### Video Transcript

In this video, we will learn how to
multiply an algebraic expression by a monomial. Letβs first think about what an
algebraic expression is. An algebraic expression is a
mathematical expression that consists of variables, numbers, and operations. π₯ plus three is an example of an
algebraic expression. It has a variable, an operation,
and a number. π₯ plus π¦ 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 π₯π¦ 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
thatβs 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, letβs look at some of
the skills weβll need to be able to multiply monomials by algebraic expressions.

The primary thing we need to
remember is the distributive property. Hereβs the distributive property
represented with variables. It says that π times π plus π is
equal to π times π plus π times π. Often, when weβre 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 π
plus π is an algebraic expression, and π is a monomial. π times π plus π is multiplying
a monomial by an algebraic expression.

When working with problems like
these, thereβs 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 π times π minus π, we remember that weβre multiplying the
variable plus its sign by both of these terms. That means it should be negative π
times π. And then for our second term, we
see that π is also negative. We could think of this subtracting
π as adding negative π so that when we do our distribution, weβre adding negative
π times negative π. Negative π times π is negative
ππ, but negative π times negative π will equal positive ππ.

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, weβre ready to look at some examples.

Calculate three times π₯ minus π₯
cubed.

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

Now, three times π₯ equals three π₯
and three times negative π₯ cubed would equal negative three π₯ cubed. Three times π₯ minus π₯ cubed is
equal to three π₯ minus three π₯ 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 π₯ minus three
π₯ cubed is in its simplest form.

Hereβs another example.

Simplify three π₯π¦ times four π₯
cubed π¦ squared.

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

Three times four is 12. We know that π₯ is equal to π₯ to
the first power. And we also know that π₯ to the π
power times π₯ to the π power is equal to π₯ to the π plus π power. That means we need to add one plus
three to get π₯ to the fourth power. And weβll do the same thing for π¦,
which will give us π¦ cubed. And we donβt need to have those
multiplication signs between them. We can write them all together as
12π₯ to the fourth power π¦ cubed.

Now, these are quite a lot of
steps. You might be wondering if you have
to write this out every time. And you wouldnβt 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 π₯ to
the first power times π₯ cubed equals π₯ to the fourth power. And finally, π¦ to the first power
times π¦ squared is π¦ cubed. Both methods show the simplified
form 12π₯ to the fourth π¦ cubed.

In our next example, weβll deal
with an expression that has many more terms.

Simplify the expression two π₯
times π₯π¦ plus π¦ minus π¦ times two π₯ minus π¦ minus π₯ squared times two π¦
minus one.

First, we copy down our
expression. In this expression, everywhere we
see parentheses, weβll have to do some distribution. So letβs start here. We need to distribute this
multiplication across both terms in the parentheses. That means weβre multiplying two π₯
times π₯π¦. And then weβll be adding two π₯
times π¦. Now, for our second set of
parentheses, weβll need to be a bit more careful because weβre multiplying by
negative π¦. And that means weβll have negative
π¦ times two π₯.

We want to write this as plus
negative π¦ times two π₯. And then weβll multiply negative π¦
by negative π¦. So weβll write that as plus
negative π¦ times negative π¦. And when we come to the third set,
we see the same thing. We have a negative π₯ squared weβre
multiplying. And weβll write that as plus
negative π₯ squared times two π¦ and then negative π₯ squared times negative one,
which we write as plus negative π₯ squared times negative one.

This step that Iβve put here is
kind of an intermediate step. Once you get really good at these
type of problems, you wonβt have to write this out all the way. Instead, you would just say two π₯
times π₯π¦ equals two π₯ squared π¦ and two π₯ times π¦ is two π₯π¦. Negative π¦ times two π₯ we would
rewrite as negative two π₯π¦. This is because we generally list
the coefficient first. And the order of π₯ and π¦ doesnβt
have to be like this, but itβs good to keep that consistent throughout the same
expression. From there, negative π¦ times
negative π¦ is positive π¦ squared. And negative π₯ squared times two
π¦ we would write as negative two π₯ squared π¦. And finally, negative π₯ squared
times negative one we write as positive π₯ squared.

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

In our next question, weβll 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 π₯ and π₯ plus π¦ is blank.

If we consider a rectangle, since
the dimensions are two π₯ and π₯ plus π¦, we can let one side be equal to two π₯ and
the other π₯ plus π¦. 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 π₯ for the length
and π₯ plus π¦ for the width, to find the area, weβll need to distribute this two π₯
over both the term π₯ and π¦. Two π₯ times π₯ equals two π₯
squared. And two π₯ times π¦ equals two
π₯π¦, which makes the area of this rectangle two π₯ squared plus two π₯π¦.

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

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

Now, we canβt 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 π₯ and
the height is two π₯ plus one. Now, even if you didnβt 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 weβll take a height
of two π₯ plus one, plug it into our formula, and a base of two π₯. And then, of course, weβll need to
bring down the one-half.

When you look at two π₯ plus one
times two π₯, you might recognize that this is in the reverse order we usually
see. When weβre multiplying a monomial
by an algebraic expression, most often, youβll list the monomial first because the
monomial is what we distribute. We need to multiply two π₯ by two
π₯ 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
π₯ by two π₯ and two π₯ by one. Two π₯ times two π₯ equals four π₯
squared, and two π₯ times one is two π₯.

However, we canβt forget our
one-half. Weβll have to do this distribution
a second time. Weβll have to multiply one-half by
four π₯ squared and by two π₯. Four times one-half equals two. So four π₯ squared times one-half
is two π₯ squared. And two π₯ times one-half will be
equal to π₯. So we can say the area of a
triangle with these dimensions is equal to two π₯ squared plus π₯.

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