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# 4.4: Multiplying Integers

Difficulty Level: At Grade Created by: CK-12

## Introduction

Temperature Change

As Cameron gets ready for his dive one morning, he notices that the temperature at the resort is already 87\begin{align*}87^\circ\end{align*}.

“It is going to be a hot one!” Jesse, the boat driver, comments heading down to the dock.

Cameron gets his suit and his gear and climbs aboard the boat. The ride out to the dive site feels wonderful with the wind blowing in his hair. The water is clear and sparkling under the hot sun.

Cameron knows that the water temperature changes about 2\begin{align*}-2^\circ\end{align*} every 10 feet. He wears a light skin suit because of this. Not only can it feel a bit chilly after the warm surface temperature, but it can also protect him against a random jellyfish sting.

If Cameron dives to a depth of 40 feet, what will the temperature be at that depth?

This lesson will teach you how to use multiplying integers to solve a problem like this one. You will need to multiply and then subtract to figure out what the temperature of the water will be, at the depth that Cameron plans to go on his dive.

What You Will Learn

By the end of this lesson, you will be able to complete the following:

• Analyze patterns of products of integers with same and different signs recognizing multiplication by zero as zero.
• Multiply integers.
• Evaluate variable expressions involving integer multiplication.
• Model and solve real-world problems using simple equations involving integer multiplication.

Teaching Time

I. Analyze Patterns of Products of Integers with Same and Different Signs Recognizing Multiplication by Zero as Zero

Now that you have learned how to add and subtract integers, it is time to tackle multiplying them. Remember that an integer is the set of whole numbers and their opposites. There are a few vocabulary words that help us when multiplying. The first is a factor. The numbers that are multiplied are called factors. The second is product. We multiply two or more factors to get a product.

Below are some multiplication facts for 5. Notice that the products show a pattern. Suppose you did not know the product of 5×0\begin{align*}5 \times 0\end{align*}. How could you use the pattern shown below to determine that product?

5×45×35×25×15×0=20=15=10=5=?\begin{align*}5 \times 4 & =20\\ 5 \times 3 & =15\\ 5 \times 2 & =10\\ 5 \times 1 & =5\\ 5 \times 0 & = ?\end{align*}

Notice that each product shown is 5 less than the previous product. So, you can subtract 5 from the previous product, 5, to find the missing product. Since 55=0\begin{align*}5-5=0\end{align*}, the product of 5×0\begin{align*}5 \times 0\end{align*} must be 0.

There are patterns that we can see when we multiply integers. Analyzing patterns like these can help us multiply positive and negative integers.

How do analyze these patterns?

First, we notice that the pattern for the multiplication facts of 5 could continue below zero. Up until this time in math, we have only looked at the positive products. But now, since we are learning about negative numbers, we can look at continuing the facts. Let's see how patterns can help us find the products of integers.

Example

Use a pattern to find the missing products below.

5×45×35×25×15×05×(1)5×(2)=20=15=10=5=0=?=?\begin{align*}5 \times 4 & = 20\\ 5 \times 3 & = 15\\ 5 \times 2 & = 10\\ 5 \times 1 & = 5\\ 5 \times 0 & = 0\\ 5 \times (-1) & = ?\\ 5 \times (-2) & = ?\end{align*}

You already know the pattern for this sequence of products. You can subtract 5 from the previous product to find the next product. Remember, subtracting 5 is the same thing as adding its opposite -5. Try adding -5 to the previous products to find the next products.

To find the product of 5×(1)\begin{align*}5 \times (-1)\end{align*}, add 0+(5)\begin{align*}0+(-5)\end{align*}

|0|=0\begin{align*}|0|=0\end{align*} and |5|=5\begin{align*}|-5|=5\end{align*}, so subtract the lesser absolute value from the greater absolute value :

50=5\begin{align*}5-0=5\end{align*}.

The integer with the greater absolute value is -5, so give the answer a negative sign.

0+(5)=5\begin{align*}0+(-5)=-5\end{align*}, so 5×(1)=5\begin{align*}5 \times (-1)=-5\end{align*}

To find the product of 5×(2)\begin{align*}5 \times (-2)\end{align*}, add -5 to the previous product, which is also -5.

In other words, add: 5+(5)\begin{align*}-5+(-5)\end{align*}

Both integers have the same sign, so add their absolute values.

|5|=5\begin{align*}|-5|=5\end{align*}, so add.

5+5=10\begin{align*}5+5=10\end{align*}.

Give that answer a negative sign.

5+(5)=10\begin{align*}-5+(-5)=-10\end{align*}, so 5×(2)=10\begin{align*}5 \times (-2) = -10\end{align*}

Now we have our completed multiplication facts

5×45×35×25×15×05×(1)5×(2)=20=15=10=5=0=5=10\begin{align*}5 \times 4 & = 20\\ 5 \times 3 & = 15\\ 5 \times 2 & = 10\\ 5 \times 1 & = 5\\ 5 \times 0 & = 0\\ 5 \times (-1) & = -5\\ 5 \times (-2) & = -10\end{align*}

Each product is 5 less than the previous product.

You may also notice that depending on what you are multiplying the sign changes.

Here are a few rules that we can conclude from the pattern.

• When 5, a positive integer, is multiplied by a positive integer, the product is positive.
• When 5, a positive integer, is multiplied by zero, the product is zero.
• When 5, a positive integer, is multiplied by a negative integer, the product is negative.

You may think that these rules apply only to the five's table but they actually apply to ANY table when you are multiplying positive and negative integers.

Let’s look at another example that can help us finish figuring out the rules for multiplying integers.

Example

Use a pattern to find the missing products below. What do you notice? Do you notice that the answer increases by 3 every time?

(3)×3(3)×2(3)×1(3)×0(3)×(1)(3)×(2)(3)×(3)=9=6=3=0=3=?=?\begin{align*}(-3) \times 3 & = -9\\ (-3) \times 2 & = -6\\ (-3) \times 1 & = -3\\ (-3) \times 0 & = 0\\ (-3) \times (-1) & = 3\\ (-3) \times (-2) & = ?\\ (-3) \times (-3) & = ?\end{align*}

To find the product of (3)×(2)\begin{align*}(-3) \times (-2)\end{align*}, add: 3+3=6\begin{align*}3+3=6\end{align*}. So, (3)×(2)=6\begin{align*}(-3) \times (-2)= 6\end{align*}.

To find the product of (3)×(3)\begin{align*}(-3) \times (-3)\end{align*}, add: 6+3=9\begin{align*}6+3=9\end{align*}. So, (3)×(3)=9\begin{align*}(-3) \times (-3) = 9\end{align*}.

This shows the completed multiplication facts.

(3)×3(3)×2(3)×1(3)×0(3)×(1)(3)×(2)(3)×(3)=9=6=3=0=3=6=9\begin{align*}(-3) \times 3 & = -9\\ (-3) \times 2 & = -6\\ (-3) \times 1 & = -3\\ (-3) \times 0 & = 0\\ (-3) \times (-1) & = 3\\ (-3) \times (-2) & = 6\\ (-3) \times (-3) & = 9\end{align*}

Here are some other rules that you may notice.

• When -3, a negative integer, is multiplied by a positive integer, the product is negative.
• When -3, a negative integer, is multiplied by zero, the product is zero.
• When -3, a negative integer, is multiplied by a negative integer, the product is positive.

We can conclude the following rules.

Take a few minutes and write these rules down in your notebooks.

II. Multiply Integers

Now that you understand the rules, we can work on applying them when actually multiplying integers.

Refer back to the rules as you work, but the key to becoming GREAT at multiplying integers is to commit these rules to memory!!

Example

(4)(3)\begin{align*}(-4)(-3)\end{align*}

Here we have negative four times a negative three. First, we multiply the terms, remember that a set of parentheses next to another set means multiplication.

4×3=12\begin{align*}4 \times 3 = 12\end{align*}

Next, we figure out the sign.

A negative times a negative is a positive.

Example

58\begin{align*}-5 \cdot 8\end{align*}

Here we have a negative five times a positive eight. Remember that a dot can also mean multiplication.

5×8=40\begin{align*}5 \times 8 = 40\end{align*}

Next, we figure out the sign.

A negative times a positive is a negative.

What about multiplying more than one term?

We can do this easily. The key is to work from left to right and remember that the sign of each product can change with each factor.

Example

(8)(3)(2)\begin{align*}(-8)(-3)(-2)\end{align*}

Here we have three negative terms being multiplied. First, let’s multiply the first two terms to get a product.

8×3=24\begin{align*}-8 \times -3 = 24\end{align*} (A negative times a negative equals a positive.)

Now we multiply that product times the factor negative two.

24×2=48\begin{align*}24 \times -2 = -48\end{align*} (A positive times a negative equals a negative.)

4J. Lesson Exercises

1. 9(3)\begin{align*}-9(-3)\end{align*}
2. (3)(12)\begin{align*}(-3)(12)\end{align*}
3. (4)(3)(2)\begin{align*}(-4)(3)(-2)\end{align*}

Take a few minutes to check your answers with a partner. Is your work accurate? Did you remember the rules for multiplying integers?

III. Evaluate Variable Expressions Involving Integer Multiplication

Do you remember what a variable expression is?

A variable expression is a number sentence using numbers, operations and variables. A variable expression can contain like terms. A like term is a term that is common between one or more terms in the equation. When you have like terms, we can combine them using addition and subtraction.

Pay attention! Here is a change!

How does this happen?

Let’s look at an example so that it will make more sense.

Example

Find the value of this expression 3z(2)\begin{align*}3z \cdot (-2)\end{align*}

The Commutative Property of Multiplication states that the order in which factors are multiplied does not matter.

3z(2)=23z\begin{align*}3z \cdot (-2) = -2 \cdot 3z\end{align*}

The Associative Property of Multiplication states that you can group the factors being multiplied in any order.

23z=23z=(23)z\begin{align*}-2 \cdot 3z = -2 \cdot 3 \cdot z = (-2 \cdot 3) \cdot z\end{align*}

Now we can multiply the integers. Since -2 and 3z\begin{align*}3z\end{align*} have different signs, the product will be negative.

23=6\begin{align*}2 \cdot 3=6\end{align*}, so (2)3=6\begin{align*}(-2)\cdot 3=-6\end{align*} and (23)z=6z\begin{align*}(-2 \cdot 3)\cdot z = -6z\end{align*}

The value of the expression is 6z\begin{align*}-6z\end{align*}.

Example

Find the value of this expression (5)(2m)(n)\begin{align*}(-5)(-2m)(n)\end{align*}

The Associative Property of Multiplication states that you can group the factors being multiplied in any order.

(5)(2m)(n)=(5)(2)mn\begin{align*}(-5)(-2m)(n) = (-5)\cdot (-2) \cdot m \cdot n\end{align*}

Now we can multiply the integers. Since -5 and -2 are both negative, the product will be positive.

52=52=10\begin{align*}-5 \cdot -2=5 \cdot 2 = 10\end{align*}, so \begin{align*}(-5) \cdot (-2) \cdot m \cdot n = 10 \cdot mn=10mn\end{align*}

The value of the expression is \begin{align*}10mn\end{align*}.

4K. Lesson Exercises

Multiply the following variable expressions.

1. \begin{align*}3x(4y)\end{align*}
2. \begin{align*}-6a(-4b)\end{align*}
3. \begin{align*}-4z(10)\end{align*}

IV. Model and Solve Real-World Problems Using Simple Equations Involving Integer Multiplication

Knowing how to multiply integers can also help us represent and solve problems in real life.

To solve a real-world problem, write an expression or an equation that can be used to solve the problem, then solve.

Example

The number of students voting in a school election has been decreasing at a rate of 15 students per year. Represent the change in the number of students voting over the next 3 years as an integer.

First, represent the decrease in the number of students voting as an integer.

The problem states that the number of students voting has been decreasing by 15 students each year. To show a decrease, use a negative integer -15.

To represent the decrease in the number of students voting over the next 3 years, multiply the number of years by the integer representing the decrease.

\begin{align*}3 \times (-15) = ?\end{align*}

Find the product to solve the problem.

\begin{align*}3 \times 15 =45\end{align*}, so \begin{align*}3 \times (-15)=-45\end{align*}

The change in the number of students voting over the next 3 years can be represented as -45.

Now let’s use this information to solve our real-world problem.

## Real-Life Example Completed

Temperature Change

Here is the original problem once again. Reread it and underline any important information.

As Cameron gets ready for his dive one morning, he notices that the temperature at the resort is already \begin{align*}\underline{87^\circ}\end{align*}.

“It is going to be a hot one!” Jesse, the boat driver, comments heading down to the dock.

Cameron gets his suit and his gear and climbs aboard the boat. The ride out to the dive site feels wonderful with the wind blowing in his hair. The water is clear and sparkling under the hot sun.

Cameron knows that the water temperature changes about \begin{align*}-2^\circ\end{align*} every 10 feet. He wears a light skin suit because of this. Not only can it feel a bit chilly after the warm surface temperature, but it can also protect him against a random jellyfish sting.

If Cameron dives to a depth of 40 feet, what will the temperature be at that depth?

To figure this out, we must first figure out the change in temperature between the surface and the depth of 40 feet. To figure this out, we multiply.

-2 degrees per 10 feet

If we are going down 40 feet, then we can multiply \begin{align*}-2 \times 4\end{align*} because 40 is \begin{align*}4 \times 10\end{align*}.

\begin{align*}-2(4) = -8\end{align*}

Now we can add this to the surface temperature of \begin{align*}87^\circ\end{align*}.

\begin{align*}87 + (-8) = 79^\circ\end{align*}

The temperature of the water at 40 feet deep will be \begin{align*}79^\circ\end{align*}.

## Vocabulary

Here are the vocabulary words in this lesson.

Integer
the set of whole numbers and their opposites.
Product
the answer in a multiplication problem
Factors
the numbers being multiplied
Variable Expression
a number sentence using numbers, operations and variables.
Commutative Property of Multiplication
states that the order that we multiply terms does not change the product.
Associative Property of Multiplication
states that changing the grouping of factors does not change the product.

## Time to Practice

Directions: Analyze the patterns below and find the missing products.

1. \begin{align*}6 \times 3 = 18\!\\ 6 \times 2 = 12\!\\ 6 \times 1 = 6\!\\ 6 \times 0 = 0\!\\ 6 \times (-1) = ?\!\\ 6 \times (-2) = ?\!\\ 6 \times (-3) = ?\end{align*}

2. \begin{align*}(-9) \times 4 = -36\!\\ (-9) \times 3 = -27\!\\ (-9) \times 2 = -18\!\\ (-9) \times 1 = -9\!\\ (-9) \times 0 = ?\!\\ (-9) \times (-1) = ?\!\\ (-9) \times (-2) = ?\end{align*}

Directions: Use integer rules to find each product.

3. \begin{align*}12 \times 8\end{align*}

4. \begin{align*}(-4)(7)\end{align*}

5. \begin{align*}(-2) \cdot (-11)\end{align*}

6. \begin{align*}3 \cdot (-25)\end{align*}

7. \begin{align*}5 \times (-6) \times (-1)\end{align*}

8. \begin{align*}(-8)(-7)\end{align*}

9. \begin{align*}(-2)(3)(-4)\end{align*}

10. \begin{align*}(-9)(1)(-1)\end{align*}

Directions: Multiply each variable expression.

11. \begin{align*}(-7k)(-6)\end{align*}

12. \begin{align*}(-8)(3a)(b)\end{align*}

13. \begin{align*}-6a(b)(c)\end{align*}

14. \begin{align*}-8a(6b)\end{align*}

15. \begin{align*}(12y)(-3x)(-1)\end{align*}

16. \begin{align*}-8x(4)\end{align*}

17. \begin{align*}-a(5)(-4b)\end{align*}

18. \begin{align*}-2ab(12c)\end{align*}

19. \begin{align*}-2ab(12c)\end{align*}

Directions: Solve this real world problem.

20. The temperature outside Fred's house is dropping at a rate of \begin{align*}2^\circ F\end{align*} each hour. Represent the total change in the temperature over the next 5 hours as an integer.

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Date Created:
Feb 22, 2012