Question

If $\frac{3x^2+10x+13}{(x-1{)}^4}=\frac{A}{(x-1{)}^2}+\frac{B}{(x-1{)}^3}+\frac{C}{(x-1{)}^4}$ then descending order of $A,B,C$

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

Answer 1

Answer: (B)

$C,B,A$

$\frac{3x^2+10x+13}{(x-1{)}^4}=\frac{A}{(x-1{)}^2}+\frac{B}{(x-1{)}^3}+\frac{C}{(x-1{)}^4}$
substitute $x-1=t\Rightarrow x=(t+1)$
$\frac{3(t+1{)}^2+10(t+1)+13}{t^4}=\frac{3t^2+16t+26}{t^4}$
$=\frac{3}{t^2}+\frac{16}{t^3}+\frac{26}{t^4}$
$=\frac{3}{(x-1{)}^2}+\frac{16}{(x-1{)}^3}+\frac{26}{(x-1{)}^4}$
So, decreasing order is
$C,B,A$