If frac{x^4+3 x+1}{(x+1)^2(x-1)}=A x+B+frac{C | vedclass

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

Question

(D) Given the expression: $\frac{x^4+3 x+1}{(x+1)^2(x-1)}=A x+B+\frac{C}{x+1}+\frac{D}{(x+1)^2}+\frac{E}{x-1}$.First, perform polynomial division of $

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(D) Given the expression: $\frac{x^4+3 x+1}{(x+1)^2(x-1)}=A x+B+\frac{C}{x+1}+\frac{D}{(x+1)^2}+\frac{E}{x-1}$.First, perform polynomial division of $x^4+3x+1$ by $(x+1)^2(x-1) = (x^2+2x+1)(x-1) = x^3+x^2-x-1$.Dividing $x^4+3x+1$ by $x^3+x^2-x-1$ gives a quotient of $(x-1)$ and a remainder of $3x^2+3x$. So, $Ax+B = x-1$, which implies $A=1$ and $B=-1$.Now, $\frac{3x^2+3x}{(x+1)^2(x-1)} = \frac{C}{x+1} + \frac{D}{(x+1)^2} + \frac{E}{x-1}$.Multiply by $(x+1)^2(x-1)$: $3x^2+3x = C(x+1)(x-1) + D(x-1) + E(x+1)^2$.For $x=1$: $3(1)^2+3(1) = E(1+1)^2 \Rightarrow 6 = 4E \Rightarrow E = 3/2$.For $x=-1$: $3(-1)^2+3(-1) = D(-1-1) \Rightarrow 0 = -2D \Rightarrow D = 0$.Comparing coefficients of $x^2$: $3 = C + E \Rightarrow 3 = C + 3/2 \Rightarrow C = 3/2$.Thus, $A=1, B=-1, C=3/2, D=0, E=3/2$.$A+B+C+D+E = 1 - 1 + 3/2 + 0 + 3/2 = 6/2 = 3$.

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

Similar Questions

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

If $\frac{1-x+6x^2}{x-x^3} = \frac{A}{x} + \frac{B}{1-x} + \frac{C}{1+x}$,then $A$ is equal to

$\frac{1}{x(x+1)(x+2) \ldots(x+n)} = \sum_{r=0}^{n} \frac{A_r}{x+r}$. Then $A_r$ is equal to:

$\frac{3 x^{2}+1}{x^{2}-6 x+8}$ is equal to

If $\frac{2x^2-3x+5}{(x-7)^3}=\frac{A}{x-7}+\frac{B}{(x-7)^2}+\frac{C}{(x-7)^3}$, then $2A-3B+C=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module