Answer :

Let, f (x) = x4+px3+2x2-3x+q be the given polynomial.

And, let g (x) = (x2 – 1) = (x – 1) (x + 1)


Clearly,


(x – 1) and (x + 1) are factors of g (x)


Given, g (x) is a factor of f (x)


(x – 1) and (x + 1) are factors of f (x)


From factor theorem


If (x – 1) and (x + 1) are factors of f (x) then f (1) = 0 and f (-1) = 0 respectively.


f (1) = 0


(1)4 + p (1)3 + 2 (1)2 – 3 (1) + q = 0


1 + p + 2 – 3 + q = 0


p + q = 0 (i)


Similarly,


f (-1) = 0


(-1)4 + p (-1)3 + 2 (-1)2 - 3 (-1) + q = 0


1 – p + 2 + 3 + q = 0


q – p + 6 = 0 (ii)


Adding (i) and (ii), we get


p + q + q – p + 6 = 0


2q + 6 = 0


2q = - 6


q = -3


Putting value of q in (i), we get


p – 3 = 0


p = 3


Hence, x2 – 1 is divisible by f (x) when p = 3 and q = - 3.


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