008A Sample Final A, Question 11

Question: Decompose into separate partial fractions ${\frac {3x^{2}+6x+7}{(x+3)^{2}(x-1)}}$

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, $x-1$, will appear once in the denominator of the decomposition. The other two denominators will be $x+3{\text{, and }}(x+3)^{2}$.
2) After writing the equality, ${\frac {3x^{2}+6x+7}{(x+3)^{2}(x-1)}}={\frac {A}{x-1}}+{\frac {B}{x+3}}+{\frac {C}{(x+3)^{2}}}$, clear the denominators, and evaluate both sides at x = 1, -3, and any third value. Each evaluation will yield the value of one of the three unknowns.

Solution:

Step 1:
From the factored form of the denominator we can observe that there will be three denominators: $x-1,x+3$, and $(x+3)^{2}$. So the final answer with have the following form: ${\frac {A}{x-1}}+{\frac {B}{x+3}}+{\frac {C}{(x+3)^{2}}}$

Step 2:
Now we have the equality ${\frac {3x^{2}+6x+7}{(x+3)^{2}(x-1)}}={\frac {A}{x-1}}+{\frac {B}{x+3}}+{\frac {C}{(x+3)^{2}}}$. Now clearing the denominators we end up with $A(x+3)^{2}+B(x-1)(x+3)+C(x-1)=3x^{2}+6x+7$.

Step 3:
To proceed we start by evaluating both sides at different x-values. We start with x = 1, since this will zero out the B and C. This leads to $A(1+3)^{2}=3(1)^{2}+6(1)+7$, $16A=3+6+7$, and finally A = 1.

Step 4:
Now evaluate at -3 to zero out both A and B. This yields the following equations $C((-3)-1)=3(-3)^{2}+6(-3)+7$, $-4C=3(9)-18+7$, $-4C=27-11$, and $C=-4$

Step 5:

To obtain the value of B we can evaluate x at any value except 1, and -3. We do not want to evaluate at 1 and -3 since both of these will zero out the B. Evaluating at x = 0 will make the arithmetic easier, and gives us $A(0+3)^{2}+B(0-1)(0+3)+C(0-1)=9A-3B-C=7$. However, we know the values of both A and C, which are 1 and -4, respectively. So $9-3B-(-4)=7$, $-3B+13=7$, $-3B=-6$, and finally $B=2$. This means the final answer is ${\frac {1}{x-1}}+{\frac {2}{x+3}}-{\frac {4}{(x+3)^{2}}}$

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

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