### Video Transcript

Simplifying Monomials: Product
Rule

In this video, we’re going to learn
what the product rule for monomials is and how we can use this to simplify
expressions involving the products of monomials. Finally, we’ll also see how this
can help us evaluate equations.

Before we start explaining the
product rule for monomials, it can help to see this in an example. Let’s start with the expression two
cubed or two raised to the power of three. We call three the power or exponent
of our number. This is also sometimes referred to
as index notation or indices. We want to remember what this
means. This means we multiply two by
itself and then again. In other words, it’s equal to two
multiplied by two multiplied by two. In this product, our factor of two
appears three times. And of course, we could evaluate
this expression. We know two multiplied by two
multiplied by two is equal to eight.

We now want to ask the question,
what happens if we multiplied two of these expressions together? For example, what if we wanted to
multiply two cubed by two squared? In this case, the power of two is
two. Remember, two squared is two
multiplied by itself. In other words, it’s equal to two
multiplied by two. So if we were asked the question,
what is two cubed multiplied by two squared, one way of answering this would be to
calculate two cubed and then calculate two squared and multiply these together. So we could answer this by saying
it’s equal to eight multiplied by four, which we know is equal to 32. And this gives us the correct
answer. And it’s one way of answering this
question.

However, we’re going to go over a
slightly different method. Instead of calculating each part
individually, let’s use their definitions. Remember two squared is two
multiplied by itself. So in the product of this, we’ll
have two appearing two times. So in our expression, we could
replace two squared with two multiplied by two. We could do exactly the same with
two cubed. Remember, two cubed is two
multiplied by itself multiplied by itself again. So by writing this into our
expression, in our first product we have two appearing three times and in our second
product we have two appearing two times. But we don’t need to think of this
as two separate products, remember, we’re multiplying these together. By just looking at this expression
as a whole, this is a product with five factors of two.

In other words, we could just write
this as two to the power of five because two is appearing five times in our
product. And we could ask the question, how
exactly did we get this number of five? Well, it’s equal to three plus
two. We added the powers together. So in this instance, when we
multiplied two cubed by two squared, we added our powers together to show that this
was equal to two to the power of three plus two, which is two to the power of
five. Now, you may be thinking this
didn’t help us actually evaluate this expression. Two to the power of five is not
easier than just calculating two cubed and then multiplying it by two squared. And this is correct. However, it does give us a useful
result we can notice from this working.

We can notice something
interesting. It doesn’t actually matter which
number we had raised to the power of three and two. For example, we could have replaced
both of these with the number seven. To calculate this, we would do
exactly the same lines of working. However, instead of saying two
cubed is two multiplied by two multiplied by two, we would say seven is seven
multiplied by seven multiplied by seven. And of course, we would have seven
squared is seven multiplied by itself. So our answer would also be seven
to the power of five. And when a result is true for any
value, we can turn it into a rule.

We’re now ready to state the
product rule for monomials. This tells us that 𝑥 to the power
of 𝑚 times 𝑥 to the power of 𝑛 is equal to 𝑥 to the power of 𝑚 plus 𝑛. And remember, whenever we’re
talking about monomials, we’re talking about our powers being nonnegative integer
values. So in particular, the product rule
for monomials only counts when 𝑚 and 𝑛 are nonnegative integers. And it’s also worth pointing out
that this formula is true for other values of 𝑚 and 𝑛. However, it’s beyond the scope of
this video to prove this. So we’ll just assume that we’re
dealing with monomials. In other words, 𝑚 and 𝑛 are
nonnegative integers.

And in fact, we can prove this
result by using exactly what we did before. 𝑚 and 𝑛 are both positive
integers, so 𝑥 to the power of 𝑚 will be 𝑥 multiplied by itself, where 𝑥 appears
𝑚 times in our product. So just like before, we start by
writing 𝑥 to the power of 𝑚 out in full. It’s 𝑥 multiplied by itself, where
𝑥 appears 𝑚 times in our product. Of course, we can then do exactly
the same for 𝑥 to the power of 𝑛. We know 𝑥 to the power of 𝑛 is 𝑥
multiplied by itself, where 𝑥 appears 𝑛 times in our product. And then multiplying these
together, we have 𝑚 factors of 𝑥 multiplied by 𝑛 factors of 𝑥. Of course, this is just 𝑚 plus 𝑛
factors of 𝑥 multiplied together. And of course, this is exactly the
same as saying 𝑥 to the power of 𝑚 plus 𝑛. It’s 𝑚 plus 𝑛 factors of 𝑥
multiplied together. Let’s now see an example of how we
might use this to simplify a monomial.

Simplify 𝑥 to the power of five
multiplied by 𝑥 squared.

We’re given an algebraic
expression, and we’re asked to simplify this. And we can see this expression is a
product. In fact, it’s the product of two
monomials. And remember, a monomial is the
product of constants and variables, where our variables are raised to nonnegative
integer values. And in this case, we can see this
is true. 𝑥 is a variable. And our powers of 𝑥, five and two,
are nonnegative integers. Therefore, because this is asking
us to simplify the product of monomials, we can answer this by using the product
rule for monomials.

Remember, this tells us 𝑥 to the
power of 𝑚 multiplied by 𝑥 to the power of 𝑛 is equal to 𝑥 to the power of 𝑚
plus 𝑛. If we raise a number to our power
and multiply it by itself raised to a different power, we can do this by just adding
together the powers. In our case, we want to use this to
multiply 𝑥 to the power of five with 𝑥 to the power of two. And since both of these are 𝑥
raised to a different power, we can just do this by adding together the
exponents. This is just equal to 𝑥 to the
power of five plus two. And we can just calculate this. Five plus two is equal to
seven. In other words, this entire
expression is just equal to 𝑥 to the power of seven. Therefore, by using the power rule
for monomials, we were able to simplify 𝑥 to the power of five multiplied by 𝑥
squared to be equal to 𝑥 to the power of seven.

Let’s now see another example
asking us to simplify an even more complicated expression.

Simplify 𝑏 to the power of four
multiplied by 𝑎 to the power of four multiplied by 𝑏 squared times 𝑎 to the power
of five.

In this question, we’re asked to
simplify a product. And we can see that each of our
factors has something interesting; it’s a variable raised to a positive integer
power. In other words, this is the product
of four monomials because, remember, a monomial is the product of constants and
variables where our variables are only raised to positive integer powers. Therefore, because this is the
product of monomials, we’re going to want to simplify this by using the product rule
for monomials. Remember, this tells us 𝑥 to the
power of 𝑚 multiplied by 𝑥 to the power of 𝑛 is equal to 𝑥 to the power of 𝑚
plus 𝑛.

So we’ll start with the expression
we want to simplify, 𝑏 to the power of four times 𝑎 to the power of four times 𝑏
squared multiplied by 𝑎 to the power five. And we can immediately see a
problem. We’re not directly multiplying 𝑎
to the power of four by 𝑎 to the power of five and 𝑏 to the power of four by 𝑏
squared. But we don’t need to worry about
this because, remember, it doesn’t matter which order we multiply our numbers; we’ll
get the same result. So we can switch our second and
third factor around. This gives us the equivalent
expression 𝑏 to the power of four times 𝑏 squared multiplied by 𝑎 to the power of
four times 𝑎 to the power of five.

Now, we can evaluate 𝑏 to the
fourth power multiplied by 𝑏 squared by using our product rule and 𝑎 to the power
of four multiplied by 𝑎 to the power five by using the same rule. All we need to do is add our powers
together. So let’s start with 𝑏 to the power
four times 𝑏 squared. We need to add our powers
together. This gives us 𝑏 to the power of
four plus two. Remember, we then need to multiply
this by 𝑎 to the fourth power multiplied by 𝑎 to the fifth power. And we can evaluate this by adding
our powers together. It’s equal to 𝑎 to the power of
four plus five. And of course, we can simplify this
expression, we know four plus two is equal to six and four plus five is equal to
nine.

And this gives us our final answer
of 𝑏 to the power of six times 𝑎 to the power of nine. Therefore, by using the product
rule for monomials, we were able to simplify 𝑏 to the power of four times 𝑎 to the
power of four multiplied by 𝑏 squared times 𝑎 to the power of five to be equal to
𝑏 to the power of six multiplied by 𝑎 to the power of nine.

Let’s now see an example of how we
might use this rule to find a specific value.

Fill in the blank: two squared
multiplied by 𝑥 cubed times two to the power of seven multiplied by 𝑥 to the power
of five is equal to two to the power of nine multiplied by 𝑥 to the power of
blank.

In this question, we’re given an
equation and we’re told to fill in the blank in our equation. We can see the blank in our
equation is the power of 𝑥 on the right-hand side of the equation. So we want to find the power of 𝑥
which makes this equation true. To do this, we need to make sure
both sides of our equation are equal. So let’s take a look at the
left-hand side of this equation. We can see we’re multiplying four
terms together. In fact, we can see something
interesting. All of these are raised to positive
integer values. And because they’re raised to
positive integer values, this means this is the product of monomials. So we can simplify this by using
the product rule for monomials.

Remember, this tells us that 𝑥 to
the power of 𝑚 multiplied by 𝑥 to the power of 𝑛 will be equal to 𝑥 to the power
of 𝑚 plus 𝑛. We just need to add the powers
together. So let’s try applying this to the
left-hand side of our equation. First, it might help to recall that
we can multiply numbers in any order. It won’t change its value. So we can switch the second and
third factor in this expression around. So we can rewrite this
expression. It’s two squared multiplied by two
to the power of seven times 𝑥 cubed times 𝑥 to the power of five.

Now, we can simplify two squared
multiplied by two to the power of seven. All we need to do is add their
powers together. Two squared multiplied by two to
the power of seven is just equal to two to the power of two plus seven. And we can do exactly the same for
𝑥 cubed multiplied by 𝑥 to the power five. To multiply these, we just add
their powers together. We get 𝑥 to the power of three
plus five. Now, we can simplify this. In the power of two, two plus seven
is equal to nine. So this simplifies to give us two
to the power of nine. And in the power of 𝑥, three plus
five is equal to eight. So we get 𝑥 to the power of
eight. And now we can see this is exactly
the form given to us in the right-hand side of our equation, where the power of 𝑥
is equal to eight, so the value of the blank should be equal to eight.

Therefore, by using the product
rule for monomials, we were able to show for two squared times 𝑥 cubed multiplied
by two to the power of seven times 𝑥 to the power of five to be equal to two to the
power of nine times 𝑥 to the power of blank, the blank must be equal to eight.

Let’s now see an example of how we
could use this to simplify an algebraic expression involving the product of two
monomials.

Fill in the blank: two 𝑥 to the
power of six times six 𝑥 squared is equal to blank.

In this question, we’re given an
equation, and we’re asked to fill in the blank. We can see that the blank is the
entire right-hand side of this equation. So it must be equal to the
left-hand side of this equation. So to find the right-hand side of
this equation, we’re going to need to simplify the left-hand side of this
equation. And to do this, we need to notice
something. Two 𝑥 to the power of six and six
𝑥 squared are both examples of monomials. Remember, a monomial is a product
of constants and variables, where our variables are only raised to positive integer
values, which is exactly what’s happening here because our exponents of 𝑥 are two
and six, which are positive integers.

So to multiply these together, we
need to recall our rule for multiplying monomials. We know 𝑥 to the power of 𝑚
multiplied by 𝑥 to the power of 𝑛 is equal to 𝑥 to the power of 𝑚 plus 𝑛. In other words, to multiply these
together, we just add their exponents. The first thing we’re going to want
to do is multiply our coefficients together. We want to multiply two by six. Of course, two multiplied by six is
equal to 12. So our new coefficient is going to
be 12.

Next, we need to multiply 𝑥 to the
power of six by 𝑥 to the power of two. And we’re going to do this by
adding the exponents together. 𝑥 to the power of six multiplied
by 𝑥 to the power of two is 𝑥 to the power of six plus two. And of course, we know that six
plus two is equal to eight. This gives us 12 times 𝑥 to the
power of eight. And this is our final answer. Therefore, by using the product
rule for monomials, we were able to show that two 𝑥 to the power of six times six
𝑥 squared is equal to 12𝑥 to the power of eight.

Let’s now see one last example of
how we might use this property.

If 𝑑 is equal to four, which of
the following is equal to the volume of this cube? Option (A) 20 to the power of six,
option (B) 20 cubed, option (C) nine to the power of 18, option (D) nine to the
power of six, or option (E) 20 to the power of 18.

In this question, we’re given the
diagram of a cube. And from this diagram, we can see
every side of our cube has length five 𝑑 all raised to the power of six. And in fact, the question tells us
the value of 𝑑. We’re told that 𝑑 is equal to
four. We need to determine the volume of
our cube. To do this, we need to start by
recalling how we find the volume of our cube. We need to recall that if a cube
has a side length we’ll call 𝑠, then we can calculate this volume as 𝑠 cubed. In other words, to find the volume
of a cube, we need to cube the length of one of its sides. And there’s several different ways
we could do this; however, we’ll only go through one of these.

Remember, we know the length of the
sides of our cube. All of the cubes have side length
five 𝑑 all raised to the power of six. But we know the value of 𝑑. In this question, we’re told that
𝑑 is equal to four. So we can substitute this to find
the length of our side. Substituting 𝑑 is equal to four
into the expression we have for the length of our side, we have the side length of
our cube 𝑠 is equal to five times four all raised to the power of six. And we can evaluate this. Inside of our parentheses, we have
five multiplied by four, and this is equal to 20. And remember, we’re raising this to
the power of six. So 𝑠 is equal to 20 to the power
of six.

But remember, this is only the
length of the sides of our cube. We need to cube this value to find
the volume. So the volume of our cube is equal
to 𝑠 cubed. And we just showed the value of 𝑠
is 20 to the power of six. So the volume is equal to 20 to the
power of six all cubed. And this is a very
complicated-looking expression. However, we can simplify this. Remember, when we cube a number, we
multiply it by itself and then multiply by itself again. So in fact, this is equal to 20 to
the power of six multiplied by 20 to the power of six multiplied by 20 to the power
of six. And in all three of these cases, we
have an integer power. This means we can simplify this by
using the product rule for monomials.

We need to recall 𝑥 to the power
of 𝑚 times 𝑥 to the power of 𝑛 is equal to 𝑥 to the power of 𝑚 plus 𝑛. In other words, to multiply these
together, all we need to do is add their powers. There’s a few different ways of
doing this. Let’s just start with simplifying
20 to the power of six multiplied by 20 to the power of six. To do this, all we need to do is
add the powers together. This is equal to 20 to the power of
six plus six. And of course, we still need to
multiply this by our other factor of 20 to the power of six. Now, we could simplify this. However, we could also notice we
can just apply our product rule again.

To multiply these together, all we
need to do is add their powers. In other words, this is just equal
to 20 to the power of six plus six plus six. And finally, we can evaluate
this. Six plus six plus six is equal to
18. So this is just equal to 20 to the
power of 18. And this was given to us as option
(E). Therefore, we were able to show if
a cube has side length five 𝑑 all raised to the power of six and 𝑑 is equal to
four, then the volume of this cube can be represented as 20 to the power of 18,
which was option (E).

Let’s now go over the key points of
this video. First, we showed for monomials 𝑥
to the power of 𝑚 multiplied by 𝑥 to the power of 𝑛 is equal to 𝑥 to the power
of 𝑚 plus 𝑛. In other words, to multiply these
together, all we need to do is add their powers. And in fact, we saw why this was
true. We saw 𝑥 to the power of 𝑚 will
be 𝑥 multiplied by itself, where 𝑥 appears 𝑚 times in our product. And 𝑥 to the power of 𝑛 will be
𝑥 multiplied by itself, where 𝑥 appears 𝑛 times in our product. And of course, multiplying these
together, we’re going to get 𝑚 plus 𝑛 factors of 𝑥 multiplied together. In other words, this would just be
equal to 𝑥 to the power of 𝑚 plus 𝑛.

We also saw that we can use this to
simplify some expressions. For example, if we had two 𝑥 to
the power of six multiplied by six 𝑥 squared, we could start by multiplying the
coefficients together to give us a new coefficient of two times six, which is equal
to 12, and then add the powers of 𝑥 together to give us a new power of 𝑥, which is
two plus six, which is equal to eight.