Question

4a²-b²-2b-1 . This is a question of factorization

Answer 1

Answer:

$\large\boxed{\sf{(2a+b+1)(2a-b-1)}}$

Step-by-step explanation:

Given an expression,

$4 {a}^{2} - {b}^{2} - 2b - 1$

To factorise the expression,

Taking out (-) sign common, we get,

$= 4 {a}^{2} - ( {b}^{2} + 2b + 1)$

But, we know that,

• ${b}^{2} + 2b + 1 = {(b + 1)}^{2}$

Therefore, we will get,

$= 4 {a}^{2} - {(b + 1)}^{2}$

But, we know that,

• $4 {a}^{2} = {(2a)}^{2}$

Therefore, we will get,

$= {(2a)}^{2} - {(b + 1)}^{2}$

But, we know that,

• ${x}^{2} - {y}^{2} = (x + y)(x - y)$

Therefore, we will get,

$= (2a + b + 1)(2a - b - 1)$

Hence, the factors are (2a+b+1)(2a-b-1)

Answer 2

Answer:

(4a²)-b(b+2)-1

Step-by-step explanation:

4a²-b²-2b-1

Take common

=(4a²) -b(b+2)-1