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Lesson Explainer: Multiplying an Algebraic Expression by a Monomial Mathematics • 9th Grade

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

We first recall that we can multiply monomials together by multiplying the coefficients and adding the exponents of the shared variables together. For example, we find the product of 5π‘₯π‘¦π‘§οŠ¨οŠ© and 3π‘₯π‘§οŠ¨ by first rearranging the product as follows: ο€Ή5π‘₯𝑦𝑧×3π‘₯𝑧=(5Γ—3)Γ—ο€Ήπ‘₯Γ—π‘₯×𝑦×𝑧×𝑧.

We can evaluate this product by multiplying the coefficients 5 and 3 and adding the exponents of the shared variables π‘₯ and 𝑧. This gives (5Γ—3)Γ—ο€Ήπ‘₯Γ—π‘₯×𝑦×𝑧×𝑧=15Γ—π‘₯×𝑦×𝑧=15π‘₯𝑦𝑧.

This process allows us to multiply any two monomials. In fact, we can use this same process to find the product of any number of monomials. It is worth noting that we use the properties of multiplication to rearrange this product. In particular, we use the associativity and commutativity properties to rearrange the product such that the like factors are multiplied together.

This idea of using the properties of multiplication to help simplify algebraic expressions can also be extended to the distributive property of multiplication over addition. We recall this tells us that if π‘Ž, 𝑏, π‘βˆˆβ„š, then π‘Ž(𝑏+𝑐)=π‘Žπ‘+π‘Žπ‘.

In other words, multiplication distributes over addition. An example of this is the area model.

We note that the smaller rectangles have areas π‘Žπ‘ and π‘Žπ‘ and the combined rectangle has area π‘Ž(𝑏+𝑐). Since the combined rectangle has area equal to the sum of its parts, we have π‘Ž(𝑏+𝑐)=π‘Žπ‘+π‘Žπ‘.

Similarly, for subtraction, we note that we can model π‘Ž(π‘βˆ’π‘) as the area of a rectangle with sides of length π‘Ž and π‘βˆ’π‘.

The combined rectangle has area π‘Žπ‘, and the smaller rectangles have area π‘Ž(π‘βˆ’π‘) and π‘Žπ‘. So, π‘Žπ‘=π‘Ž(π‘βˆ’π‘)+π‘Žπ‘.

We then subtract π‘Žπ‘ from both sides of the equation to see π‘Žπ‘βˆ’π‘Žπ‘=π‘Ž(π‘βˆ’π‘).

The same will be true for algebraic expressions. For example, we can multiply π‘₯ by ο€Ήπ‘₯+1ο…οŠ¨ by distributing over the addition. This gives

This works since monomials are the product of constants and variables raised to integer exponents. Since our variables are rational, we can say that these monomials are output rational numbers. Hence, all of the properties of the rational numbers must apply to them.

Let’s see an example of distributing over subtraction.

Example 1: Multiplying a Binomial by a Constant

Calculate 3ο€Ήπ‘₯βˆ’π‘₯ο…οŠ©.

Answer

We first recall that we can multiply an algebraic expression by a monomial by using the distributive property of multiplication over addition. Of course, we are working with a subtraction, but we can also distribute over subtraction as follows: 3ο€Ήπ‘₯βˆ’π‘₯=(3Γ—π‘₯)βˆ’ο€Ή3Γ—π‘₯=3π‘₯βˆ’3π‘₯.

In our next example, we will distribute a monomial over a binomial.

Example 2: Multiplying a Monomial by a Binomial

Calculate βˆ’2π‘₯ο€Ήπ‘₯+π‘₯ο…οŠ¨οŠͺ.

Answer

We first recall that we can multiply an algebraic expression by a monomial by using the distributive property of multiplication over addition. This means we need to multiply each term in the binomial by βˆ’2π‘₯ separately. This gives βˆ’2π‘₯ο€Ήπ‘₯+π‘₯=ο€Ήο€Ήβˆ’2π‘₯×π‘₯+ο€Ήο€Ήβˆ’2π‘₯×π‘₯.οŠͺοŠͺ

We then recall that π‘₯Γ—π‘₯=π‘₯οŠο‰οŠοŠ°ο‰, where π‘š and 𝑛 are integers. Hence, ο€Ήο€Ήβˆ’2π‘₯×π‘₯+ο€Ήο€Ήβˆ’2π‘₯×π‘₯=βˆ’2π‘₯βˆ’2π‘₯=βˆ’2π‘₯βˆ’2π‘₯.οŠͺοŠͺοŠͺ

It is worth noting that the distributive property of multiplication over addition (or subtraction) works for any number of terms. For example, if π‘Ž, 𝑏, 𝑐, and π‘‘βˆˆβ„š, then π‘Ž(𝑏+𝑐+𝑑)=π‘Žπ‘+π‘Žπ‘+π‘Žπ‘‘.

Extending this to algebraic expressions allows us to multiply every term in the sum by the first factor. Let’s see an example of applying this property to multiply a trinomial by a monomial.

Example 3: Multiplying a Monomial by a Trinomial

Expand and simplify the expression ο€Ή2π‘₯3π‘₯+6π‘₯+8οŠͺ.

Answer

We first recall that we can multiply an algebraic expression by a monomial by using the distributive property of multiplication over addition. This means we need to multiply every term in the trinomial by 2π‘₯οŠͺ. This gives us

We then recall that π‘₯Γ—π‘₯=π‘₯οŠο‰οŠοŠ°ο‰. Hence, ο€Ή2π‘₯Γ—3π‘₯+ο€Ή2π‘₯Γ—6π‘₯+ο€Ή2π‘₯Γ—8=6π‘₯+12π‘₯+16π‘₯=6π‘₯+12π‘₯+16π‘₯.οŠͺοŠͺοŠͺοŠͺοŠͺοŠͺοŠͺ

In our next example, we will see how we can apply the multiplication process to determine an expression for an area of a triangle whose base and height are given as expressions in π‘₯.

Example 4: Finding the Area of a Triangle by Multiplying an Algebraic Expression by a Monomial

A triangle has a height of (2π‘₯+1) and a base of 2π‘₯. Find the area of the triangle in terms of π‘₯.

Answer

We first recall that the area of a triangle is equal to one-half of the length of its base multiplied by its perpendicular height. This means its area is given by the expression 12Γ—(2π‘₯)Γ—(2π‘₯+1). We can simplify this by first noting that 12Γ—(2π‘₯)=π‘₯, giving us 12Γ—(2π‘₯)Γ—(2π‘₯+1)=π‘₯(2π‘₯+1).

We can rewrite this as a binomial by distributing π‘₯ over the parentheses. We have

We then recall that π‘₯Γ—π‘₯=π‘₯οŠο‰οŠοŠ°ο‰ and π‘₯=π‘₯. Hence, (π‘₯Γ—2π‘₯)+(π‘₯Γ—1)=2π‘₯+π‘₯=2π‘₯+π‘₯.

In our next example, we will determine an expression for a region given in a diagram by using the distribution of monomials.

Example 5: Finding the Area of a Compound Shape by Multiplying Algebraic Expressions and Monomials

Find the area of the shaded region.

Answer

We begin by noting that the area of the shaded region will be the area of the larger rectangle, with the area of the smaller rectangle removed. We can determine each of these areas separately.

First, let’s isolate the larger rectangle.

The area of a rectangle is length times width. So, its area is 2π‘₯Γ—(π‘₯+2). We can distribute 2π‘₯ over the parentheses to get 2π‘₯Γ—(π‘₯+2)=2π‘₯(π‘₯+2)=(2π‘₯Γ—π‘₯)+(2π‘₯Γ—2).

We then recall that π‘₯Γ—π‘₯=π‘₯οŠο‰οŠοŠ°ο‰, so π‘₯Γ—π‘₯=π‘₯Γ—π‘₯=π‘₯. Hence, (2π‘₯Γ—π‘₯)+(2π‘₯Γ—2)=2π‘₯+4π‘₯.

Second, let’s isolate the smaller rectangle.

Multiplying its length by its width gives us an area of 4π‘₯.

We need to subtract the area of the smaller rectangle from the area of the larger rectangle to find an expression for the area of the shaded region.

This gives us 2π‘₯+4π‘₯βˆ’4π‘₯. Finally, we can combine like terms, so we get 2π‘₯+4π‘₯βˆ’4π‘₯=2π‘₯+(4βˆ’4)π‘₯=2π‘₯.

In our next example, we will apply this process to multiply a trinomial by a monomial with multiple variables.

Example 6: Multiplying a Trinomial by a Monomial with Multiple Variables

Expand and simplify the expression βˆ’2π‘Žπ‘π‘ο€Ήπ‘Žπ‘+4π‘Žπ‘βˆ’π‘π‘ο…οŠ¨οŠ©οŠ¨οŠ¨.

Answer

We first recall that we can multiply an algebraic expression by a monomial by using the distributive property of multiplication over addition and subtraction. This means we need to multiply every term in the trinomial by βˆ’2π‘Žπ‘π‘οŠ¨οŠ©. This gives us

We can then evaluate each product by recalling that we multiply the coefficients and add the exponents of the shared variables together. We get ο€Ήβˆ’2π‘Žπ‘π‘ο…Γ—ο€Ήπ‘Žπ‘ο…+ο€Ήβˆ’2π‘Žπ‘π‘ο…Γ—(4π‘Žπ‘)βˆ’ο€Ήβˆ’2π‘Žπ‘π‘ο…Γ—ο€Ήπ‘π‘ο…=βˆ’2π‘Žπ‘π‘βˆ’8π‘Žπ‘π‘+2π‘Žπ‘π‘=βˆ’2π‘Žπ‘π‘βˆ’8π‘Žπ‘π‘+2π‘Žπ‘π‘.οŠͺοŠͺ

In our final example, we will use this process of multiplying algebraic expressions by monomials to determine the missing value of a constant that equates two expressions.

Example 7: Finding an Unknown by Comparing the Result after the Simplification of the Multiplication Problem

Find the value of the constant π‘˜, given that π‘₯𝑦(12π‘₯+π‘˜π‘¦) is equivalent to 12π‘₯𝑦+2π‘₯π‘¦οŠ©οŠ¨οŠ¨.

Answer

Since both expressions are equivalent, we can start by distributing the monomial in the first expression in order to have it in a form that can be compared with the second expression. We do this by multiplying every term by π‘₯π‘¦οŠ¨. We get π‘₯𝑦(12π‘₯+π‘˜π‘¦)=ο€Ήπ‘₯𝑦×12π‘₯+ο€Ήπ‘₯π‘¦Γ—π‘˜π‘¦ο….

We note that π‘˜ is a constant, and to multiply variables, we add their exponents. We get ο€Ήπ‘₯𝑦×12π‘₯+ο€Ήπ‘₯π‘¦Γ—π‘˜π‘¦ο…=12π‘₯𝑦+π‘˜π‘₯𝑦=12π‘₯𝑦+π‘˜π‘₯𝑦.

Since this expression needs to be equal to 12π‘₯𝑦+2π‘₯π‘¦οŠ©οŠ¨οŠ¨, we can equate the coefficients. Doing this, we get π‘˜=2.

Let’s finish by recapping some of the important points from this explainer.

Key Points

  • We can multiply polynomials by a monomial expression by distributing the monomial. This means we multiply every term of the polynomial by the monomial.

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