Question

The number of partial fraction of $\frac{3x^2+70x+93}{(x-1{)}^4}$ is

• A. $3$
• B. $4$
• C. $5$
• D. $2$

Answer 1

Answer: (A)

$3$

by subsitution method
$x-1=t\Rightarrow x=t+1$

$\frac{3(t+1{)}^2+70(t+1)+93}{t^4}$
$\Rightarrow \frac{3t^2+76t+166}{t^4}=\frac{3}{t^2}+\frac{76}{t^3}+\frac{166}{t^4}$
$\Rightarrow \frac{3}{(x-1{)}^2}+{\frac{76}{(x-1)}}^3+\frac{166}{(x-1{)}^4}$
So, $3$ partial fractions.