004 Sample Final A, Problem 10

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

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

Solution:

Step 1:
We set ${\frac {6x^{2}+27x+31}{(x+3)^{2}(x-1)}}={\frac {A}{x-1}}+{\frac {B}{x+3}}+{\frac {C}{{(x+3)}^{2}}}$.

Step 2:
Multiplying both sides of the equation by $(x+3)^{2}(x-1)$, we get
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 6x^2+27x+31=A(x+3)^2+B(x+3)(x-1)+C(x-1)} .

Step 3:
If we set $x=1$ in the above equation, we get $16A=64$ and $A=4$.
If we set $x=-3$ in the above equation, we get $-4C=4$ and $C=-1$.

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

Final Answer:
${\frac {4}{x-1}}+{\frac {2}{x+3}}+{\frac {-1}{{(x+3)}^{2}}}$

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