Question

$\frac{x^2+3x+5}{(x+1)(x+2)(x+3)}=$
$\frac{A}{(x+1)}+\frac{B}{(x+1)(x+2)}+\frac{c}{(x+1)(x+2)(x+3)}$ then ascending order of A, B, C is

• A. B, A, C
• B. A, B, C
• C. C, A, B
• D. B, C, A

Answer 1

Answer: (A)

B, A, C

$\frac{x^2+3x+5}{(x+1)(x+2)(x+3)}=A(x+2)(x+3)+B(x+3)+C(x+1)(x+2)(x+3)$
$f\left(x\right)=x^2+3x+5$
$\left[A\right(x+2)+B](x+3)+C^{\prime }{C}^{\prime }$ is the remainder
when $f\left(x\right)$ is divided by $x+3$
So, $c=f(-3)=-1$
$f(-2)=B+C=3\Rightarrow B=4$
and comparing $x^2$ terms $A=1$
So, $A=1;B=4;C=-1$
decreasing order
B,A,C