0Question:
(x + a)(x + 1991) + 1 can be factored as (x + b)(x + c) where a, b, c are integers. Find all the possible values of 'a'.
Solution:
Method 1:
x^2+(1991+a)x+(1991a +1) = 0
From here find the two roots. You will have the discriminant part having two factors 1989 and 1993. b & c being integers
you need to have the discriminant part either=0 or a perfect square. Perfect square wont be possible you will see.
So zero, which gives us a as 1989 & 1993.
Method 2:
-b and - c is roots of the equation.
Hence ( a-b) ( 1991 - b ) = 1
So (a-b) = 1 or -1 and (1991 - b ) = -1 or 1
Solving we get 1989 and 1993
You will have its factors which in turn have been given to be equal to b and c.
