The partial fraction decomposition of frac{3x | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let $u = x - 1$,so $x = u + 1$. Substituting this into the numerator: $3(u + 1)^3 - 8(u + 1)^2 + 10$. $= 3(u^3 + 3u^2 + 3u + 1) - 8(u^2 + 2u + 1)

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let $u = x - 1$,so $x = u + 1$. Substituting this into the numerator: $3(u + 1)^3 - 8(u + 1)^2 + 10$. $= 3(u^3 + 3u^2 + 3u + 1) - 8(u^2 + 2u + 1) + 10$. $= 3u^3 + 9u^2 + 9u + 3 - 8u^2 - 16u - 8 + 10$. $= 3u^3 + u^2 - 7u + 5$. Now,divide by $u^4$: $\frac{3u^3 + u^2 - 7u + 5}{u^4} = \frac{3}{u} + \frac{1}{u^2} - \frac{7}{u^3} + \frac{5}{u^4}$. Substituting $u = x - 1$ back: $\frac{3}{(x - 1)} + \frac{1}{(x - 1)^2} - \frac{7}{(x - 1)^3} + \frac{5}{(x - 1)^4}$.

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

Similar Questions

$\begin{aligned} & \text{If } \frac{x^4}{(x-a)(x-b)(x-c)}=P(x)+\frac{A}{x-a}+\frac{B}{x-b} \\ & +\frac{C}{x-c} \text{, then } P(0)+A(a-b)(a-c)= \end{aligned}$

If $\frac{x^4}{(x-1)(x-2)(x-3)}=p(x)+\frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x-3}$ then $p\left(\frac{3}{2}\right)+C=$

If $\frac{x^4}{(x-1)(x-2)(x-3)}=Ax+B \cdot \frac{1}{x-1}+C \cdot \frac{1}{x-2}+D \cdot \frac{1}{x-3}+E$,then $A+B+C+D+E=$

If $\frac{8}{(x+3)^2(x-2)}=\frac{Ax+B}{(x+3)^2}+\frac{C}{x-2}$,then $25(B+8C-A)=$

The partial fractions of $\frac{x^4 + 24x^2 + 28}{(x^2 + 1)^3}$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module