Question

Resolve into partial fractions
$\frac{3x^3-8x^2+10}{{(x-1)}^4}$

Answer 1

$\frac{3x^3-8x^2+10}{{(x-1)}^4}$ can be resolved into partial fraction as $\frac{A}{x-1}+\frac{B}{{(x-1)}^2}+\frac{C}{{(x-1)}^3}+\frac{D}{{(x-1)}^4}$

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

Put $x=1$ , we get $3-8+10=D$

$D=5$

Comparing the coefficient of $x^3$ on both sides we get , $A=3$ .

So equation reduces to $3x^3-8x^2+10=3(x-1{)}^3+B(x-1{)}^2+C(x-1)+5$

Now compare the constant on both sides , $10=-3+B-C+5$

$=>8=B-C$ - (1)

Comparing the coefficent of $x^2$ we get $-8=-9+B$

$=>B=1$

So , put $B=1$ in (1) , we get $C=-7$

Therefore , $\frac{3x^3-8x^2+10}{{(x-1)}^4}=\frac{3}{x-1}+\frac{1}{{(x-1)}^2}-\frac{7}{{(x-1)}^3}+\frac{5}{{(x-1)}^4}$