For R.H.S
if we expand the terms; the last term will be abcd
for L.H.S.
we can take it as
the ( a - 1) ^ 2 - 25 ; here the last term in the bracekts will be +1 always.
so 1 - 25 = -24
what are the options given.
Let me know if its wrong.
value of abcd
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Agree with Suyog OA should be -24
1st approach (fastest)
(x^2 - 5x - 1)^2 - 25
here when the term (x^2 - 5x - 1)^2 is expanded the result will be all multiples of x & only 1 as a constant term (without x coz it is a square)
when (x-a) * (x-b) * (x-c) * (x-d) is expanded then the constant term will be abcd
so some multiples of x + 1-25 = some multiples of x + abcd
so abcd =-24
2nd approach
(x^2 - 5x - 1)^2 - 25
= (x^2 - 5x - 1)^2 - 5^2
= (x^2 - 5x + 4)(x^2 - 5x - 6)
= (x-1)(x-4)(x-6)(x+1)
abcd =-24
1st approach (fastest)
(x^2 - 5x - 1)^2 - 25
here when the term (x^2 - 5x - 1)^2 is expanded the result will be all multiples of x & only 1 as a constant term (without x coz it is a square)
when (x-a) * (x-b) * (x-c) * (x-d) is expanded then the constant term will be abcd
so some multiples of x + 1-25 = some multiples of x + abcd
so abcd =-24
2nd approach
(x^2 - 5x - 1)^2 - 25
= (x^2 - 5x - 1)^2 - 5^2
= (x^2 - 5x + 4)(x^2 - 5x - 6)
= (x-1)(x-4)(x-6)(x+1)
abcd =-24
Regards
Samir
Samir