One of our topics, as Axie explained in her presentation, was partial fractions.
In this lesson, 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 LC
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: Distinct Linear Factors
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:Repeated Linear Factors
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: Distinct Linear Factors
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: Distinct Linear and Quadratic Factors
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
When the exponent of the numerator is greater than or equal to the exponent of the denominator, use long division. Once you have found the quotient, take the remainder and put it over the original denominator, using it as your new partial fraction.
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