Question

Factorise a^4+2 (x^2+y^2)a^2b^2+(x^2-y^2)^2​

Answer 1

A little mistype in the question:-
Check it:-
${a}^{4} + 2( {x}^{2} + {y}^{2} ) {a}^{2} {b}^{2} + ( {x}^{2} - {y}^{2} )^ {2}{b}^{4}$

Now to solve this you have to see the terms and rearrange it carefully

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

= ${a}^{4} +[ ( {x + y)}^{2} + {(x - y)}^{2} ] {a}^{2} {b}^{2} + {(x - y)}^{2} {(x + y)}^{2} {b}^{4}$

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

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

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

This is the required steps for the factorization.

Answer 2

A little mistype in the question:

Factorise : $a^4+2 (x^2+y^2)a^2b^2+(x^2-y^2)^2b^4$

Answer :

$a^4+2 (x^2+y^2)a^2b^2+(x^2-y^2)^2b^4\\\\\implies a^4+[(x+y)^2+(x-y)^2]a^2b^2+(x^2-y^2)^2b^4\\\\\implies a^4+a^2b^2(x+y)^2+a^2b^2(x-y)^2+b^4(x+y)^2(x-y)^2\\\\\implies a^2[a^2+b^2(x+y)^2]+b^2(x-y)^2[a^2+b^2(x-y)^2]\\\\\implies [a^2+b^2(x+y)^2][a^2+b^2(x-y)^2]$

Step-by-step explanation :

Start by middle term break . We know that (a+b)² can be written as a²+b²+2ab and also (a-b)² can be written as a²+b²-2ab . Hence we have (a+b)²+(a-b)² = a²+b²+2ab + a²+b²-2ab = 2a²+2b² . Thus we can factorise the equation using the expansion . After middle splitting group the terms and take commons . Then we will get the final result . Note that a factor is a polynomial which divides the polynomial .