CCSS Math: 3.OA.B.5
Practice decomposing the factors in multiplication problems and see how it affects the product.

Breaking up multiplication

This array is made up of 33 rows with 66 dots in each row. The dots show 3×6=183 \times 6 = 18.
If we add a line dividing the dots into two groups, the total number of dots does not change.
The top group has 11 row with 66 dots. The dots show 1×61 \times 6.
The bottom group has 22 rows with 66 dots in each row. The dots show 2×62 \times 6.
We still have a total of 1818 dots.

Distributive property

The math rule that allows us to break up multiplication problems is called the distributive property.
The distributive property says that in a multiplication problem, when one of the factors is rewritten as the sum of two numbers, the product does not change.
55 could be rewritten as 2+32 + 3 or 4+14 +1 because 2+3=52 + 3 = 5 and 4+1=54 + 1 = 5.
Using the distributive property allows us to solve two simpler multiplication problems.
In the example with the dots we started with 3×6\greenD{3} \times \purpleD{6}.
We broke the 3\greenD{3} down into 1+2\greenD{1 + 2}. We can do this because 1+2=3\greenD{1 + 2 = 3}
We used the distributive property to change the problem from 3×6\greenD{3} \times \purpleD{6} to (1+2)×6(\greenD{1 + 2}) \times \purpleD{6}.
The 6\purpleD{6} gets distributed to the 1\greenD{1} and 2\greenD{2} and the problem changes to:
(1×6)+(2×6)(\greenD{1} \times \purpleD{6}) + (\greenD{2} \times \purpleD{6})
Now we need to find the two products:
6+126 + 12
And finally, the sum:
6+12=186 + 12 = 18
3×6=18\greenD{3} \times \purpleD{6} = 18 and
(1+2)×6=18(\greenD{1 + 2}) \times \purpleD{6} =18
Practice problem 1
Which expressions are the same as 4×94 \times 9?
Choose all answers that apply:
Choose all answers that apply:

Small numbers

Some numbers like 1,2,51, 2, 5, and 1010 are easier to multiply. The distributive property allows us to change a multiplication problem so that we can use these numbers as one of the factors.
For example, we can change 4×124 \times 12 into 4×(10+2)4 \times (\tealD{10} + \greenC{2}).
The array of dots on the left shows (4×10)(\tealD{4 \times 10}). The array of dots on the right shows (4×2)(\greenC{4 \times 2}).
Now we can add the expressions to find the total.
(4×10)+(4×2)(\tealD{4 \times 10}) + (\greenC{4 \times 2})
=40+8= \tealD{40} + \greenC{8}
=48=48
Since 1010 and 22 are both easy to multiply, using the distributive property for this problem made finding the product easier.

Practice problem 2

The dots represent 9×49 \times 4.
Problem 2, part A
Which expression shows the dots above the dotted line?
Choose 1 answer:
Choose 1 answer:
Problem 2, Part B
Which expression shows the dots below the dotted line?
Choose 1 answer:
Choose 1 answer:
Problem 2, Part C
(5×4)(5 \times 4)
(4×4)= total number of dots(4 \times 4) = \text { total number of dots}

More practice

Problem 3A
The dots represent 3×83 \times 8.
Which expression can we use to calculate the total number of dots?
Choose 1 answer:
Choose 1 answer:

Working with large numbers

The distributive property is very helpful when multiplying larger numbers. Look at how we can use the distributive property to simplify 15×815 \times 8.
We will start by breaking 15\blueD{15} into 10+5\blueD{10 +5}. Then we will distribute the 88 to both of these numbers.
15×8=(10×8)+(5×8)\blueD{15}\times 8 = (\blueD{10}\times 8)+{(\blueD{5} \times 8)}
15×8\phantom{15\times 8}= =~80+4080 + 40
15×8\phantom{15\times 8}= =~120120
Problem 4
Use the distributive property to find the product.
18×3=(10×3)+( \blueD{18}\times 3 = (\blueD{10} \times 3) +(~
×3) \times 3)
18×3= 30+\phantom{18\times 3}=~30 +
18×3= \phantom{18\times 3}=~