Question

If $\frac{3x+5}{2x^2+x-6}=\frac{A}{2x-3}+\frac{B}{x+2}$ identically then

• A. $A=\frac{1}{7},B=\frac{19}{7}$
• B. $A=\frac{-19}{7},B=\frac{1}{7}$
• C. $A=\frac{19}{7},B=\frac{1}{7}$
• D. none of these

Answer 1

The correct option is C $A=\frac{19}{7},B=\frac{1}{7}$

$\frac{3x+5}{2x^2+x-6}=\frac{A}{2x-3}+\frac{B}{x+2}$
$\frac{3x+5}{2x^2+4x-3x-6}=\frac{3x+5}{(2x-3)(x+2)}$
Resolving into partial fractions
$\frac{3x+5}{(2x-3)(x+2)}=\frac{A}{2x-3}+\frac{B}{x+2}$
$\Rightarrow 3x+5=A(x+2)+B(2x-3)$
$\Rightarrow 3x+5=(A+2B)x+(2A-3B)$
$\Rightarrow A+2B=3,2A-3B=5$
Solving these equations, we get
$B=\frac{1}{7},A=\frac{19}{7}$