# 004 Sample Final A, Problem 10

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Decompose into separate partial fractions.      ${\displaystyle {\frac {6x^{2}+27x+31}{(x+3)^{2}(x-1)}}}$

Foundations
1) What is the form of the partial fraction decomposition of ${\displaystyle {\frac {3x-37}{(x+1)(x-4)}}}$?
2) What is the form of the partial fraction decomposition of ${\displaystyle {\frac {4x^{2}}{(x-1){(x-2)}^{2}}}}$?
1) ${\displaystyle {\frac {A}{x+1}}+{\frac {B}{x-4}}}$
2)${\displaystyle {\frac {A}{x-1}}+{\frac {B}{x-2}}+{\frac {C}{{(x-2)}^{2}}}}$

Solution:

Step 1:
We set ${\displaystyle {\frac {6x^{2}+27x+31}{(x+3)^{2}(x-1)}}={\frac {A}{x-1}}+{\frac {B}{x+3}}+{\frac {C}{{(x+3)}^{2}}}}$.
Step 2:
Multiplying both sides of the equation by ${\displaystyle (x+3)^{2}(x-1)}$, we get
${\displaystyle 6x^{2}+27x+31=A(x+3)^{2}+B(x+3)(x-1)+C(x-1)}$.
Step 3:
If we set ${\displaystyle x=1}$ in the above equation, we get ${\displaystyle 16A=64}$ and ${\displaystyle A=4}$.
If we set ${\displaystyle x=-3}$ in the above equation, we get ${\displaystyle -4C=4}$ and ${\displaystyle C=-1}$.
Step 4:
In the equation ${\displaystyle 6x^{2}+27x+31=A(x+3)^{2}+B(x+3)(x-1)+C(x-1)}$, we compare the constant terms of both sides. We must have
${\displaystyle 9A-3B-C=31}$. Substituting ${\displaystyle A=4}$ and ${\displaystyle C=-1}$, we get ${\displaystyle B=2}$.
Thus, the partial fraction decomposition is ${\displaystyle {\frac {4}{x-1}}+{\frac {2}{x+3}}+{\frac {-1}{{(x+3)}^{2}}}}$
${\displaystyle {\frac {4}{x-1}}+{\frac {2}{x+3}}+{\frac {-1}{{(x+3)}^{2}}}}$