BINOMIAL EXPANSION USING ALGEBRAIC IDENTITIES

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Problem 1 :

(x + 7)²

Solution :

By comparing (x + 7)² with (a + b)², we get a  =  x and b  =  7

(x + 7)²  =  x² + 2(x)(7) + 7²

=  x² + 14x + 49

Problem 2 :

(2x + 5)²

Solution :

Given , (2x + 5)²

Here a  =  2x and b  =  5.

(2x + 5)²  =  (2x)² + 2(2x)(5) + 5²

=  4x² + 20x + 25

Problem 3 :

(6 + 5x)²

Solution :

Given , (6 + 5x)²

Here a  =  6 and b  =  5x

(6 + 5x)²  =  6² + 2(6)(5x) + (5x)²

= 25x² + 60x + 36

Problem 4 :

(3a + 1)²

Solution :

Given , (3a + 1)²

Here a  =  3a and b  =  1

(3a + 1)²  =  (3a)² + 2(3a)(1) + 1²

=  9a² + 6a + 1 

Problem 5 :

(1 + 2b)²

Solution :

Given , (1 + 2b)²

Here a  =  1 and b  =  2b

(1 + 2b)²  =  1² + 2(1)(2b) + (2b)²

=  4b² + 4b + 1 

Problem 6 :

(3 – n)²

Solution :

Given , (3 – n)²

Here a  =  3 and b  =  n

(3 – n)²  =  3² - 2(3)(n) + (-n)²

=  n² – 6n + 9

Problem 7 :

(3x – 2)²

Solution :

Given , (3x – 2)²

Here a  =  3x and b  =  2

(3x – 2)² =  (3x)² - 2(3x)(2) + (-2)²

=  9x² – 12x + 4

Problem 8 :

(2x – 5)²

Solution :

Given , (2x – 5)²

Here a  =  2x and b  =  5

(2x – 5)² =  (2x)² – 2(2x)(5) + (-5)²

=  4x² – 20x + 25

Problem 9 :

(4 – 3a)²

Solution :

Given , (4 – 3a)²

Here a  =  4 and b  =  3a

(4 – 3a)² = 4² – 2(4)(3a) + (-3a)²

=  9a² – 24a + 16 

Problem 10 :

(2x -3y)²

Solution :

Given , (2x -3y)²

Here a  =  2x and b  =  3y

(2x -3y)²  =  (2x)² – 2(2x)(3y) + (-3y)²

=  4x² – 12xy + 9y² 

Problem 11 :

(2x + 13) (2x – 13)

Solution :

Given , (2x + 13) (2x – 13)

We can compare (2x + 13) (2x – 13) with (a+b)(a-b).

(a+b)(a-b)  =  a²-b²

Here a  =  1 and b  =  2b

=  (2x)² – (13)²

=  4x² – 169

Problem 12 :

(m + n) (m – n)

Solution :

Given , (m + n) (m – n)

We can compare (m + n) (m – n) with (a+b)(a-b).

(a+b)(a-b)  =  a²-b²

=  m² – n²

=  m² – n²

Problem 13 :

(7x + 1) (7x – 1)

Solution :

Given , (7x + 1) (7x – 1)

We can compare (7x + 1) (7x – 1) with (a+b)(a-b).

(a+b)(a-b)  =  a²-b²

=  (7x)²  – 1²

=  49x² - 1

Problem 14 :

(a + 8) (a – 8)

Solution :

Given , (a + 8) (a – 8)

We can compare (a + 8) (a – 8) with (a+b)(a-b).

=  a² – 8²

=  a² - 64

Problem 15 :

Expand : 

(p + 2q)³ 

Solution :

Comparing (a + b)³ and (p + 2q)³, we get

a  =  p and b  =  2q

Substitute p for a and 2q for b. 

(p + 2q)³  =  p³ + 3(p²)(2q) + 3(p)(2q)² + (2q)³

(p + 2q)³  =  p³ + 6p²q + 3(p)(4q²) + 8q³

(p + 2q)³  =  p³ + 6p²q + 12pq² + 8q³

So, the expansion of (p + 2q)³ is

p³ + 6p²q + 12pq² + 8q³

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