005 Sample Final A, Question 10

Question Write the partial fraction decomposition of the following,

${\frac {x+2}{x^{3}-2x^{2}+x}}$

Step 1:
First, we factor the denominator. We have $x^{3}-2x^{2}+x=x(x^{2}-2x+1)=x(x-1)^{2}$

Step 2:
Since we have a repeated factor in the denominator, we set ${\frac {x+2}{x(x-1)^{2}}}={\frac {A}{x}}+{\frac {B}{x-1}}+{\frac {C}{(x-1)^{2}}}$.

Step 3:
Multiplying both sides of the equation by the denominator $x(x-1)^{2}$, we get
$x+2=A(x-1)^{2}+B(x)(x-1)+Cx$.

Step 4:
If we let $x=0$, we get $2=A$. If we let $x=1$, we get $3=C$.

Step 5:
To solve for $B$, we plug in $A=2$ and $C=3$ and simplify. We have
$x+2=2(x-1)^{2}+B(x)(x-1)+3x=2x^{2}-4x+2+Bx^{2}-Bx+3x.So,<math>x+2=(2+B)x^{2}+(-1-B)x+2$. Since both sides are equal,
we must have $2+B=0$ and $-1-B=1$. So, $B=2$. Thus, the decomposition is ${\frac {x+2}{x(x-1)^{2}}}={\frac {2}{x}}+{\frac {2}{x-1}}+{\frac {3}{(x-1)^{2}}}$.

Final Answer:
${\frac {x+2}{x(x-1)^{2}}}={\frac {2}{x}}+{\frac {2}{x-1}}+{\frac {3}{(x-1)^{2}}}$