# Distributive Property of Multiplication

Content Distributive Property of Multiplication
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## Distributive Property

What is the distributive property of multiplication? We can explain distributive property of multiplication breaking down one factor into smaller parts, to multiply the broken down parts by the other whole factor, and then adding them together to get the product of the original factor pair. To understand and apply the distributive property of multiplication, read on to see an example of distributive property.

Now you have the answer to ‘what does the distributive property of multiplication mean’, why is it useful? It is useful because it can help break down more challenging multiplication problems into smaller ones.

## Distributive Property – Examples

Let’s have a go at using the distributive property of multiplication! Below, we have the factor pair four and six.

First, we will break down the six into three and three.

Then, we will distribute the four to be threes, to make four times three and four times three.

Next, we find the product of both smaller factor pairs. Since they are both four times three, the product for both is twelve, so write the solution.

Finally, add the two products together. Twelve plus twelve gives a sum of twenty-four. The product of four times six is twenty-four!

## Distributive Property Multiplication – Summary

To use the distributive property of multiplication, remember these steps:

Step # What to do
1 Break down one of the factor pairs
2 Distribute the whole factor to the
broken down factor pairs
3 Multiply both smaller factor pairs
4 Add the product of both to find the
product of the original factor pair

After watching the video, you will find more interactive exercises for practice of distributive property and a distributive property of multiplication worksheet.

## Distributive Property of Multiplication exercise

Would you like to apply the knowledge you’ve learnt? You can review and practice it with the tasks for the video Distributive Property of Multiplication.
• ### How can we find Big 'Arry's area?

Hints

Break down one of the original factors into two smaller factors and multiply both of these by the other original factor. Here, 7 has been broken into 2 and 5, then both multiplied by 5.

Solve both of the calculations within the brackets. Do both products total the original caculation of 7 x 5?

Solution

The original array was 7 x 5.

• Break down the bigger factor (7) into two parts.
• Then multiply both of these by 5.
• We could break down 7 into 2 and 5: (2 x 5) + (5 x 5). 10 + 25 = 35.
• We could break down 7 into 3 and 4: (4 x 5) + (3 x 5). 20 + 15 = 35.

• ### Use the distributive law.

Hints

To find the factors that 6 can be broken into, think about what 6 is made up of.

Once you know that 6 is broken into 4 and 2, what are the next steps?

Solution

This array has been made with 6 squares across and 5 squares down.

Instead of solving 6 x 5 we can break down the factor, 6, into 4 + 2.

Then we solve: 4 x 5 = 20 and 2 x 5 = 10.

Finally, we can add together the two products, 20 and 10 to get the total of 30.

We can check this by solving 6 x 5 which also equals 30.

• ### Find the equations that match each array's area.

Hints

This array is 8 x 4. Can the 8 be broken down into two other factors?

In this array, 8 x 4 could be broken down into (5 x 4) + (3 x 4).

Solution

Array measuring 7 x 3.

• 7 can be broken into 5 and 2.
• (5 x 3) + (2 x 3) = 21.
• This is the same as 7 x 3 = 21.

Array measuring 9 x 4.

• 9 can be broken into 5 and 4.
• (5 x 4) + (4 x 4) = 36.
• This is the same as 9 x 4 = 36.

Array measuring 8 x 5.

• 8 can be broken into 5 and 3.
• (5 x 5) + (3 x 5) = 40.
• This is the same as 8 x 5 = 40.

Array measuring 9 x 5.

• 9 can be broken into 6 and 3.
• (6 x 5) + (3 x 5) = 45.
• This is the same as 9 x 5 = 45.

• ### Breaking down factors.

Hints

When there are two factors multiplied together, look at the larger factor. Can you see another equation where this factor has been split into two smaller factors?

An example of a larger factor being broken into two smaller factors would be: 15 x 7. 15 can be split into 10 + 5 to become (10 x 7) + (5 x 7).

Solution
• (3 x 5) + (5 x 5) = 8 x 5. Both equations = 40
• (6 x 6) + (3 x 6) = 9 x 6. Both equations = 54
• (10 x 7) + (2 x 7) = 12 x 7. Both equations = 84
• (6 x 4) + (6 x 4) = 12 x 4. Both equations = 48
• ### Finding the array to match the equation.

Hints

What two factors can 7 be split into? Both of these are then multiplied by 3.

Solve the separate multiplication equations (5 x 3) + (2 x 3) then combine these products to find the total. This is how many squares are in the array.

7 x 3 = 21. Can you see an array with 21 squares?

Solution
• This image shows how the array 7 x 3 can be broken into two parts to make the multiplication simpler.
• An array of 7 x 3 is the same as an array of 2 x 3 and 5 x 3.
• (2 x 3) = 6 and (5 x 3) = 15.
• 6 + 15 = 21.
• 7 x 3 = 21.
• ### Finding factors.

Hints

First find the equations where two factors multiply together to make the product.

Where an equation has been broken down into two parts, multiply the equations within the brackets, then add these products together.

Solution

36

• (3 x 4) + (6 x 4)
• (2 x 6) + (4 x 6)
• (9 x 3) + (3 x 3)
• 9 x 4

42

• (5 x 6) + (2 x 6)
• (8 x 3) + (6 x 3)
• 7 x 6

25

• (2 x 5) + (3 x 5)
• 5 x 5