Find the value is of a and b, if x2 - 4 is a factor of ax4 + 2x3 - 3

1.	Find the value is of a and b, if x2 - 4 is a factor of ax4 + 2x3 - 3x2 + bx - 4.
Answer» Let, f (x) = ax⁴ + 2x³ - 3x² + bx - 4 and g (x) = x² – 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 x (-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 x (-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 x 1 + b = 0 b = -8 Hence, a = 1 and b = -8.

Find the value is of a and b, if x2 - 4 is a factor of ax4 + 2x3 - 3x2 + bx - 4.

Answer»

Let, f (x) = ax⁴ + 2x³ - 3x² + bx - 4 and g (x) = x² – 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 x (-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 x (-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 x 1 + b = 0

b = -8

Hence,

a = 1 and b = -8.