Q. 144.4( 9 Votes )

# Find the value is

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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