Question

factorise a⁴-b⁴+2b²-1​

Answer 1

Answer:

(a^2+b^2-1)(a^2-b^2+1).

Step-by-step explanation:

Given a polynomial such that,

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

To factorise it.

Taking negative sign common, we get,

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

Solving further, we will get,

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

But, we know that,

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

Therefore, we will get,

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

Solving further, we get,

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

But, we know that,

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

Therefore, we will get,

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

Hence, the required factors are (a^2+b^2-1) and (a^2-b^2+1).

Answer 2

a⁴ - b⁴ + 2b² - 1

a⁴ - (b⁴ - 2b² + 1)

Using Identity — (x - y)² = x² + y² - 2xy

Here, x = b² and y = 1

Therefore,

a⁴ - (b² - 1)²

(a²)² - (b² - 1)²

Using Identity — x² - y² = (x+y) (x-y)

Here, x = a² and y = b² - 1

(a² + b² - 1) (a² - b² + 1)