Question

Number of Partial Fractions of $\frac{x^3+x^2+1}{x^4+x^2+1}$ is

• A. 2
• B. 3
• C. 4
• D. 7

Answer 1

The correct option is A 2

$\frac{x^3+x^2+1}{x^4+x^2+1}=\frac{x^3+x^2+1}{(x^2+x-1)(x^2-x+1)}$
$=\frac{Ax+B}{x^2+x+1}+\frac{Cx+D}{x^2-x+1}$
$x^2+x+1;x^2-x+1\Rightarrow$ non reducable
So, no. of partial fractions 2