What is Distributive Property? – Definition, Formula, Examples

The distributive property is also referred to as the distributive law of multiplication over addition and subtraction. The term “distribute” refers to dividing, sharing, or giving a portion of something. According to the distributive property, multiplying a number by the sum of two or more addends is equivalent to multiplying each addend separately by the number and then combining the products collectively. The distributive property is applicable to both addition and subtraction and it aids in simplifying difficult problems.

The distributive property of multiplication over addition	The distributive property of multiplication over subtraction
A × (B + C) = AB + AC	A × (B – C) = AB – AC

For example, let us solve (2 + 7 + 9 + 5) × 6 and (28 – 15) × 4 normally and also using the distributive property. From the outcomes, we can observe that the outcomes did not change in the case of addition or subtraction when it was solved normally and when it was solved using the distributive property. 

	Solved without using Distributive property	Solved with using Distributive property
Addition	(2 + 7 + 9 + 5) × 6 = 23 × 6  = 138	(2 + 7 + 9 + 5) × 6 = 2 × 6 + 7 × 6 + 9 × 6 + 5 × 6 = 12 + 42 +54 + 30  = 138
Subtraction	(28 – 15) × 4 = 13 × 4  = 52	(28 – 15) × 4 = 28 × 4 – 15 × 4 = 112 – 60 = 52

Distributive property of multiplication over addition 

We use the distributive principle of multiplication over addition when we need to multiply a number by the sum of two or more addends. For instance, let us multiply 5 by the sum of 9 + 25, which is mathematically represented as [5 × (9 + 25)].

Example: Solve the expression [5 × (9 + 25)] using the distributive property.

Solution:

To solve the given expression [5 × (9 + 25)] using the distributive property, first, we have to multiply both addends (9 and 25) by 7. This is called distributing the number between the two addends. Now, add the obtained products; i.e., the multiplication of 5 (9), and 5 (25) will be carried out before performing the addition. 5 × (9) + 5 × (25) = 45 + 125 = 170.

Distributive property of multiplication over subtraction

We use the distributive principle of multiplication over subtraction when we need to multiply the difference between two or more numbers by a number. It is similar to the distributive property of multiplication over addition except for the operation of addition and subtraction. For instance, let us multiply the difference between 45 and 21 by 3, which is mathematically represented as [3 × (45 – 21)].

Example: Solve the expression [3 × (45 – 21)] using the distributive property.

Solution:

To solve the given expression  [3 × (45 – 21)] using the distributive property, first, we have to multiply (45 and 21) by 3. This is called distributing the number between the two numbers. Now, subtract the obtained products; i.e., the multiplication of 3×(45), and 3×(21) will be carried out before performing the subtraction. 3 × (45) – 3 × (21) = 135 + 63 = 72.

Verification of the Distributive Property

Let’s try to justify how the distributive property functions for various operations by applying the distributive property separately to the two fundamental operations, addition, and subtraction.

Distributive property of multiplication over addition: The distributive property of addition is expressed as a × (b + c) = a × b + a × c. Now, let us verify it by using an example.

Example: Solve the expression [2 × (19 + 6)] using the distributive property.

Solution:

Given expression: [2 × (19 + 6)]

First, let us solve the given expression using the BODMAS rule, i.e., we have to add the numbers in brackets first, then the obtained result is multiplied by the number outside the brackets. 2 × (19 + 6) =  2 × (25) = 50.

Now, let us solve the given expression using the distributive property. 

2 × (19 + 6) = (2 × 19) + (2 × 6)

= 38 + 12 = 50.

 Thus, the result in both the methods is the same.

Distributive property of multiplication over subtraction: The distributive property of subtraction is expressed as a×(b–c) = (a × b)–(a × c). Now, let us verify it by using an example.

Example: Solve the expression [4 × (23 – 7)] using the distributive property.

Solution:

Given expression: [4 × (23 – 7)]

First, let us solve the given expression using the BODMAS rule, i.e., we have to subtract the numbers in brackets first, then the obtained result is multiplied by the number outside the brackets. 4 × (23 – 7) =  4× (16) = 64.

Now, let us solve the given expression using the distributive property. 

4 × (23 – 7) = (4 × 23) – (4 × 7)

= 92 – 28 = 64.

Hence, the result in both the methods is the same.

Distributive Property of Division

The distributive property helps in dividing larger numbers easily by breaking the number into two or more smaller factors and then distributing the divide operation between them. Let us understand this concept with the help of an example.

Example: Divide 76 ÷ 4 using the distributive property of division.

Solution: 

Given expression: 76 ÷ 4

We can write 76 as 64 + 12

So, 76 ÷ 4 = (64 + 12) ÷ 4

Now, let us distribute the division operation for each factor (64 and 12) in the bracket.

= (64 ÷ 4) + (12 ÷ 4)

= 16 + 3 = 19

Therefore, the answer is 19.

Sample Problems

Problem 1: Solve equation 5 (y + 8) = 120 using the distributive property.

Solution:

Given, 5 (y + 8) = 120

We have a set of two parentheses inside the bracket. So, distribute 5.

5 × y + 5 × 8 = 120

5y + 40 = 120

5y = 120 – 40 = 80

y = 80/5 = 16

Hence, y = 16.

Problem 2: Solve 3x + 4(x – 6) + 17 = 28 using the distributive property.

Solution:

Given, 3x + 4(x – 6) + 17 = 28

We have a set of two parentheses inside the bracket. So, distribute 4.

3x + 4 × x – 4 × 6 + 17 = 28

3x + 4x – 24 + 17 = 28

7x – 7 = 28

7x = 28 + 7 = 35

x = 35/7 = 5

Hence, x = 5.

Problem 3: Solve the equation (a + 3b) (2a + b) using the distributive property.

Solution:

Given, (a + 3b) (2a + b)

From the distributive property, we have

(p + q) × r=(p × r)+(q × r)      

So, (a + 3b) (2a + b) =  a × (2a + b) + 3b × (2a + b)

= 2a² + ab + 6ab + 3b²

= 2a² + 7ab + 3b²

Thus, (a + 3b) (2a + b) = 2a² + 7ab + 3b².

Problem 4: Solve the equation (3x – 2y) (x + 4y) using the distributive property.

Solution:

Given, (3x – 2y) (x + 4y)

From the distributive property, we have

(p + q) × r=(p × r)+(q × r)      

So, (3x – 2y) (x + 4y)=  3x × (x + 4y) – 2y × (x + 4y)

= 3x² + 12xy – 2xy – 8y²

= 3x² + 10xy – 8y²

Thus, (3x – 2y) (x + 4y) = 3x² + 10xy – 8y².

Problem 5: Divide 108 ÷ 9 using the distributive property of division.

Solution:

Given expression: 108 ÷ 9

We can write 108 as 81 + 27

So, 108 ÷ 9 = (81 + 27) ÷ 9

Now, let us distribute the division operation for each factor (64 and 12) in the bracket.

= (81 ÷ 9) + (27 ÷ 9)

= 9 + 3 = 12

Therefore, the answer is 12.