Question

Factorize it partially
$\frac{2x^2-5x+1}{x^2(x^2-1)}$

• A. $\frac{5}{x}-\frac{1}{x^2}-\frac{1}{(x-1)}-\frac{4}{(x+1)}$
• B. $\frac{1}{x}+\frac{1}{x^2}-\frac{1}{(x-1)}-\frac{4}{(x+1)}$
• C. $\frac{5}{x}-\frac{1}{x^2}+\frac{1}{(x-1)}+\frac{4}{(x+1)}$
• D. $\frac{1}{x}+\frac{1}{x^2}-\frac{1}{(x-1)}+\frac{4}{(x+1)}$

Answer 1

The correct option is A $\frac{5}{x}-\frac{1}{x^2}-\frac{1}{(x-1)}-\frac{4}{(x+1)}$

Let $\frac{2x^2-5x+1}{x^2(x^2+1)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x-1}+\frac{D}{x+1}$ .....(1)

Multiply equation with $x^2(x^2-1)$, we get

$2x^2-5x+1=A.x.(x-1)(x+1)+B(x-1)(x+1)+C.x^2(x+1)+D{x}^2.(x-1)$

We put $x=0$ to find value of B

$\therefore 1=B(-1)\Rightarrow B=-1$

Now put $x=1$, to find value of C

$\therefore 2-5+1=C\left(2\right)\Rightarrow C=-1$

Now put $x=-1$, to find value of D

$\therefore 2+5+1=D(-2)\Rightarrow D=-4$

Further to find A, we need to solve whole equation and compare the like terms.

$2x^2-5x+1=A{x}^3-Ax-1(x^2-1)-1(x+1)x-4x^2(x-1)$ ... Putting values of B,C,D

$=A{x}^3-Ax-x^2+1-x^3-x^2-4x^3+4x^2$

$=x^3(A-5)-Ax+2x^2+1$

Now comparing coefficients of $x$, we get $-5=-A$

$\therefore A=5$

Thus putting all these values of A,B,C and D in equation (1), we get

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