# 008A Sample Final A, Question 11

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

Foundations:
1) How many fractions will this decompose into? What are the denominators?
2) How do you solve for the numerators?
1) Since each of the factors are linear, and one has multipliclity 2, there will be three denominators. The linear term, ${\displaystyle x-1}$, will appear once in the denominator of the decomposition. The other two denominators will be ${\displaystyle x+3{\text{, and }}(x+3)^{2}}$.
2) After writing the equality, ${\displaystyle {\frac {3x^{2}+6x+7}{(x+3)^{2}(x-1)}}={\frac {A}{x-1}}+{\frac {B}{x+3}}+{\frac {C}{(x+3)^{2}}}}$, 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: ${\displaystyle x-1,x+3}$, and ${\displaystyle (x+3)^{2}}$. So the final answer with have the following form: ${\displaystyle {\frac {A}{x-1}}+{\frac {B}{x+3}}+{\frac {C}{(x+3)^{2}}}}$
Step 2:
Now we have the equality ${\displaystyle {\frac {3x^{2}+6x+7}{(x+3)^{2}(x-1)}}={\frac {A}{x-1}}+{\frac {B}{x+3}}+{\frac {C}{(x+3)^{2}}}}$. We clear the denominators and end up with ${\displaystyle A(x+3)^{2}+B(x-1)(x+3)+C(x-1)=3x^{2}+6x+7}$.
Step 3:
By evaluation both sides by using x = 1, we will zero out the B and C. This leads to ${\displaystyle A(1+3)^{2}=3(1)^{2}+6(1)+7,16A=3+6+7}$ , and finally ${\displaystyle A=1}$.
Step 4:
Now evaluating at x = -3 to zero out both A and B. This yields the following equations ${\displaystyle C((-3)-1)=3(-3)^{2}+6(-3)+7}$, ${\displaystyle -4C=3(9)-18+7}$, ${\displaystyle -4C=27-11}$, and ${\displaystyle C=-4}$
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 ${\displaystyle A(0+3)^{2}+B(0-1)(0+3)+C(0-1)=9A-3B-C=7}$. However, we know the values of both A and C, which are 1 and -4, respectively. So ${\displaystyle 9-3B-(-4)=7}$, ${\displaystyle -3B+13=7}$, ${\displaystyle -3B=-6}$, and finally ${\displaystyle B=2}$.
${\displaystyle {\frac {1}{x-1}}+{\frac {2}{x+3}}-{\frac {4}{(x+3)^{2}}}}$