## Section5.4Products of Binomials

### SubsectionProducts of Variables

In Section 5.1, we learned to simplify the sum of two algebraic expressions by combining like terms. In this section we see how to multiply algebraic expressions.

Recall that we can simplify a product such as $3(2x)$ because

\begin{equation*} 3(2x) = 2x+2x+2x = 6x~~~~~~~~~~~~\blert{\text{Three terms}} \end{equation*}

This calculation is actually an application of the associative property:

\begin{equation*} 3(2x) = (3 \cdot 2)x = 6x \end{equation*}

In a similar way, we can simplify the product $(3b)(4b)$ by applying the commutative and associative properties:

\begin{equation*} (3b)(4b) = 3 \cdot b \cdot 4 \cdot b = 3 \cdot 4 \cdot b \cdot b = (3 \cdot 4) \cdot (b \cdot b) = 12b^2 \end{equation*}
###### Look Closer.

You can convince yourself that $(3b)(4b)$ is equivalent to $12b^2$ by substituting some values for $b\text{;}$ for example, if $b=\alert{2}\text{,}$ then

The commutative and associative properties tell us that we can multiply the factors of a product in any order.

###### Example5.27.

Simplify the product or power.

1. $(5a)(-3a)$
2. $(2x)^3$
3. $(xy^2)(4x^2)$
Solution
1. We apply the commutative property:

\begin{equation*} (5a)(-3a) = 5(-3) \cdot a \cdot a = -15a^2 \end{equation*}
2. To cube an expression means to multiply three copies of the expression together:

\begin{equation*} (2x)^3 = (2x)(2x)(2x) = 2 \cdot 2 \cdot 2 \cdot x \cdot x \cdot x = 8x^3 \end{equation*}
3. We rearrange the factors to group each variable together:

\begin{equation*} (xy^2)(4x^2) = x \cdot y \cdot y \cdot 4 \cdot x \cdot x = 4 \cdot x \cdot x \cdot x \cdot y \cdot y = 4x^3y^2 \end{equation*}
###### Caution5.28.

When we add like terms, we do not change the variable in the terms; we combine the coefficients. For example,

\begin{equation*} 3a+2a=5a \end{equation*}

When we multiply expressions, we multiply the coefficients and we multiply the variables:

\begin{equation*} 3a(2a) = 3(2) \cdot a \cdot a = 6a^2 \end{equation*}

###### 1.

Explain the difference between $5x-2x$ and $5x(-2x)\text{.}$

### SubsectionUsing the Distributive Law

We can use the areas of rectangles to investigate products of algebraic expressions. Recall that we find the area of a rectangle by multiplying its length times its width, $A=lw\text{.}$ We have already used rectangles to visualize the distributive law. Here are some examples.

###### Example5.29.

Calculate the area of the rectangle by adding the areas of each piece. Then use the distributive law to find the product of the algebraic expressions.

1. $\blert{\text{Area} = 15x+20}$

$5(3x+4)=5(3x)+5(4) = 15x + 20$

2. $\blert{\text{Area} = 2x^2+18x}$

$2x(x+9)=2x(x)+2x(9)=2x^2+18x$

3. $\blert{\text{Area} = 12b^2+21b}$

$3b(4b+7)=3b(4b)+3b(7)=12b^2+21b$

###### 2.

State the distributive law, and explain what it means.

At this stage it will be helpful to introduce some terminology.

• An algebraic expression with only one term, such as $2x^3\text{,}$ is called a monomial.
• An expression with two terms, such as $x^2-16\text{,}$ is called a binomial.
• An expression with three terms is a trinomial.
• The expression $ax^2+bx+c$ is thus called a quadratic trinomial, because it involves the square of the variable.
###### Caution5.30.

Notice the difference between $(3a)(2a)$ and $3a(2+a)\text{:}$

• $(3a)(2a)$ is the product of two monomials, and we use the commutative property to simplify it:

\begin{equation*} (3a)(2a)= 3 \cdot a \cdot 2 \cdot a =3 \cdot 2 \cdot a \cdot a = 6a^2 \end{equation*}
• $3a(2+a)$ is the product of a monomial and a binomial, and we use the distributive law to simplify it:

\begin{equation*} 3a(2+a)=3a(2)+3a(a)=6a+3a^2 \end{equation*}

###### 3.

Explain the terms monomial, binomial, and trinomial.

### SubsectionMultiplying Binomials

Consider the rectangle shown at right. As you can see, it is divided into four smaller rectangles. You can verify that we get the same answer when we compute its area in two different ways: We can add up the areas of the four smaller rectangles, or we can find the length and width of the entire large rectangle and then find their product: $\blert{\text{Sum of four sub-rectangles:}}$

\begin{align*} \text{Area} \amp = 8(5)+8(15)+4(5)+4(15)\\ \amp = 40+120+20+60\\ \amp = 240 \end{align*}

$\blert{\text{One large rectangle:}}$

\begin{align*} \text{Area} \amp =(8+4)(5+15)~~~~~~~~\\ \amp = (12)(20)\\ \amp = 240 \end{align*}

Our goal in this Lesson is to understand products of binomials. We can use rectangles to illustrate, or model, the product of two binomials. The rectangles do not have to be drawn exactly to scale; they are merely tools for visualizing products. With a small stretch of the imagination, we can use rectangles to represent negative numbers as well.

###### Example5.31.
1. Use a rectangle to represent the product $(x-4)(x+6)\text{.}$
2. Write the product as a quadratic trinomial.
Solution
1. We let the first factor, $(x-4)\text{,}$ represent the width of the rectangle, and the second factor, $(x+6)\text{,}$ represent its length.
2. We find the area of each sub-rectangle, as shown in the figure. Then we add the areas together. \begin{align*} \text{Area} \amp =x^2+6x-4x-24\\ \amp = x^2+2x-24 \end{align*}

We say that $(x+6)(x-4)$ is the factored form of the product, and $x^2+2x-24$ is the expanded form.

###### 4.

When we use a rectangle to model the product of two binomials, what do the two binomials represent? What does their product represent?

### SubsectionThe Four Terms in a Binomial Product

Using a rectangle to multiply binomials illustrates how the distributive law works. Analyzing the rectangle method will help us in a later Lesson, when we reverse the process to factor a quadratic trinomial. Let us take a closer look at the example above.

###### Look Closer.

In Example 5.31 we computed the product $(x-4)(x+6)\text{.}$ The top row of the rectangle corresponds to

\begin{equation*} x(x+6)=x^2+6x \end{equation*}

and the bottom row corresponds to

\begin{equation*} -4(x+6)=-4x-24 \end{equation*} Thus, we multiply each term of the first binomial by each term of the second binomial, resulting in four multiplications in all:

\begin{equation*} (x-4)(x+6)=x^2+6x-4x-24 \end{equation*}

Each term of the product corresponds to the area of one of the four sub-rectangles, as shown below. The letters $\blert{\text{F, O, I, L}}$ indicate the four steps in computing the product:

1. $\blert{\text{F}}$ stands for the product of the $\blert{\text{First}}$ terms in each binomial.
2. $\blert{\text{O}}$ stands for the product of the $\blert{\text{Outer}}$ terms.
3. $\blert{\text{I}}$ stands for the product of the $\blert{\text{Inner}}$ terms.
4. $\blert{\text{L}}$ stands for the product of the $\blert{\text{Last}}$ terms.

Note how each term of the trinomial product arises from the binomial factors:

\begin{align*} (x+6)(x-4)= \amp x^2+2x-24\\ \amp \blert{\text{F}~~~~~~~\text{O+I}~~~~~~~\text{L}} \end{align*}
• The quadratic term in the product comes from the First terms.
• The linear or $x$-term of the product is the sum of Outer and Inner.
• The constant term of the product comes from the Last terms.

###### 5.

What does the acronym FOIL stand for?

###### Example5.32.

Compute the product $(3x-5)(4x+2)\text{,}$ and write your answer as a quadratic trinomial.

Solution

We multiply each term of the first factor by each term of the second factor, using the "FOIL" template to keep track of the products.

\begin{align*} (3x-5)(4x+2) \amp =3x(4x)+3x(2)-5(4x)-5(2)\\ \amp ~~~~~~~~~\blert{\text{F}~~~~~~~~~~~~~~~~~\text{O}~~~~~~~~~~~~~~~~\text{I}~~~~~~~~~~~~~~\text{L}}\\ \amp = 12x^2+6x-20x-10 \amp \amp \blert{\text{Combine like terms.}}\\ \amp = 12x^2-14x-10 \end{align*}

###### 6.

In the "FOIL" representation of a binomial product, which terms are like terms?

### SubsectionSkills Warm-Up

#### ExercisesExercises

Find the area and perimeter of each figure.

###### 1. ###### 2. ###### 3. ###### 4. ###### 5. ###### 6. ### ExercisesHomework 5.4

For Problems 1–3, simplify each product or power.

###### 1.
1. $3(4n)$
2. $3n(4n)$
3. $(4n)^3$
###### 2.
1. $6x(-5x^2)$
2. $-6x^2(-5x)$
3. $(-5x)^2$
###### 3.
1. $(4p)(-4p)$
2. $-(4p)^2$
3. $-4(-p)^2$

For Problems 4–6, simplify each expression.

###### 4.
1. $-8x(5t)$
2. $-8x(5+t)$
3. $-8x(-5-t)$
###### 5.
1. $3n(-4n)$
2. $3n-4n$
3. $(3n-4)n$
###### 6.
1. $2x(-5x^2)$
2. $2(x-5x^2)$
3. $2x-(5x)^2$

For Problems 7–8,

1. Write a product (length $\times$ width) for the area of the rectangle.
2. Use the distributive law to compute the product.
###### 7. ###### 8. For Problems 9–14, compute the product.

###### 9.
$-2b(6b-2)$
###### 10.
$-(6a-5)(3a)$
###### 11.
$3v(5v-2v^2)$
###### 12.
$-4x^2(2x+3y)$
###### 13.
$(y^3+3y-2)(2y)$
###### 14.
$-xy(2x^2-xy+3y^2)$

For Problems 15–18, simplify.

###### 15.
$2a(x+3)-3a(x-3)$
###### 16.
$2x(3-x)+2(x^2+1)-2x$
###### 17.
$ax(x^2+2x-3)-a(x^3+2x^2)$
###### 18.
$3ab^2(2+3a)-2ab(3ab+2b)$

For Problems 19–20, write two different expressions for the area of the rectangle:

1. as the sum of four small areas,
2. as one large rectangle, using the formula Area $=$ length $\times$ width.
###### 19. ###### 20. For Problems 21–23,

1. Use a rectangle to represent each product.
2. Write the product as a quadratic trinomial.
###### 21.
$(a-5)(a-3)$
###### 22.
$(y+1)(3y-2)$
###### 23.
$(5x-2)(4x+3)$

For Problems 24–26, use rectangles to help you multiply these binomials in two variables.

###### 24.
$(x+2y)(x-y)$
###### 25.
$(3s+t)(2s+3t)$
###### 26.
$(2x-a)(x-3a)$

For Problems 27–30, compute the product: Multiply the binomials together first, then multiply the result by the numerical coefficient.

###### 27.
$2(3x-1)(x-3)$
###### 28.
$-3(x+4)(x-1)$
###### 29.
$-(4x+3)(x-2)$
###### 30.
$5(2x+1)(2x-1)$

For Problems 31–32, find the product without a calculator by using rectangles.

###### 31.

$36 \times 42$ ###### 32.

$82 \times 16$

For Problems 33–34,

1. Find the linear term in the product.
2. Shade the sub-rectangles that correspond to the linear term.
###### 33.

$(x+6)(x-9)$ ###### 34.

$(2x-5)(x+4)$ For Problems 35–37, compute the product. What do you notice? Explain why this happens.

###### 35.
$(x+3)(x-3)$
###### 36.
$(x-2a)(x+2a)$
###### 37.
$(3x+1)(3x-1)$

For Problems 38–40, compute the product.

###### 38.
$(w+4)(w+4)$
###### 39.
$(z-6)(z-6)$
###### 40.
$(3a-2c)(3a-2c)$
###### 41.
1. Complete the table below.
2. Decide whether the two expressions, $(a-b)^2$ and $a^2-b^2\text{,}$ are equivalent.

 $a$ $b$ $a-b$ $(a-b)^2$ $a^2$ $b^2$ $a^2-b^2$ $5$ $3$ $\hphantom{0000}$ $\hphantom{0000}$ $\hphantom{0000}$ $\hphantom{0000}$ $\hphantom{0000}$ $2$ $6$ $\hphantom{0000}$ $\hphantom{0000}$ $\hphantom{0000}$ $\hphantom{0000}$ $\hphantom{0000}$ $-4$ $-3$ $\hphantom{0000}$ $\hphantom{0000}$ $\hphantom{0000}$ $\hphantom{0000}$ $\hphantom{0000}$

###### 42.

Explain why $(x-y)^2$ cannot be simplified to $x^2-y^2\text{.}$

For Problems 43–44, write the area of the square in two different ways:

1. as the sum of four smaller areas,
2. as one large square, using the formula Area $= (\text{length})^2\text{.}$
###### 43. ###### 44. ###### 45.

Is $(x+4)^2$ equivalent to $x^2+4^2\text{?}$ Explain why or why not, and give a numerical example to justify your answer.

For Problems 46–48, compute the product.

###### 46.
$(x-2)^2$
###### 47.
$(2x+1)^2$
###### 48.
$(3x-4y)^2$

For Problems 49–50, use the Pythagorean theorem to write an equation about the sides of the right triangle.

###### 49. ###### 50. 