# Intro to distributive property

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 $3$ rows with $6$ dots in each row. The dots show $3 \times 6 = 18$.
If we add a line dividing the dots into two groups, the total number of dots does not change.
The top group has $1$ row with $6$ dots. The dots show $1 \times 6$.
The bottom group has $2$ rows with $6$ dots in each row. The dots show $2 \times 6$.
We still have a total of $18$ 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.
$5$ could be rewritten as $2 + 3$ or $4 +1$ because $2 + 3 = 5$ and $4 + 1 = 5$.
Using the distributive property allows us to solve two simpler multiplication problems.
In the example with the dots we started with $\greenD{3} \times \purpleD{6}$.
We broke the $\greenD{3}$ down into $\greenD{1 + 2}$. We can do this because $\greenD{1 + 2 = 3}$
We used the distributive property to change the problem from $\greenD{3} \times \purpleD{6}$ to $(\greenD{1 + 2}) \times \purpleD{6}$.
The $\purpleD{6}$ gets distributed to the $\greenD{1}$ and $\greenD{2}$ and the problem changes to:
$(\greenD{1} \times \purpleD{6}) + (\greenD{2} \times \purpleD{6})$
Now we need to find the two products:
$6 + 12$
And finally, the sum:
$6 + 12 = 18$
$\greenD{3} \times \purpleD{6} = 18$ and
$(\greenD{1 + 2}) \times \purpleD{6} =18$
Practice problem 1
Which expressions are the same as $4 \times 9$?

### Small numbers

Some numbers like $1, 2, 5$, and $10$ 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 $4 \times 12$ into $4 \times (\tealD{10} + \greenC{2})$.
The array of dots on the left shows $(\tealD{4 \times 10})$. The array of dots on the right shows $(\greenC{4 \times 2})$.
Now we can add the expressions to find the total.
$(\tealD{4 \times 10}) + (\greenC{4 \times 2})$
$= \tealD{40} + \greenC{8}$
$=48$
Since $10$ and $2$ are both easy to multiply, using the distributive property for this problem made finding the product easier.

### Practice problem 2

The dots represent $9 \times 4$.
Problem 2, part A
Which expression shows the dots above the dotted line?

Problem 2, Part B
Which expression shows the dots below the dotted line?

Problem 2, Part C
$(5 \times 4)$
$(4 \times 4) = \text { total number of dots}$

### More practice

Problem 3A
The dots represent $3 \times 8$.
Which expression can we use to calculate the total number of dots?

## 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 $15 \times 8$.
We will start by breaking $\blueD{15}$ into $\blueD{10 +5}$. Then we will distribute the $8$ to both of these numbers.
$\blueD{15}\times 8 = (\blueD{10}\times 8)+{(\blueD{5} \times 8)}$
$\phantom{15\times 8}$$=~$$80 + 40$
$\phantom{15\times 8}$$=~$$120$
Problem 4
Use the distributive property to find the product.
$\blueD{18}\times 3 = (\blueD{10} \times 3) +(~$
$\times 3)$
$\phantom{18\times 3}=~30 +$
$\phantom{18\times 3}=~$