Question

Factorize it partially
$\frac{x+2}{2x^2-7x+3}$

• A. $\frac{1}{(2x-3)}+\frac{1}{(2x-1)}$
• B. $\frac{1}{(2x-3)}-\frac{1}{(2x-1)}$
• C. $\frac{1}{(x-3)}+\frac{1}{(2x-1)}$
• D. $\frac{1}{(x-3)}-\frac{1}{(2x-1)}$

Answer 1

The correct option is D $\frac{1}{(x-3)}-\frac{1}{(2x-1)}$

Given, $\frac{x+2}{2x^2-7x+3}=\frac{x+2}{(x-3)(2x-1)}$

Let $\frac{x+2}{2x^2-7x+3}=\frac{A}{x-3}+\frac{B}{2x-1}$ ....(1)

Multiply both sides by $(x-3)(2x-1)$, we get

$x+2=A(2x-1)+B(x-3)$

Put $x=3$ to find value of A

$\therefore 3+2=A(6-1)\Rightarrow A=1$

Now put $x=\frac{1}{2}$ to find value of B

$\therefore \frac{1}{2}+2=B(\frac{1}{2}-3)\Rightarrow B=-1$

Putting the value of A and B in equation (1), we get

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