# Lesson Video: Simplifying Monomials: Quotient Rule Mathematics • 9th Grade

In this video, we will learn how to divide monomials involving single and multiple variables.

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### 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, letβs 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. Letβs 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, itβs important to note that π₯ represents a nonzero number. Otherwise, the quotient would be undefined. The reason π₯ canβt equal zero is because we know that we canβt divide by zero. And in the same way, we canβt 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, letβs 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.

Letβs 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.