Doubt by Saksham
Solution :
(x²+4)-2a-a²-5
=x²+4-2a-a²-5
=x²-2a-a²-1
=x²-[2a+a²+1]
=x²-[a²+2a+1]
=x²-[(a)²+2(a)(1)+(1)²]
=x²-(a+1)²
=x²-2a-a²-1
=x²-[2a+a²+1]
=x²-[a²+2a+1]
=x²-[(a)²+2(a)(1)+(1)²]
=x²-(a+1)²
[∵(a²+2ab+b²)=(a+b)²]
=(x+a+1)=(x-[a+1])
=(x+a+1)(x-a-1)
=(x+a+1)=(x-[a+1])
=(x+a+1)(x-a-1)
Hence, (x+a+1) and (x-a-1) are the required factors of the given question.