Partial Fraction Decomposition worksheet with ten problems for practice.

Problem: Partial Fraction Decomposition

The task is to decompose the given rational functions into partial fractions. We will solve each problem step by step.

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#### Problem 1:
$$
\frac{-7x - 15}{x^2 + 6x + 9}
$$

1. Factor the denominator:
$$
x^2 + 6x + 9 = (x + 3)^2
$$

2. Set up the partial fraction decomposition:
Since the denominator has a repeated linear factor, we write:
$$
\frac{-7x - 15}{(x + 3)^2} = \frac{A}{x + 3} + \frac{B}{(x + 3)^2}
$$

3. Combine the fractions on the right-hand side:
$$
\frac{A}{x + 3} + \frac{B}{(x + 3)^2} = \frac{A(x + 3) + B}{(x + 3)^2}
$$

4. Equate the numerators:
$$
-7x - 15 = A(x + 3) + B
$$

5. Expand and simplify:
$$
-7x - 15 = Ax + 3A + B
$$
$$
-7x - 15 = (A)x + (3A + B)
$$

6. Match coefficients:
- Coefficient of \(x\): \(A = -7\)
- Constant term: \(3A + B = -15\)

7. Solve for \(B\):
$$
3(-7) + B = -15
$$
$$
-21 + B = -15
$$
$$
B = 6
$$

8. Write the partial fraction decomposition:
$$
\frac{-7x - 15}{x^2 + 6x + 9} = \frac{-7}{x + 3} + \frac{6}{(x + 3)^2}
$$

Answer:
$$
\boxed{\frac{-7}{x + 3} + \frac{6}{(x + 3)^2}}
$$

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#### Problem 2:
$$
\frac{3x + 10}{x^2 + 9x + 20}
$$

1. Factor the denominator:
$$
x^2 + 9x + 20 = (x + 4)(x + 5)
$$

2. Set up the partial fraction decomposition:
$$
\frac{3x + 10}{(x + 4)(x + 5)} = \frac{A}{x + 4} + \frac{B}{x + 5}
$$

3. Combine the fractions on the right-hand side:
$$
\frac{A}{x + 4} + \frac{B}{x + 5} = \frac{A(x + 5) + B(x + 4)}{(x + 4)(x + 5)}
$$

4. Equate the numerators:
$$
3x + 10 = A(x + 5) + B(x + 4)
$$

5. Expand and simplify:
$$
3x + 10 = Ax + 5A + Bx + 4B
$$
$$
3x + 10 = (A + B)x + (5A + 4B)
$$

6. Match coefficients:
- Coefficient of \(x\): \(A + B = 3\)
- Constant term: \(5A + 4B = 10\)

7. Solve the system of equations:
$$
\begin{cases}
A + B = 3 \\
5A + 4B = 10
\end{cases}
$$
- From \(A + B = 3\), solve for \(B\): \(B = 3 - A\)
- Substitute into \(5A + 4B = 10\):
$$
5A + 4(3 - A) = 10
$$
$$
5A + 12 - 4A = 10
$$
$$
A + 12 = 10
$$
$$
A = -2
$$
- Substitute \(A = -2\) into \(B = 3 - A\):
$$
B = 3 - (-2) = 5
$$

8. Write the partial fraction decomposition:
$$
\frac{3x + 10}{x^2 + 9x + 20} = \frac{-2}{x + 4} + \frac{5}{x + 5}
$$

Answer:
$$
\boxed{\frac{-2}{x + 4} + \frac{5}{x + 5}}
$$

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#### Problem 3:
$$
\frac{20x + 9}{25x^2 + 20x + 4}
$$

1. Factor the denominator:
$$
25x^2 + 20x + 4 = (5x + 2)^2
$$

2. Set up the partial fraction decomposition:
$$
\frac{20x + 9}{(5x + 2)^2} = \frac{A}{5x + 2} + \frac{B}{(5x + 2)^2}
$$

3. Combine the fractions on the right-hand side:
$$
\frac{A}{5x + 2} + \frac{B}{(5x + 2)^2} = \frac{A(5x + 2) + B}{(5x + 2)^2}
$$

4. Equate the numerators:
$$
20x + 9 = A(5x + 2) + B
$$

5. Expand and simplify:
$$
20x + 9 = 5Ax + 2A + B
$$
$$
20x + 9 = (5A)x + (2A + B)
$$

6. Match coefficients:
- Coefficient of \(x\): \(5A = 20\)
- Constant term: \(2A + B = 9\)

7. Solve for \(A\) and \(B\):
$$
5A = 20 \implies A = 4
$$
$$
2A + B = 9 \implies 2(4) + B = 9 \implies 8 + B = 9 \implies B = 1
$$

8. Write the partial fraction decomposition:
$$
\frac{20x + 9}{25x^2 + 20x + 4} = \frac{4}{5x + 2} + \frac{1}{(5x + 2)^2}
$$

Answer:
$$
\boxed{\frac{4}{5x + 2} + \frac{1}{(5x + 2)^2}}
$$

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#### Problem 4:
$$
\frac{3x + 5 - 6x^2}{x(x^2 - 1)}
$$

1. Factor the denominator:
$$
x(x^2 - 1) = x(x - 1)(x + 1)
$$

2. Set up the partial fraction decomposition:
$$
\frac{3x + 5 - 6x^2}{x(x - 1)(x + 1)} = \frac{A}{x} + \frac{B}{x - 1} + \frac{C}{x + 1}
$$

3. Combine the fractions on the right-hand side:
$$
\frac{A}{x} + \frac{B}{x - 1} + \frac{C}{x + 1} = \frac{A(x - 1)(x + 1) + Bx(x + 1) + Cx(x - 1)}{x(x - 1)(x + 1)}
$$

4. Equate the numerators:
$$
3x + 5 - 6x^2 = A(x - 1)(x + 1) + Bx(x + 1) + Cx(x - 1)
$$

5. Expand and simplify:
$$
A(x - 1)(x + 1) = A(x^2 - 1)
$$
$$
Bx(x + 1) = Bx^2 + Bx
$$
$$
Cx(x - 1) = Cx^2 - Cx
$$
$$
3x + 5 - 6x^2 = A(x^2 - 1) + Bx^2 + Bx + Cx^2 - Cx
$$
$$
3x + 5 - 6x^2 = (A + B + C)x^2 + (B - C)x - A
$$

6. Match coefficients:
- Coefficient of \(x^2\): \(A + B + C = -6\)
- Coefficient of \(x\): \(B - C = 3\)
- Constant term: \(-A = 5\)

7. Solve the system of equations:
- From \(-A = 5\), solve for \(A\): \(A = -5\)
- Substitute \(A = -5\) into \(A + B + C = -6\):
$$
-5 + B + C = -6 \implies B + C = -1
$$
- From \(B - C = 3\), solve for \(B\) and \(C\):
$$
\begin{cases}
B + C = -1 \\
B - C = 3
\end{cases}
$$
Add the equations:
$$
2B = 2 \implies B = 1
$$
Substitute \(B = 1\) into \(B + C = -1\):
$$
1 + C = -1 \implies C = -2
$$

8. Write the partial fraction decomposition:
$$
\frac{3x + 5 - 6x^2}{x(x^2 - 1)} = \frac{-5}{x} + \frac{1}{x - 1} + \frac{-2}{x + 1}
$$

Answer:
$$
\boxed{\frac{-5}{x} + \frac{1}{x - 1} + \frac{-2}{x + 1}}
$$

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Final Answers:

1. $\boxed{\frac{-7}{x + 3} + \frac{6}{(x + 3)^2}}$
2. $\boxed{\frac{-2}{x + 4} + \frac{5}{x + 5}}$
3. $\boxed{\frac{4}{5x + 2} + \frac{1}{(5x + 2)^2}}$
4. $\boxed{\frac{-5}{x} + \frac{1}{x - 1} + \frac{-2}{x + 1}}$

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(Note: The remaining problems can be solved similarly. If you need solutions for the rest, please let me know!)