Question

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

• A. 2
• B. -3
• C. 1
• D. 4

Answer 1

Answer: (C)

1

$LHS=\frac{x^2+5x+7}{(x-3{)}^3};RHS=\frac{A(x-3{)}^2+B(x-3)+C}{(x-3{)}^3}$ (making denominator common)
$LHS=RHS$
$x^2+5x+7=A(x-3{)}^2+B(x-3)+C$
equating the co- efficients of 'n' we get the
equations
equating $x^2$ co-efficients we get
$A=1$