Question

multiply positive and negative integers
study the example problem showing how to use repeated
addition to multiply a positive and a negative integer. then
solve problems 1 - 6.
example
dues for the art club are $2 per week. noriko pays $2 each
week from her bank account for her art club dues. at the end
of 5 weeks, what is the change in the amount of money in
norikos account?
you can think of this as 5 groups of (-2).
5 groups of (-2) = (-2) + (-2) + (-2) + (-2) + (-2).
you can start at 0 and make 5 jumps of -2.
(-2) + (-2) + (-2) + (-2) + (-2) = -10
so, norikos account changes by -$10.
1 complete the table and describe the pattern
in the products.

2 rewrite the repeated addition expression as a
multiplication expression and find the product.
a. (-2) + (-2) + (-2) + (-2) + (-2) = ____·(-2) = ____
b. (-3) + (-3) + (-3) + (-3) + (-3) + (-3) = __________
c. (-5) + (-5) + (-5) = __________

multiplication	product
-2·2	-4
-2·1	-2
-2·0
-2·(-1)
-2·(-2)
-2·(-3)

Explanation:

Problem 1: Complete the table and describe the pattern

Step 1: Calculate the missing products

• For \(-2 \cdot 0\): Any number multiplied by 0 is 0, so \(-2 \cdot 0 = 0\).
• For \(-2 \cdot (-1)\): The product of two negative numbers is positive, so \(-2 \cdot (-1) = 2\).
• For \(-2 \cdot (-2)\): \(-2 \cdot (-2) = 4\) (negative times negative is positive).
• For \(-2 \cdot (-3)\): \(-2 \cdot (-3) = 6\) (negative times negative is positive).

Step 2: Describe the pattern

• When the second factor (the number being multiplied by \(-2\)) increases by 1 (e.g., from 3 to 2 to 1 to 0 to \(-1\) to \(-2\) to \(-3\)), the product increases by 2 each time.
• For positive second factors (3, 2, 1), the product is negative and becomes less negative (increases) as the second factor decreases.
• When the second factor is 0, the product is 0.
• For negative second factors (\(-1\), \(-2\), \(-3\)), the product is positive and increases as the second factor becomes more negative (decreases).

Problem 2: Rewrite repeated addition as multiplication and find the product

Part a:

• The repeated addition \((-2) + (-2) + (-2) + (-2) + (-2)\) has 5 terms of \(-2\). So it can be written as \(5 \cdot (-2)\).
• Calculating the product: \(5 \cdot (-2) = -10\).

Part b:

• The repeated addition \((-3) + (-3) + (-3) + (-3) + (-3) + (-3)\) has 6 terms of \(-3\). So it can be written as \(6 \cdot (-3)\).
• Calculating the product: \(6 \cdot (-3) = -18\).

Part c:

• The repeated addition \((-5) + (-5) + (-5)\) has 3 terms of \(-5\). So it can be written as \(3 \cdot (-5)\).
• Calculating the product: \(3 \cdot (-5) = -15\).

Filled Table for Problem 1:

Multiplication	Product
\(-2 \cdot 2\)	\(-4\)
\(-2 \cdot 1\)	\(-2\)
\(-2 \cdot 0\)	\(0\)
\(-2 \cdot (-1)\)	\(2\)
\(-2 \cdot (-2)\)	\(4\)
\(-2 \cdot (-3)\)	\(6\)

Final Answers:

Problem 1 Table:

Multiplication	Product
\(-2 \cdot 2\)	\(-4\)
\(-2 \cdot 1\)	\(-2\)
\(-2 \cdot 0\)	\(0\)
\(-2 \cdot (-1)\)	\(2\)
\(-2 \cdot (-2)\)	\(4\)
\(-2 \cdot (-3)\)	\(6\)

Pattern: As the second factor increases by 1, the product increases by 2. Negative second factors give positive products, positive second factors give negative products, and 0 gives 0.

Problem 2:

• a. \(5 \cdot (-2) = -10\)
• b. \(6 \cdot (-3) = -18\)
• c. \(3 \cdot (-5) = -15\) (rewritten from \((-5) + (-5) + (-5)\))

Answer:

Problem 1: Complete the table and describe the pattern

Step 1: Calculate the missing products

• For \(-2 \cdot 0\): Any number multiplied by 0 is 0, so \(-2 \cdot 0 = 0\).
• For \(-2 \cdot (-1)\): The product of two negative numbers is positive, so \(-2 \cdot (-1) = 2\).
• For \(-2 \cdot (-2)\): \(-2 \cdot (-2) = 4\) (negative times negative is positive).
• For \(-2 \cdot (-3)\): \(-2 \cdot (-3) = 6\) (negative times negative is positive).

Step 2: Describe the pattern

• When the second factor (the number being multiplied by \(-2\)) increases by 1 (e.g., from 3 to 2 to 1 to 0 to \(-1\) to \(-2\) to \(-3\)), the product increases by 2 each time.
• For positive second factors (3, 2, 1), the product is negative and becomes less negative (increases) as the second factor decreases.
• When the second factor is 0, the product is 0.
• For negative second factors (\(-1\), \(-2\), \(-3\)), the product is positive and increases as the second factor becomes more negative (decreases).

Problem 2: Rewrite repeated addition as multiplication and find the product

Part a:

• The repeated addition \((-2) + (-2) + (-2) + (-2) + (-2)\) has 5 terms of \(-2\). So it can be written as \(5 \cdot (-2)\).
• Calculating the product: \(5 \cdot (-2) = -10\).

Part b:

• The repeated addition \((-3) + (-3) + (-3) + (-3) + (-3) + (-3)\) has 6 terms of \(-3\). So it can be written as \(6 \cdot (-3)\).
• Calculating the product: \(6 \cdot (-3) = -18\).

Part c:

• The repeated addition \((-5) + (-5) + (-5)\) has 3 terms of \(-5\). So it can be written as \(3 \cdot (-5)\).
• Calculating the product: \(3 \cdot (-5) = -15\).

Filled Table for Problem 1:

Multiplication	Product
\(-2 \cdot 2\)	\(-4\)
\(-2 \cdot 1\)	\(-2\)
\(-2 \cdot 0\)	\(0\)
\(-2 \cdot (-1)\)	\(2\)
\(-2 \cdot (-2)\)	\(4\)
\(-2 \cdot (-3)\)	\(6\)

Final Answers:

Problem 1 Table:

Multiplication	Product
\(-2 \cdot 2\)	\(-4\)
\(-2 \cdot 1\)	\(-2\)
\(-2 \cdot 0\)	\(0\)
\(-2 \cdot (-1)\)	\(2\)
\(-2 \cdot (-2)\)	\(4\)
\(-2 \cdot (-3)\)	\(6\)

Pattern: As the second factor increases by 1, the product increases by 2. Negative second factors give positive products, positive second factors give negative products, and 0 gives 0.

Problem 2:

• a. \(5 \cdot (-2) = -10\)
• b. \(6 \cdot (-3) = -18\)
• c. \(3 \cdot (-5) = -15\) (rewritten from \((-5) + (-5) + (-5)\))