Week 2, Day 1

Last week, we learned how to perform partial fraction decomposition, which involves writing a rational expression as the sum of two or more rational expressions. For instance, (x+7) / (x^2-x-6) can be written as the sum of two fractions with linear denominators. That is, (x+7) / (x^2-x-6) = (2 / x-3) + (-1 / x+2). Each fraction on the right side of the equation is a partial fraction, and together they make up the partial fraction decomposition of the left side. The following are steps to carry out partial fraction decomposition: 

1. Multiply by LCD

2. Distribute

3. Collect like terms

4. Factor out the variable

5. Equate the coefficients

6. Solve the system of equations

7. Write the answer as a partial fraction

When the denominator of a partial fraction is, for example, (x+1)^2, it is called a repeated factor.

When the denominator of a partial fraction is, for example, (x^2+1), it is called a quadratic factor. 

Here are a few examples:

Example 1:

5/(x^2+x-6)--> 5/(x+3)(x-2) = A/(x+3) + B/(x-2)

1. 5 = A(x-2) + B(x+3)

2. 5 = Ax - 2A + Bx + 3B

3. 5 = Ax + Bx - 2A + 3B 

4. 5 = (A + B)x - 2A + 3B 

5. 0 = A + B, 5 = -2A + 3B

6. 5 = 5B--> B = 1, A = -1

7. -1/(x+3) + 1/(x-2)

Example 2:

2x - 3/(x-1)^2 

In this case, a factor (x-1) repeats, and thus the steps to solve it are as follows: 

1. Multiply by LCD

2. Distribute

3. Collect like terms

4. Equate the coefficients

5. Solve the system of equations

6. Write the answer as a partial fraction

2x - 3/(x-1)^2 = A/(x-1) + B/(x-1)^2 Every exponent gets a factor!

1. 2x-3 = A(x-1) + B

2. 2x-3 = Ax - A + B

3. 2x-3 = Ax + (-A + B)

4. A = 2, -3 = -A + B

5. -3 + 2 = B = -1

6. 2/(x-1) - 1/(x-1)^2

Example 3:

4x^2 + 2x - 1 / x^2(x +1) = A/x + B/x^2 + C/x+1

1. 4x^2 + 2x - 1 = Ax(x+1) + B(x+1) + Cx^2

2. 4x^2 + 2x - 1 = Ax^2 + Ax + Bx + B + Cx^2

3. 4x^2 + 2x - 1 = Ax^2 + Cx^2 + Ax + Bx + B

4. 4x^2 + 2x - 1 = (A + C)x^2 + (A + B)x + B

5. A + C = 4, A + B = 2, B = -1

6. A = 3, B = -1, C = 1

7. 3/x -1/x^2 + 1/x+1

Example 4:

x^2 - 1 / x(x^2 + 1) 

In this case, the denominator contains an irreducible quadratic factor, so the set up is as follows:

x^2 - 1 / x(x^2 + 1) = A/x + Bx + C / x^2 + 1

1. x^2 - 1 = A(x^2 + 1) + (Bx + C)x

2. x^2 - 1 = Ax^2 + A + Bx^2 + Cx

3. x^2 - 1 = Ax^2 + Bx^2 + Cx + A

4. x^2 - 1 = (A + B)x^2 + Cx + A

5. A + B = 1, C = 0, A = -1

6. A = -1, B = 2, C = 0

7. -1/x + 2x / x^2 + 1