# Intro to distributiveÂ 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.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

$(\greenD{1} \times \purpleD{6}) + (\greenD{2} \times \purpleD{6})$

*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:

$6 + 12$

$6 + 12$

And finally, the sum:

$6 + 12 = 18$

$6 + 12 = 18$

$\greenD{3} \times \purpleD{6} = 18$ and

$(\greenD{1 + 2}) \times \purpleD{6} =18$

$(\greenD{1 + 2}) \times \purpleD{6} =18$

### 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.

$(\tealD{4 \times 10}) + (\greenC{4 \times 2})$

$= \tealD{40} + \greenC{8}$

$=48$

$(\tealD{4 \times 10}) + (\greenC{4 \times 2})$

$= \tealD{40} + \greenC{8}$

$=48$

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 $9 \times 4$.

### More practice

## 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$

$\phantom{15\times 8}$$=~$$80 + 40$

$\phantom{15\times 8}$$=~$$120$