Question

Factorise 16x⁴-(z-x)⁴​

Answer 1

Answer:

Factorization of 16x^2 - ( z - x )^4 is ( 3x - z )( z + x )( 5x^2 + z^2 - 2zx )

Step-by-step explanation:

Given equation : 16x^4 - ( z - x )^4

= > 4^2 x^4 - ( z - x )^4

= > ( 4x^2 )^2 - { ( z - x )^2 }^2

From the properties of factorization we know that a^2 - b^2 can be written as ( a + b ) ( a - b ) .

= > [ 4x^2 - ( z - x )^2 ][ 4x^2 + ( z - x )^2 ]

= > [ ( 2x )^2 - ( z - x )^2 ] [ 4x^2 + ( z - x )^2 ]

Again by the same property ; and using ( a - b )^2 = a^2 + b^2 - 2ab

= > [ 2x - ( z - x ) ][ 2x + ( z - x ) ][ 4x^2 + z^2 + x^2 - 2zx ]

= > [ 2x - z + x ] [ 2x + z - x ][ 5x^2 + z^2 - 2zx ]

= > ( 3x - z )( z + x )( 5x^2 + z^2 - 2zx )

Hence,

Factorization of 16x^2 - ( z - x )^4 is ( 3x - z )( z + x )( 5x^2 + z^2 - 2zx )  

Answer 2

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