004 Sample Final A, Problem 10

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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 ${\frac {4x^{2}}{(x-1){(x-2)}^{2}}}$ ?
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
$6x^{2}+27x+31=A(x+3)^{2}+B(x+3)(x-1)+C(x-1)$ .
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 ${\frac {4}{x-1}}+{\frac {2}{x+3}}+{\frac {-1}{{(x+3)}^{2}}}$ ${\frac {4}{x-1}}+{\frac {2}{x+3}}+{\frac {-1}{{(x+3)}^{2}}}$ 