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