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 ${\frac {3x^{2}+6x+7}{(x+3)^{2}(x-1)}}={\frac {A}{x-1}}+{\frac {B}{x+3}}+{\frac {C}{(x+3)^{2}}}$ . We clear the denominators and end up with $A(x+3)^{2}+B(x-1)(x+3)+C(x-1)=3x^{2}+6x+7$ .

Step 3:
By evaluation both sides by using x = 1, we will zero out the B and C. This leads to $A(1+3)^{2}=3(1)^{2}+6(1)+7,16A=3+6+7$ , and finally $A=1$ .

Step 4:
Now evaluating at x = -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