑎 times 𝑏 plus 𝑐 is equal to 𝑎 times 𝑏 plus 𝑎 times 𝑐. Often when we’re working with the distributive property, we represent it with arrows. It means we need to multiply three by 𝑥. And we should be careful with our second term. It’s helpful to rewrite this subtraction as plus and negative as we need to multiply three by negative 𝑥 cubed. So we get something like this.

Now, three times 𝑥 equals three 𝑥 and three times negative 𝑥 cubed would equal negative three 𝑥 cubed. Three times 𝑥 minus 𝑥 cubed is equal to three 𝑥 minus three 𝑥 cubed. We need to be really careful here. Some people might see these two three coefficients and think that they can do some subtraction, but that is not possible. Remember, we cannot add or subtract values like this when their exponents are different. And that means three 𝑥 minus three 𝑥 cubed is in its simplest form.

Here’s another example.

Simplify three 𝑥𝑦 times four 𝑥 cubed 𝑦 squared.

We need to think about what’s happening when we have a whole number and a list of variables behind it. We know that this means we’re multiplying three times 𝑥 times 𝑦. And inside the parentheses, we’re multiplying four times 𝑥 cubed times 𝑦 squared. What would it mean to multiply these values together? What would it look like? Well, it would look something like this: Three times 𝑥 times 𝑦 times four times 𝑥 cubed times 𝑦 squared. But if we wanted to simplify, we can do some regrouping. We could group the whole numbers together, three times four, and the 𝑥- terms together, 𝑥 times 𝑥 cubed, and then the 𝑦-terms together, 𝑦 times 𝑦 squared.

Three times four is 12. We know that 𝑥 is equal to 𝑥 to the first power. And we also know that 𝑥 to the 𝑎 power times 𝑥 to the 𝑏 power is equal to 𝑥 to the 𝑎 plus 𝑏 power. That means we need to add one plus three to get 𝑥 to the fourth power. And we’ll do the same thing for 𝑦, which will give us 𝑦 cubed. And we don’t need to have those multiplication signs between them. We can write them all together as 12𝑥 to the fourth power 𝑦 cubed.

Now, these are quite a lot of steps. You might be wondering if you have to write this out every time. And you wouldn’t need to. This is just to show you where we get this from. Usually, if I was going to solve something like this, I would say that we need to multiply like terms. Three times four is 12 and 𝑥 to the first power times 𝑥 cubed equals 𝑥 to the fourth power. And finally, 𝑦 to the first power times 𝑦 squared is 𝑦 cubed. Both methods show the simplified form 12𝑥 to the fourth 𝑦 cubed.

In our next example, we’ll deal with an expression that has many more terms.

Simplify the expression two 𝑥 times 𝑥𝑦 plus 𝑦 minus 𝑦 times two 𝑥 minus 𝑦 minus 𝑥 squared times two 𝑦 minus one.

First, we copy down our expression. In this expression, everywhere we see parentheses, we’ll have to do some distribution. So let’s start here. We need to distribute this multiplication across both terms in the parentheses. That means we’re multiplying two 𝑥 times 𝑥𝑦. And then we’ll be adding two 𝑥 times 𝑦. Now, for our second set of parentheses, we’ll need to be a bit more careful because we’re multiplying by negative 𝑦. And that means we’ll have negative 𝑦 times two 𝑥.

We want to write this as plus negative 𝑦 times two 𝑥. And then we’ll multiply negative 𝑦 by negative 𝑦. So we’ll write that as plus negative 𝑦 times negative 𝑦. And when we come to the third set, we see the same thing. We have a negative 𝑥 squared we’re multiplying. And we’ll write that as plus negative 𝑥 squared times two 𝑦 and then negative 𝑥 squared times negative one, which we write as plus negative 𝑥 squared times negative one.

This step that I’ve put here is kind of an intermediate step. Once you get really good at these type of problems, you won’t have to write this out all the way. Instead, you would just say two 𝑥 times 𝑥𝑦 equals two 𝑥 squared 𝑦 and two 𝑥 times 𝑦 is two 𝑥𝑦. Negative 𝑦 times two 𝑥 we would rewrite as negative two 𝑥𝑦. This is because we generally list the coefficient first. And the order of 𝑥 and 𝑦 doesn’t have to be like this, but it’s good to keep that consistent throughout the same expression. From there, negative 𝑦 times negative 𝑦 is positive 𝑦 squared. And negative 𝑥 squared times two 𝑦 we would write as negative two 𝑥 squared 𝑦. And finally, negative 𝑥 squared times negative one we write as positive 𝑥 squared.

We’ve done all of our expanding and multiplying, but our instructions tell us to simplify. When we look at the term two 𝑥 squared 𝑦, we also see another like term negative two 𝑥 squared 𝑦. And when we add these two values together they equal zero. We also recognize we have the term two 𝑥𝑦 and negative two 𝑥𝑦. When we add those together, we get zero, which means our remaining values are 𝑦 squared and 𝑥 squared. So we could write the simplified form of this expression as 𝑥 squared plus 𝑦 squared.

In our next question, we’ll use what we know about multiplying expressions to solve for the area of a rectangle.

Fill in the blank: The area of the rectangle with dimensions two 𝑥 and 𝑥 plus 𝑦 is blank.

If we consider a rectangle, since the dimensions are two 𝑥 and 𝑥 plus 𝑦, we can let one side be equal to two 𝑥 and the other 𝑥 plus 𝑦. Then we need to think about how we find the area of a rectangle. The area of a rectangle is length times width. If we plug in two 𝑥 for the length and 𝑥 plus 𝑦 for the width, to find the area, we’ll need to distribute this two 𝑥 over both the term 𝑥 and 𝑦. Two 𝑥 times 𝑥 equals two 𝑥 squared. And two 𝑥 times 𝑦 equals two 𝑥𝑦, which makes the area of this rectangle two 𝑥 squared plus two 𝑥𝑦.

Our final example will again be dealing with area, but this time for a triangle.

A triangle has a height of two 𝑥 plus one and a base of two 𝑥. Find the area of the triangle in terms of 𝑥.

Now, we can’t know exactly what this triangle looks like. But if you wanted to sketch a triangle, you could do that. We know that the base is two 𝑥 and the height is two 𝑥 plus one. Now, even if you didn’t sketch a triangle, the most important thing that we remember here is how we find the area of a triangle. The area of a triangle is equal to one-half times the height times the base. And that means we’ll take a height of two 𝑥 plus one, plug it into our formula, and a base of two 𝑥. And then, of course, we’ll need to bring down the one-half.

When you look at two 𝑥 plus one times two 𝑥, you might recognize that this is in the reverse order we usually see. When we’re multiplying a monomial by an algebraic expression, most often, you’ll list the monomial first because the monomial is what we distribute. We need to multiply two 𝑥 by two 𝑥 and by one. If you feel more comfortable doing it with the monomial first, you can rewrite it. Either way, we have to multiply two 𝑥 by two 𝑥 and two 𝑥 by one. Two 𝑥 times two 𝑥 equals four 𝑥 squared, and two 𝑥 times one is two 𝑥.

However, we can’t forget our one-half. We’ll have to do this distribution a second time. We’ll have to multiply one-half by four 𝑥 squared and by two 𝑥. Four times one-half equals two. So four 𝑥 squared times one-half is two 𝑥 squared. And two 𝑥 times one-half will be equal to 𝑥. So we can say the area of a triangle with these dimensions is equal to two 𝑥 squared plus 𝑥.

Before we finish, let’s review our key points. To multiply a monomial by an algebraic expression, use the distributive property, which tells us that 𝑎 times 𝑏 plus 𝑐 is equal to 𝑎 times 𝑏 plus a times 𝑐. And finally, be sure to check for sign mistakes when distributing a monomial that is negative.