 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.

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.