Question

The partial fractions of $\frac{3x^{3} - 8x^{2} + 10}{(x - 1)^{4}}$ is

• A. $\frac{3}{(x - 1)} + \frac{1}{(x - 1)^{2}} + \frac{7}{(x - 1)^{3}} + \frac{5}{(x - 1)^{4}}$
• B. $\frac{3}{(x - 1)} + \frac{1}{(x - 1)^{2}} - \frac{7}{(x - 1)^{3}} + \frac{5}{(x - 1)^{4}}$
• C. $\frac{3}{(x - 1)} + \frac{1}{(x - 1)^{2}} - \frac{7}{(x - 1)^{3}} + \frac{5}{(x - 1)^{4}}$
• D. None of these

Answer 1

Answer: (C)

$\frac{3}{(x - 1)} + \frac{1}{(x - 1)^{2}} - \frac{7}{(x - 1)^{3}} + \frac{5}{(x - 1)^{4}}$

Answer 2

$\frac{3x^{3} - 8x^{2} + 10}{(x - 1)^{4}} = \frac{A}{x - 1} + \frac{B}{(x - 1)^{2}} + \frac{C}{(x - 1)^{3}} + \frac{D}{(x - 1)^{4}}$

⇒$3x^{3} - 8x^{2} + 10 = A(x - 1)^{3} + B(x - 1)^{2} + C(x - 1) + D$

Equating coefficients of different powers of $x$, $3 = A$

$- 8 = - 3A + B \Rightarrow B = 1$

0 = $3A - 2B + C \Rightarrow C = - 7$

$10 = - A + B - C + D \Rightarrow D = 5$

$\therefore$ Given expression

= $\frac{3}{x - 1} + \frac{1}{(x - 1)^{2}} - \frac{7}{(x - 1)^{3}} + \frac{5}{(x - 1)^{4}}$.