Intro to distributive property

This array is made up of $3$3 rows with $6$6 dots in each row. The dots show $3 \times 6 = 18$3, times, 6, equals, 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$1 row with $6$6 dots. The dots show $1 \times 6$1, times, 6.

The bottom group has $2$2 rows with $6$6 dots in each row. The dots show $2 \times 6$2, times, 6.

We still have a total of $18$18 dots.

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 $\greenD{3} \times \purpleD{6}$start color #1fab54, 3, end color #1fab54, times, start color #7854ab, 6, end color #7854ab.

We broke the $\greenD{3}$start color #1fab54, 3, end color #1fab54 down into $\greenD{1 + 2}$start color #1fab54, 1, plus, 2, end color #1fab54. We can do this because $\greenD{1 + 2 = 3}$start color #1fab54, 1, plus, 2, equals, 3, end color #1fab54

We used the distributive property to change the problem from $\greenD{3} \times \purpleD{6}$start color #1fab54, 3, end color #1fab54, times, start color #7854ab, 6, end color #7854ab to $(\greenD{1 + 2}) \times \purpleD{6}$left parenthesis, start color #1fab54, 1, plus, 2, end color #1fab54, right parenthesis, times, start color #7854ab, 6, end color #7854ab.

The $\purpleD{6}$start color #7854ab, 6, end color #7854ab gets distributed to the $\greenD{1}$start color #1fab54, 1, end color #1fab54 and $\greenD{2}$start color #1fab54, 2, end color #1fab54 and the problem changes to:
$(\greenD{1} \times \purpleD{6}) + (\greenD{2} \times \purpleD{6})$left parenthesis, start color #1fab54, 1, end color #1fab54, times, start color #7854ab, 6, end color #7854ab, right parenthesis, plus, left parenthesis, start color #1fab54, 2, end color #1fab54, times, start color #7854ab, 6, end color #7854ab, right parenthesis

Now we need to find the two products:
$6 + 12$6, plus, 12

And finally, the sum:
$6 + 12 = 18$6, plus, 12, equals, 18

$\greenD{3} \times \purpleD{6} = 18$start color #1fab54, 3, end color #1fab54, times, start color #7854ab, 6, end color #7854ab, equals, 18 and
$(\greenD{1 + 2}) \times \purpleD{6} =18$left parenthesis, start color #1fab54, 1, plus, 2, end color #1fab54, right parenthesis, times, start color #7854ab, 6, end color #7854ab, equals, 18

Some numbers like $1, 2, 5$1, comma, 2, comma, 5, and $10$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$4, times, 12 into $4 \times (\tealD{10} + \greenC{2})$4, times, left parenthesis, start color #01a995, 10, end color #01a995, plus, start color #74cf70, 2, end color #74cf70, right parenthesis.

The array of dots on the left shows $(\tealD{4 \times 10})$left parenthesis, start color #01a995, 4, times, 10, end color #01a995, right parenthesis. The array of dots on the right shows $(\greenC{4 \times 2})$left parenthesis, start color #74cf70, 4, times, 2, end color #74cf70, right parenthesis.

Now we can add the expressions to find the total.
$(\tealD{4 \times 10}) + (\greenC{4 \times 2})$left parenthesis, start color #01a995, 4, times, 10, end color #01a995, right parenthesis, plus, left parenthesis, start color #74cf70, 4, times, 2, end color #74cf70, right parenthesis
$= \tealD{40} + \greenC{8}$equals, start color #01a995, 40, end color #01a995, plus, start color #74cf70, 8, end color #74cf70
$=48$equals, 48

Since $10$10 and $2$2 are both easy to multiply, using the distributive property for this problem made finding the product easier.

The dots represent $9 \times 4$9, times, 4.

The distributive property is very helpful when multiplying larger numbers. Look at how we can use the distributive property to simplify $15 \times 8$15, times, 8.

We will start by breaking $\blueD{15}$start color #11accd, 15, end color #11accd into $\blueD{10 +5}$start color #11accd, 10, plus, 5, end color #11accd. Then we will distribute the $8$8 to both of these numbers.