Lesson Explainer: Multiplying Binomials | Nagwa Lesson Explainer: Multiplying Binomials | Nagwa

Lesson Explainer: Multiplying Binomials Mathematics • First Year of Preparatory School

In this explainer, we will learn how to multiply two binomials using different methods such as FOIL or the vertical method.

Algebraic expressions are an integral part in representing relationships between unknown values. As mathematicians, we want to be able to manipulate all types of algebraic expressions to help us rewrite these relationships into different forms. One thing we may want to do is simplify the product of polynomial expressions.

We can tackle this problem in steps. First, let’s start with the product of two single-term polynomials (called monomials). Say we want to simplify 5𝑥𝑦×3𝑥𝑦. We reorder the multiplication using commutativity and associativity to get 5𝑥𝑦×3𝑥𝑦=(5×3)×𝑥×𝑥×𝑦×𝑦.

Then, we apply the product rule for exponents, which states that for any nonnegative integers 𝑚 and 𝑛 and rational number 𝑥, we have 𝑥×𝑥=𝑥. This gives us (5×3)×𝑥×𝑥×𝑦×𝑦=15𝑥𝑦=15𝑥𝑦.

We can use this result to simplify the process of multiplying monomials. We can extend this process to the product of a monomial and a binomial by using the distributive property of multiplication over addition which states that for any rational numbers 𝑎, 𝑏, and 𝑐, 𝑎×(𝑏+𝑐)=(𝑎×𝑏)+(𝑎×𝑐).

We know that multiplication is commutative, so we also have (𝑏+𝑐)×𝑎=(𝑏×𝑎)+(𝑐×𝑎).

We can substitute monomials into this rule to see how we can simplify algebraic expressions of this form. For example, if 𝑎=𝑥, 𝑏=2𝑥, and 𝑐=𝑥, we get 𝑥×2𝑥+𝑥=𝑥×2𝑥+𝑥×𝑥.

We say that we distribute the factor of 𝑥 over the sum. This means we just need to multiply every term in the sum by 𝑥. We can then simplify using the power rule for exponents: 𝑥×2𝑥+𝑥×𝑥=2𝑥+𝑥=2𝑥+5𝑥.

This allows us to multiply a binomial by a monomial; we just multiply both terms by the monomial. We can extend this even further by considering the product of two binomials. We can do this directly using the properties of addition and multiplication; however, it is easier to see this by considering a geometric model.

Let’s construct a rectangle with side lengths 𝑎+𝑏 and 𝑐+𝑑. Then, the product of these two expressions is the area of the rectangle.

We can also find an expression for this area by splitting the rectangle into smaller rectangles with side lengths 𝑎, 𝑏, 𝑐, and 𝑑 as shown.

The area of each rectangle is given by the product of its side lengths. This allows us to find expressions for the area of each of the four smaller rectangles.

The sum of the areas of the four smaller rectangles must give the area of the combined rectangle. Hence, (𝑎+𝑏)×(𝑐+𝑑)=𝑎𝑐+𝑏𝑐+𝑎𝑑+𝑏𝑑.

We see that the product of two binomials is equal to the sum of four multiplication terms; namely, each term of the first binomial is multiplied by each term of the second binomial.

Let’s use this geometric approach to find the product of binomials involving only one variable. For example, let’s say we want to evaluate (𝑥+2)(𝑥+3). We can draw a rectangle of length (𝑥+3) and width (𝑥+2). Its area is then given by (𝑥+2)(𝑥+3).

We find the area of each rectangle by multiplying their widths by their heights. We have 𝑥×𝑥=𝑥,3×𝑥=3𝑥,𝑥×2=2𝑥,3×2=6.

We can then add the areas together to find the area of the larger rectangle. We get (𝑥+2)(𝑥+3)=𝑥+3𝑥+2𝑥+6.

We can then simplify by combining like terms to get 𝑥+3𝑥+2𝑥+6=𝑥+5𝑥+6.

Let’s see now an example of using this geometric approach to multiply two binomials.

Example 1: Multiplying Two Binomials

Expand the product (𝑥+4)(𝑥+6).

Answer

We see that we are asked to expand the product of two binomials. We can do this by using the grid method. We can think of a rectangle with side lengths 𝑥+4 and 𝑥+6 split into smaller rectangles as shown.

The outer rectangle has an area of (𝑥+4)(𝑥+6) and we can find the areas of the smaller rectangles by multiplying their widths by their lengths.

Adding the areas of the smaller rectangles gives (𝑥+4)(𝑥+6)=𝑥+4𝑥+6𝑥+24.

Combining like terms then gives us 𝑥+4𝑥+6𝑥+24=𝑥+(4+6)𝑥+24=𝑥+10𝑥+24.

Hence, (𝑥+4)(𝑥+6)=𝑥+10𝑥+24.

Let’s now look at (𝑥2)𝑥+1. We can represent this product as the area of a rectangle of dimensions 𝑥2 and 𝑥+1 as shown in the following diagram.

Since 𝑥=(𝑥2)+2, we can now draw a rectangle of dimensions 𝑥 and 𝑥+1 and see that it is made of this rectangle plus a rectangle of dimensions 2 and 𝑥+1.

We find that the area (𝑥2)𝑥+1, highlighted in blue, is given by the area of the larger rectangle 𝑥𝑥+1 minus the area 2𝑥+1, highlighted in green: (𝑥2)𝑥+1=𝑥𝑥+12𝑥+1.

This example illustrates how multiplying a difference by a number is the same as multiplying the difference of the products of both terms of the subtraction by the number. Note that the distributive property of multiplication over addition and subtraction applies as well with negative numbers, even if in this case we cannot represent each number as a length. This is because it is possible to use the properties of addition and multiplication to show that, in general, (𝑎+𝑏)×(𝑐+𝑑)=𝑎𝑐+𝑏𝑐+𝑎𝑑+𝑏𝑑.

The equation above, (𝑥2)𝑥+1=𝑥𝑥+12𝑥+1, shows a way to calculate the product of binomials; it is called the horizontal method. In general form, we can write it as (𝑎+𝑏)(𝑐+𝑑)=𝑎(𝑐+𝑑)+𝑏(𝑐+𝑑).

This is then an easier expression to evaluate, since we are no longer multiplying two binomials.

Before we move on to the next example, there is another way of multiplying binomials we can use. In the above example, we saw that (𝑥2)𝑥+1=𝑥×𝑥+(𝑥×1)+2×𝑥+(2×1).

We are multiplying each term in the first factor by each term in the second factor and adding the results together.

One way of remembering which terms we multiply is to use the FOIL method. This tells us that we start with the first term in each factor. Then, we move on to the outer terms in each factor. Next, we multiply the inner terms in each factor. Finally, we multiply the last terms in each factor.

It is personal preference which method to use; however, it is useful to be comfortable with both methods.

In our next example, we will use the FOIL method to determine the value of an unknown coefficient.

Example 2: Multiplying Two Different Binomials to Find an Unknown in an Expanded Expression

If (4𝑦3)(5𝑦+6)=20𝑦+𝑘𝑦18, what is the value of 𝑘?

Answer

To determine the value of 𝑘, we want to expand the left-hand side of the equation, and we recall that we can do this by using the FOIL method. We multiply the first term in each factor, then the outer terms, followed by the inner terms, and finally the last terms. The sum of these four expressions will then give us the product.

The first terms in each factor are 4𝑦 and 5𝑦, so we get 4𝑦×5𝑦=20𝑦.

Next, the outer terms are 4𝑦 and 6, so we find 4𝑦×6=24𝑦.

The inner terms are 3 and 5𝑦; multiplying these, we get 3×5𝑦=15𝑦.

Finally, the last terms are 3 and 6, and we calculate 3×6=18.

Adding these expressions gives us the following.

We can combine like terms to get (4𝑦3)(5𝑦+6)=20𝑦+(2415)𝑦18=20𝑦+9𝑦18.

We are told that this is equal to 20𝑦+𝑘𝑦18; we could solve this by setting the expressions equal to each other and rearranging: 20𝑦+9𝑦18=20𝑦+𝑘𝑦18.

However, it is easier to equate coefficients. We note that the 𝑦 and constant terms are the same, so the 𝑦-terms must be the same to make the equation true for any value of 𝑦.

Therefore, 9𝑦=𝑘𝑦.

Hence, we must have 𝑘=9.

Before we move on to our next example, let’s look at the vertical method.

This method takes inspiration from one of the methods we may have seen to multiply large numbers. For example, we can multiply 13 by 12 vertically as shown: 13×1226+130156.

This process works by rewriting 12 as 2+10 and 13 as 3+10. We have 12×13=(2+10)(3+10).

We can think of this as the product of binomials. Expanding using the horizontal method, we have (2+10)(3+10)=2(3+10)+10(3+10).

Evaluating these products gives 26 and 130; we can then add these values to find the product of 13 and 12.

In the same way, we can multiply binomials using this vertical method. Let’s say we want to multiply 𝑥+2 by 3𝑥+1. We start by writing the expressions vertically: 𝑥+2×3𝑥+1

Next, we need to multiply every term of 𝑥+2 by 1 and write the result below the line. We have 1×2=2 and 1×𝑥=𝑥. So, we get the following.

For the second step, we now follow the same process with 3𝑥. We have 3𝑥×2=6𝑥 and 3𝑥×𝑥=3𝑥. We can write the sum of these terms below the line to get the following.

For the final step, we find the sum of these four terms, which gives us the product: 𝑥+2×3𝑥+1𝑥+2+3𝑥+6𝑥3𝑥+7𝑥+2.

Hence, (𝑥+2)(3𝑥+1)=3𝑥+7𝑥+2.

Let’s now see an example of using the vertical method to expand a product of binomials.

Example 3: Multiplying Two Different Binomials Where the Result Is a Difference of Two Squares

Expand the product (2𝑚+𝑛)(2𝑚𝑛).

Answer

We are asked to expand the product of two binomials and there are many different ways of doing this. We are going to use the vertical method, which involves writing one factor below the other and then finding the product of each pair of terms: 2𝑚+𝑛×2𝑚𝑛

We start by multiplying each term of 2𝑚+𝑛 by 𝑛. We have (𝑛)×2𝑚=2𝑚𝑛 and (𝑛)×𝑛=𝑛. Adding these together gives the following.

We now multiply each term of 2𝑚+𝑛 by 2𝑚. We have 2𝑚×𝑛=2𝑚𝑛 and 2𝑚×2𝑚=4𝑚. Adding these together gives the following.

We now add these four terms together to find the product: 2𝑚+𝑛×2𝑚𝑛2𝑚𝑛𝑛+4𝑚+2𝑚𝑛4𝑚𝑛.

We see that 4𝑚+2𝑚𝑛2𝑚𝑛𝑛=4𝑚𝑛.

Hence, (2𝑚+𝑛)(2𝑚𝑛)=4𝑚𝑛.

In the previous example, it is worth noting that we have (2𝑚+𝑛)(2𝑚𝑛)=4𝑚𝑛=(2𝑚)𝑛.

This is called a difference of squares. Let’s represent 𝑥𝑦 graphically with a square of area 𝑥 from which a smaller square of area 𝑦 is removed.

The shaded region has an area of 𝑥𝑦. We can also find the area of the remaining shaded region by splitting the shape into two rectangles.

The shaded region has an area of 𝑥(𝑥𝑦)+𝑦(𝑥𝑦).

By rotating the rectangle of area 𝑦(𝑥𝑦) and placing it on top of the other, we get a combined rectangle with length 𝑥+𝑦 and width 𝑥𝑦.

We can multiply these expressions to get the area of the rectangle, or we can add the areas of each rectangle. Equating these expressions gives us (𝑥+𝑦)(𝑥𝑦)=𝑥(𝑥𝑦)+𝑦(𝑥𝑦).

We recognize here the horizontal method for the multiplication of two binomials.

Equating the expressions for the shaded areas gives 𝑥𝑦=(𝑥+𝑦)(𝑥𝑦).

This equation shows that the difference of the squares of two numbers equals the product of their sum and their difference.

In our next example, we will expand and simplify the square of a binomial.

Example 4: Squaring a Binomial

Expand and simplify (𝑥+2)(𝑥+2).

Answer

In this question, we are asked to expand the product of two binomials. We can note that the two binomials are the same, so, in fact, this is equivalent to squaring the binomial: (𝑥+2)(𝑥+2)=(𝑥+2).

We can use many methods to expand the product of binomials. We will use the grid method since there is a nice geometric interpretation when dealing with the squares of binomials.

Usually, we sketch a rectangular grid with sides of lengths equal to each factor. However, in this case, the factors are the same, so the side lengths will be the same. We get a square.

We split the square into four to help us evaluate the product.

We find the area of each of the smaller squares and rectangles by multiplying their lengths by their widths.

The sum of these areas then gives us the area of the square, which we know is (𝑥+2)(𝑥+2). Therefore, (𝑥+2)(𝑥+2)=𝑥+2𝑥+2𝑥+4.

Combining like terms gives us 𝑥+2𝑥+2𝑥+4=𝑥+(2+2)𝑥+4=𝑥+4𝑥+4.

Hence, (𝑥+2)(𝑥+2)=𝑥+4𝑥+4.

In the previous example, we found the square of a binomial. We can follow this process in general to show that (𝑎+𝑏)=(𝑎+𝑏)(𝑎+𝑏)=𝑎+𝑎𝑏+𝑏𝑎+𝑏=𝑎+2𝑎𝑏+𝑏.

In our final example, we will expand the product of binomials to find an expression for the area of a rectangle whose sides are given as binomial expressions.

Example 5: Finding an Expression for the Area of a Rectangle by Multiplying Binomials

Find the area of a rectangle that has a width of 𝑥5𝑦 cm and a length of 𝑥+4𝑦 cm.

Answer

We note that the area of a rectangle is the product of its dimensions: (𝑥5𝑦)(𝑥+4𝑦). One way of answering this question is to use the horizontal method to multiply the two binomials. We do this by multiplying each term in the second binomial by the entire first binomial: (𝑥5𝑦)(𝑥+4𝑦)=𝑥(𝑥+4𝑦)5𝑦(𝑥+4𝑦).

We can then distribute each factor of the parentheses to get 𝑥(𝑥+4𝑦)5𝑦(𝑥+4𝑦)=(𝑥×𝑥)+(𝑥×4𝑦)+(5𝑦×𝑥)+(5𝑦×4𝑦)=𝑥4𝑥𝑦+5𝑥𝑦20𝑦=𝑥+(54)𝑥𝑦20𝑦=𝑥+𝑥𝑦20𝑦.

Hence, the area of the rectangle is 𝑥+𝑥𝑦20𝑦 cm2.

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

Key Points

  • We can evaluate the product of two binomial expressions by combining the distributive property of multiplication over addition and subtraction with the power rule for exponents.
  • There are many different methods we can use to expand the product of two binomial expressions including the FOIL method, the horizontal method, and the vertical method.
  • The difference between two squares tells us that (𝑎+𝑏)(𝑎𝑏)=𝑎𝑏.
  • We can square a binomial using the fact that (𝑎+𝑏)=𝑎+2𝑎𝑏+𝑏.

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