Question

If $\frac{3x^3-8x^2+10}{(x-1{)}^4}=\frac{A}{x-1}+\frac{B}{(x-1{)}^2}+\frac{C}{(x-1{)}^3}+\frac{D}{(x-1{)}^4}$, then $A+B+C+D$ is equal to

Answer 1

$\frac{3x^3-8x^2+10}{(x-1{)}^4}=\frac{A}{x-1}+\frac{B}{(x-1{)}^2}+\frac{C}{(x-1{)}^3}+\frac{D}{(x-1{)}^4}$

Multiply both sides by $(x-1{)}^4$

$3x^3-8x^2+10=A(x-1{)}^3+B(x-1{)}^2+C(x-1)+D$

Expand and collect in terms of power of $x$

$3x^3-8x^2+10=A{x}^3+(B-3A)x^2+(3A-2B+C)x-A+B-C+D$

Gives

$A=3$

$B-3A=-8\Rightarrow B=-8+3\left(3\right)=1$

$3A-2B+C=0\Rightarrow C=2B-3A=2-9=-7$

$-A+B-C+D=10\Rightarrow D=10+C-B+A=10-7-1+3=5$

$\therefore A=3,B=1,C=-7,D=5$

Hence $A+B+C+D=2$