Question

The coefficient of \(x^3\) in the expansion of \(\frac{x^4 + 1}{(x^2 + 1)(x - 1)}\) when it is expressed in terms of positive integral powers of \(x\), is

• A. 0
• B. 1
• C. 16
• D. 24

Hint:

For polynomial division, always match the highest degree term first and proceed step by step.

Answer 1

The Correct Option is A

To determine the coefficient of \(x^3\) in the expansion of \(\frac{x^4 + 1}{(x^2 + 1)(x - 1)}\), we first simplify the denominator:
\[ (x^2 + 1)(x - 1) = x^3 - x^2 + x - 1. \]
Now, divide \(x^4 + 1\) by \(x^3 - x^2 + x - 1\) using polynomial division:
1. Divide \(x^4\) by \(x^3\), giving \(x\).
2. Multiply \(x(x^3 - x^2 + x - 1)\) to get \(x^4 - x^3 + x^2 - x\).
3. Subtract: \(x^4 + 1 - (x^4 - x^3 + x^2 - x) = x^3 - x^2 + x + 1\).
4. Divide \(x^3\) by \(x^3\), giving \(1\).
5. Multiply \(1(x^3 - x^2 + x - 1)\) to get \(x^3 - x^2 + x - 1\).
6. Subtract: \(x^3 - x^2 + x + 1 - (x^3 - x^2 + x - 1) = 2\).
The quotient is \(x + 1 + \frac{2}{x^3 - x^2 + x - 1}\). Since no term in the expansion contains \(x^3\), the coefficient of \(x^3\) is 0.