005 Sample Final A, Question 10

Question Write the partial fraction decomposition of the following,

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

Foundations
1) How many fractions will this decompose into? What are the denominators?
2) How do you solve for the numerators?
Answer:
1) Since each of the factors are linear, and one has multipliclity 2, there will be three denominators. The linear term, x, will appear once in the denominator of the decomposition. The other two denominators will be x - 1, and $(x-1)^{2}$.
2) After writing the equality, ${\frac {x+2}{x(x-1)^{2}}}={\frac {A}{x}}+{\frac {B}{x-1}}+{\frac {C}{(x-1)^{2}}}$, clear the denominators, and use the cover up method to solve for A, B, and C. After you clear the denominators, the cover up method is to evaluate both sides at x = 1, 0, and any third value. Each evaluation will yield the value of one of the three unknowns.

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,  $x+2=(2+B)x^{2}+(-1-B)x+2$. 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:
${\frac {x+2}{x(x-1)^{2}}}={\frac {2}{x}}+{\frac {2}{x-1}}+{\frac {3}{(x-1)^{2}}}$

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