Answer :

Let, f (x) = ax4+2x3-3x2+bx-4 and g (x) = x 2 – 4

We have,


g (x) = x2 – 4


= (x – 2) (x + 2)


Given,


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


(x – 2) and (x + 2) are factors of f (x).


From factor theorem if (x – 2) and (x + 2) are factors of f (x) then f (2) = 0 and f (-2) = 0 respectively.


f (2) = 0


a * (-2)4 + 2 (2)3 – 3 (2)2 + b (2) – 4 = 0


16a – 16 – 12 + 2b – 4 = 0


16a + 2b = 0


2 (8a + b) = 0


8a + b = 0 (i)


Similarly,


f (-2) = 0


a * (-2)4 + 2 (-2)3 – 3 (-2)2 + b (-2) – 4 = 0


16a – 16 – 12 - 2b – 4 = 0


16a - 2b – 32 = 0


16a – 2b – 32 = 0


2 (8a - b) = 32


8a – b = 16 (ii)


Adding (i) and (ii), we get


8a + b + 8a – b = 16


16a = 16


a = 1


Put a = 1 in (i), we get


8 * 1 + b = 0


b = -8


Hence, a = 1 and b = -8.


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