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# 4.10: Integer Multiplication

Difficulty Level: At Grade Created by: CK-12
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Practice Integer Multiplication

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Have you ever calculated temperature? Take a look at this dilemma.

As Cameron gets ready for his dive one morning, he notices that the temperature at the resort is already \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 ride out to the dive site is 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 you against a random jellyfish.

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

This Concept 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 is on Cameron’s dive.

### Guidance

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 \begin{align*}5 \times 0\end{align*}. How could you use the pattern shown below to determine that product?

\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 \begin{align*}5-5=0\end{align*}, the product of \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 can notice that the pattern for the multiplication facts of 5 continues beyond zero. Up until this time in math, we have only looked at the positive products. But now, we know about the negative numbers, so we can look at continuing the facts. Let's see how patterns can help us find the products of integers.

Use a pattern to find the missing products below.

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

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

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

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

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

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

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

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

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

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

Give that answer a negative sign.

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

Now we have our completed multiplication facts

\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.

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!!

\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.

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

Next, we figure out the sign.

A negative times a negative is a positive.

\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.

\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.

\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.

\begin{align*}-8 \times -3 = 24\end{align*}

Now we multiply that product times the factor negative two.

\begin{align*}24 \times -2 = -48\end{align*}

Now it's time to try a few on your own. Find each product.

#### Example A

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

Solution: \begin{align*}27\end{align*}

#### Example B

\begin{align*}(-3)(12)\end{align*}

Solution: \begin{align*}-36\end{align*}

#### Example C

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

Solution: \begin{align*}24\end{align*}

Here is the original problem once again.

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 ride out to the dive site is 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 you against a random jellyfish.

If Cameron dives to 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 will be \begin{align*}79^\circ .\end{align*}

### Vocabulary

Here are the vocabulary words in this Concept.

Integer
the set of whole numbers and their opposites.
Product
the answer in a multiplication problem
Factors
the numbers being multiplied

### Guided Practice

Here is one for you to try on your own.

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.

### Video Review

Here is a video for review.

### Practice

Directions: Use integer rules to find each product.

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

2. \begin{align*}8(-9)\end{align*}

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*}

11. \begin{align*}(-9)(2)(-1)\end{align*}

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

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

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

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

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