Video Transcript
In this video, we will learn how to
divide monomials involving single and multiple variables. In particular, we will focus on
simplifying the quotient of monomial terms.
To begin, letβs recall the
definition of a monomial. A monomial is a single algebraic
term where every variable is raised to a nonnegative integer power. To see how to simplify the quotient
of monomials, we can start with an example. Letβs say we want to simplify the
quotient of π₯ cubed and π₯ squared. We simplify this quotient by
recalling that exponents or powers are defined as repeated multiplication. So, π₯ cubed equals π₯ times π₯
times π₯, and π₯ squared equals π₯ times π₯. We can use this to rewrite the
quotient.
At this point, itβs important to
note that π₯ represents a nonzero number. Otherwise, the quotient would be
undefined. The reason π₯ canβt equal zero is
because we know that we canβt divide by zero. And in the same way, we canβt
cancel the shared factors of π₯ if π₯ is zero. But when π₯ is nonzero, we can
cancel the shared factors of π₯. This leaves us with π₯ over one, or
just π₯.
We can generalize this process to
take the quotient of any two monomials. To do this, letβs start with the
quotient of the monomials π₯ to the power of π and π₯ to the power of π for
positive integers π and π, with π greater than or equal to π. Then we write the product out with
π factors in the numerator and π factors in the denominator. Since π is greater than or equal
to π, the leftover noncanceled factors will be in the numerator. The number of noncanceled factors
will therefore be π minus π. Therefore, we have π₯ to the power
of π divided by π₯ to the power of π equals π₯ to the power of π minus π. This is known as the quotient rule
for exponents.
The quotient rule for exponents
states that for any integers π and π with π greater than or equal to π and any
nonzero value of π₯, we have π₯ to the power of π divided by π₯ to the power of π
equals π₯ to the power of π minus π. We can apply this rule to find the
quotient of any two monomials, as we will see in our first example.
Simplify π₯ to the power of
seven over π₯ to the power of six.
Since we are asked to simplify
the quotient of two monomials, we can start by recalling the quotient rule for
exponents, which states that for any integers π and π such that π is greater
than or equal to π and any nonzero value of π₯, we have π₯ to the power of π
divided by π₯ to the power of π equals π₯ to the power of π minus π.
In this case, we have π equals
seven and π equals six. So, when we substitute these
values into the quotient rule for exponents, we get π₯ to the power of seven
minus six, which equals π₯ to the power of one. Finally, we recall that raising
a number to the first power leaves it unchanged. Thus, π₯ to the power of seven
over π₯ to the power of six equals π₯.
It is worth noting that we can
also solve this problem from first principles by writing the product out in full
with seven factors in the numerator and six factors in the denominator. We can then cancel the six
shared factors in the numerator and denominator, which leaves us with a single
factor of π₯.
In our next example, we will see
how this rule can be applied to answer a question with an area model.
Find the missing side length of
the following rectangle.
We start by recalling that the
area of a rectangle is given by its length multiplied by its width. From the diagram, we have the
area of the rectangle, 24π₯ squared, and the width of the rectangle, four
π₯. If we call the length of the
rectangle π, we must have 24π₯ squared equals π times four π₯.
Note that π₯ cannot equal zero
since the width and area of the rectangle are nonzero. So, we can divide both sides of
the equation by four π₯. On the right-hand side of the
equation, we cancel the shared factors of four π₯ to get 24π₯ squared over four
π₯ equals π. We see that the numerator and
denominator of the resulting fraction each contain a monomial. A monomial is a single-variable
algebraic term where every variable is raised to a nonnegative integer
power.
On the left-hand side, we are
taking a quotient of monomials, which can be simplified using the quotient rule
for exponents. This rule states that for any
integers π and π such that π is greater than or equal to π and any nonzero
value of π₯, we have π₯ to the power of π divided by π₯ to the power of π
equals π₯ to the power of π minus π. It can be helpful to separate
the quotient into the product of two quotients, where the factors with a base of
π₯ are grouped together.
Since π₯ equals π₯ to the power
of one, we can apply the quotient rule with π equal to two and π equal to
one. And we find that π₯ to the
power of two divided by π₯ to the power of one equals π₯. Then we simplify 24 over four,
which equals six. Therefore, the missing length
of the rectangle is given by six π₯.
Thus far, we have only dealt with
the quotient of single-variable monomials. However, we can apply the exact
same process to find the quotient of multivariable monomials, as we will see in our
next example.
Simplify negative 12π₯ to the
power of six times π¦ to the power of five over six π₯ to the power of four
times π¦.
We notice that both the
numerator and denominator contain multivariable monomials. We recall that a monomial is a
single algebraic term where every variable is raised to a nonnegative integer
power. In this case, the monomial
terms include π₯- and π¦-variables. To simplify the given quotient,
we can rewrite it as negative 12 times π₯ to the power of six times π¦ to the
power of five over six times π₯ to the power of four times π¦.
Then we can use properties of
the multiplication of rational numbers to separate the division into factors
with the same base. The first quotient simplifies
to negative two. Then we can simplify the final
two quotients by using the quotient rule for exponents, which states that for
any two integers π and π such that π is greater than or equal to π and any
nonzero value of π₯, we have π₯ to the power of π divided by π₯ to the power of
π equals π₯ to the power of π minus π.
To simplify these quotients, we
must assume that π₯ and π¦ are nonzero. And we note that π¦ equals π¦
to the power of one. Then, according to the quotient
rule, π₯ to the power of six over π₯ to the power of four equals π₯ to the power
of six minus four. And π¦ to the power of five
over π¦ to the power of one equals π¦ to the power of five minus one.
In conclusion, negative 12π₯ to
the power of six times π¦ to the power of five over six π₯ to the power of four
times π¦ simplifies to negative two π₯ squared times π¦ to the power of
four.
In our final example, we will use
this process to the simplify the product and quotient of monomials.
Simplify π₯ to the power of
four times π¦ to the power of four multiplied by π₯ squared times π¦ to the
power of four over π₯ to the power of four times π¦ cubed.
We first notice that the
numerator of this expression is the product of two monomials. A monomial is a single
algebraic term where every variable is raised to a nonnegative integer
power. To simplify the whole
expression, we will need to first simplify the numerator using the product rule
for exponents, which states that for any rational number π₯ and nonnegative
integers π and π, we have π₯ to the power of π multiplied by π₯ to the power
of π equals π₯ to the power of π plus π.
First, using the commutative
property of multiplication, we rewrite the numerator as π₯ to the power of four
times π₯ squared multiplied by π¦ to the power of four times π¦ to the power of
four. According to the product rule
for exponents, we can add the exponents of the two factors with a base of π₯ and
then add the exponents of the factors with a base of π¦. Meanwhile, the denominator is
still π₯ to the power of four times π¦ cubed.
Then, after simplifying the
numerator using the commutative property and the product rule, we have π₯ to the
power of six times π¦ to the power of eight over π₯ to the power of four times
π¦ cubed. The result is the quotient of
two monomials. We recall that we can simplify
the quotient of two monomials by first rearranging the quotient to separate the
factors with the same base.
We can then simplify the final
two quotients by using the quotient rule for exponents, which states that for
any integers π and π such that π is greater than or equal to π and any
nonzero value of π₯, we have π₯ to the power of π divided by π₯ to the power of
π equals π₯ to the power of π minus π. We assume that π₯ and π¦ are
nonzero and then apply this rule to get π₯ to the power of two times π¦ to the
power of five. This is the simplification of
the original product and quotient of monomials.
Letβs finish by recapping some key
points from this video. First, we must be able to recognize
a monomial as a single algebraic term where every variable is raised to a
nonnegative integer power. When a quotient of monomials needs
to be simplified, we use the quotient rule for exponents. This rule states that for any
integers π and π such that π is greater than or equal to π and any nonzero value
of π₯, we have π₯ to the power of π divided by π₯ to the power of π equals π₯ to
the power of π minus π.