Question

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

• A. 0
• B. 2
• C. 3
• D. 5

Answer 1

The correct option is B 2

$LHS=\frac{3x+2}{(x+1)(2x^2+3)};RHS=\frac{A}{x+1}+\frac{Bx+C}{(2x^2+3)}$

$=\frac{A(2x^2+3)+(Bx+C)(x+1)}{(x+1)(2x^2+3)}$

$LHS=RHS$

$3x+2=A(2x^2+3)+(Bx+C)(x+1)$

for finding the values of A, B, C compare the

co-efficients of $x^2,x,1$

comparing co-efficients of $x^2$

$O=2A+B--\left(1\right)$

comparing co-efficients of $x$

$3=B+C--\left(2\right)$

comparing co-efficient of 1

$2=3A+C--\left(3\right)$

(3)-(1)

eqn. 3 - eqn. 1

we get

$(3A+C)$

$-(2A+B)$

$=2-0$

$A-B+C=2$