Students who would like to practice the problems on Expansion of Powers of Binomials and Trinomials can get them here. We have provided the Expansion of Powers of Binomials and Trinomials in the below section. There are many benefits for the students by referring to our Problems on Expansion of Powers of Binomials and Trinomials Worksheet pdf. Relate the concept to real-world problems to understand the topic in depth. This will help you to improve your math skills and also score top in the class.

Read Similar:

- Application Problems on Expansion of Powers of Binomials and Trinomials
- Worksheet on Expansion of (a ± b)^2 and its Corollaries
- Worksheet on Expanding of (a ± b ± c)^2 and its Corollaries

## Application Problems on Expansion of Powers of Binomials and Trinomials Worksheet

**Example 1.**

If the sum of two numbers is 5 and the difference between the two numbers is 3. Find the squares of difference of two numbers.

## Solution:

Given,

The sum of two numbers is 5 and

The difference between the two numbers is 3

a + b = 5

a – b = 3

By using the formula (a + b)(a – b) = a² – b²

a² – b² = (5)(3)

a² – b² = 15

**Example 2.**

Evaluate the following using (x + y)(x – y) = x² – y²

i. 13 × 7

ii. 12 × 8

iii. 23 × 17

## Solution:

i. 13 × 7

By using the formula (a + b)(a – b) = a² – b²

(10 + 3) (10 – 3) = 10² – 3²

= 100 – 9

= 91

ii. 12 × 8

By using the formula (a + b)(a – b) = a² – b²

(10 + 2)(10 – 2) = 10² – 2²

= 100 – 4

= 96

iii. 23 × 17

By using the formula (a + b)(a – b) = a² – b²

(20 + 3)(20 – 3) = 20² – 3²

= 400 – 9

= 391

**Example 3.**

If the sum of the squares of the two numbers is 10 and the sum of the two terms is 5. Find the product of the two constant terms.

## Solution:

Given,

The sum of the squares of the two numbers is 10 and

The sum of the two terms is 5.

x + y = 5

x² + y² = 10

This is in the form of (x + y)² = x² + 2xy + y²

(5)² = 10 + 2xy

25 = 10 + 2xy

2xy = 25 – 10

2xy = 15

xy = 15/2

xy = 7.5

Thus the product of the two constant terms is 7.5

**Example 4.**

Evaluate the following using (x ± y)² = x² ± 2xy + y²

i. 37²

ii. (4.03)²

iii. 16²

## Solution:

i. 37²

We can write it as (40 – 3)²

This is in the form of (x – y)² = x² – 2xy + y²

(40 – 3)² = 40² – 2(40)(3) + 3²

= 1600 – 240 + 9

= 1369

ii. (4.03)²

We can write it as (4 + 0.03)²

This is in the form of (x + y)² = x² + 2xy + y²

(4 + 0.03)² = 4² + 2(4)(0.03) + (0.03)²

= 16 + 0.24 + 0.0009

= 16.2409

iii. 16²

We can write it as (20 – 4)²

This is in the form of (x – y)² = x² – 2xy + y²

(20 – 4)² = 20² – 2(20)(4) + 4²

= 400 – 160 + 16

= 256

**Example 5.**

Evaluate the number 6.12 × 5.88

## Solution:

Given,

6.12 × 5.88

We can write it as, (6 + 0.12) × (6 – 0.12)

By using the formula (a + b)(a – b) = a² – b²

(6 + 0.12) (6 – 0.12) = 6² – 0.12²

= 36 – 0.0144

= 35.98

**Example 6.**

If the sum of the three numbers a, b, c is 4 and the sum of their squares is 20 then find the sum of the product of the three numbers taking two at a time.

## Solution:

Given that,

The sum of the three numbers a, b, c is 4

a + b + c = 4

The sum of their squares is 20

a² + b² + c² = 20

We need to find the value of ab + bc + ca

We know that

(a + b + c)² = a² + b² + c² + 2(ab + bc + ca)

(a + b + c)² – a² + b² + c² = 2(ab + bc + ca)

4² – 20 = 2( ab + bc + ca)

16 – 20 = 2( ab + bc + ca)

-4 = 2(ab + bc + ca)

ab + bc + ca = -4/2

Thus ab + bc + ca = -2

**Example 7.**

If the sum of two numbers is 7 and the sum of their cubes is 28, find the sum of their squares.

## Solution:

Given,

The sum of two numbers is 7 and

The sum of their cubes is 28

a + b = 7

a³ + b³ = 28

a³ + b³ = (a + b)³ – 3ab(a + b)

28 = (7)³ – 3ab(7)

28 = 343 – 21ab

21ab = 343 – 28

21ab = 315

ab = 15

Now a² + b² = (a + b)² – 2ab

a² + b² = (6)² – 2(15)

a² + b² = 36 – 30

a² + b² = 6

**Example 8.**

Use (x ± y)² = x² ± 2xy + y² to evaluate (2.06)²

## Solution:

Given

(2.06)²

We can write the given number as (2 + 0.06)²

This is in the form of (x + y)² = x² + 2xy + y²

(2 + 0.06)² = 2² + 2(2)(0.06) + (0.06)²

= 4 + 0.24 + 0.0036

= 4.2436

Thus the value of (2.06)² is 4.2436

**Example 9.**

Find the value of (3.98)²

## Solution:

Given,

(3.98)²

We can write the given number as (4 – 0.2 )²

This is in the form of (x – y)² = x² – 2xy + y²

(4 – 0.2 )² = 4² – 2(4)(0.2) + (0.2)²

= 16 – 0.16 + 0.04

= 15.88

Thus the value of (3.98)² is 15.88

**Example 10.**

If the sum of two numbers x and y is 15 and the sum of their squares is 35 find the product of the numbers.

## Solution:

Given,

The sum of two numbers x and y is 15 and

The sum of their squares is 35

x + y = 15

x² + y² = 35

This is in the form of (x + y)² = x² + 2xy + y²

(15)² = 35 + 2xy

225 = 35 + 2xy

225 – 35 = 2xy

190 = 2xy

xy = 190/2

xy = 95

Therefore the product of the numbers is 95.