Find the values of a and b so that x4 + x3 + 8x2 +ax + b is divisible by 4x2+ 3x – 2.
Asked by miniprasad | 11th May, 2018, 12:32: PM
According to the question,
Let a1x2 + b1x + c1 be the quotient when x4 + x3 + 8x2 +ax + b is divisible by 4x2 + 3x – 2.
x4 + x3 + 8x2 +ax + b = (4x2 + 3x – 2)(a1x2 + b1x + c1)
After simplifying we get
x4 + x3 + 8x2 +ax + b = 4a1x4 + 4b1x3 + 4c1x2 + 3a1x3 + 3b1x2 + 3c1x - 2a1x2 - 2b1x - 2c1
x4 + x3 + 8x2 +ax + b = 4a1x4 + (4b1 + 3a1)x3 + (4c1 + 3b1 - 2a1)x2 + (3c1 - 2b1) x - 2c1
Comparing coefficients on both sides we get,
4a1 = 4 → a1 = 1
4b1 + 3a1 = 1 → 4b1 + 3 = 1 → b1 = -1/2
Similarly using 4c1 + 3b1 - 2a1 = 8 find c1 we get c1 = 23/8
Put the values of a1, b1, c1 in 3c1 - 2b1 = a and - 2c1 = b to find values of a and b.
a = 3×23/8 - 2×(-1/2)
a = 77/8
- 2c1 = b
b = -2 × (23/8) = -23/4
Answered by Sneha shidid | 11th May, 2018, 03:44: PM
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