In multiplication, just like addition, we assign positive numbers to balloons and negative numbers to sand bags. However, this time we will use the balloons and sand bags to think of multiplication as shown here.

Consider the following problem statements:
Situation 1  Attach two sets of three balloons to the basket. What will happen to the basket? 
$(+2)\,\times~(+3)\,=\,+6$ $+6$ means the basket will rise 6 units. 
$(+2)\,\times~(+3)\,=\,+6$

Situation 2  Attach two sets of three sand bags to the basket. What will happen to the basket? 
$(+2)\,\times~(3)\,=\,6$ $6$ means the basket will drop 6 units. 
$(+2)\,\times~(3)\,=\,6$

The previous cases involved adding balloons or sand bags to the basket. Now let's look at two cases where items are being removed from the basket.
Situation 3  Remove three sets of four balloons from the basket. What will happen to the basket? 
$ (3)\,\times~ (+4)\,=\,12$ $12$ means the basket will drop 12 units. 
$ (3)\,\times~ (+4)\,=\,12$

Situation 4  Remove three sets of two sand bags. What will happen to the basket? 
$ (3)\,\times~ (2)\,=\,+6$ $+6$ means the basket will rise 6 units. 
$ (3)\,\times~ (2)\,=\,+6$

Let's look closely at the four examples we've studied thus far.
$ (+2) \cdot~ (3) = 6$ 
$(+2) \cdot~ (+3) = +6$ 
$(3) \cdot~ (+4) = 12$ 
$(3) \cdot~ (2) = +6$ 
The two examples in the left column both have negative products, while the two examples in the right column both have positive products. The rule for multiplying without the use of balloons and sand bags is simple.
When multiplying two numbers:

Question 1 $(4)\cdot~(3)=$

[show answer] 
Question 2 $(3)\cdot~(+9)=$

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Question 3 $(+8)\cdot~(5)=$

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Question 4 $(4)\cdot~(+3)\cdot~(2)=$

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