1. Partial Fractions (8.10) How to decompose a rational expression into a rich compost of algebra and fractions

2. Start with a POD The sum of this expression is a. 1 b. 0 c. 1/x d. 2/(x-1) e. 2/(x 2 -1) Everyone to the board!

3. Start with a POD So, answer e. Today we’re learning that process in reverse and obtaining the partial fraction decomposition.

4. Review A rational expression is a polynomial divided by another polynomial where f(x) and g(x) are polynomials.

5. New idea Where In other words, rational expressions can be broken down (decomposed) into fractions of constants over linear expressions or linear expressions over quadratic expressions. (Remember that all polynomials can be factored into linear or quadratic factors with real coefficients.)

6. New idea Where Critical condition: f(x) must be a lower degree than g(x). We will use division to get there.

7. The Method 1. Initial test: is f(x) a lower degree than g(x)? If not, use long division and decompose the remainder over the divisor. 2. Factor g(x)-- the denominator. 3. Split according to linear or quadratic factors. 4. Solve for the numerators-- A, B, C and so forth– by multiplying both sides by g(x).

8. Use it Step 1. Is the numerator a lower degree than the denominator? In this case, yes.

9. Use it Step 2. Factor the denominator.

10. Use it Step 3. Break into parts using the factors. Use A, B, and C as the numerators.

11. Use it Step 4. Solve for A, B, and C by multiplying each side by x 3 + 2x 2 - 3x. Now what?

12. Use it Substitute values of x that make each factor equal 0 in order to find A, B, and C. If x = 0, then A = 3. If x = 1, then C = 2. If x = -3, then B = -1.

13. Use it If x = 0, then A = 3. If x = 1, then C = 2. If x = -3, then B = -1. And the final breakdown is Test it.

14. Use it a second time This time you try the steps. Notice how you have to do long division first.

15. Use it a second time Work with that remainder. Notice the linear over quadratic fraction.

16. Use it a second time Solve for A, B, and C. If x = 1/2, then C = -5. If x = 0, then B = 1 (we can use the value of C we just got). Why did I choose x=0? Then choose any other value for x to find A. I like x = 1. If x = 1, and we use the values of B and C, then A = 3.

17. Use it a second time Solve for A, B, and C. A = 3, B = 1, C = -5 The final breakdown:

18. Repeated linear factors With a repeated linear factor, we have a special case. Notice how we added a factor with the linear denominator to a second power.

19. Repeated linear factors In the case of a repeated linear factor, we have a special case. Here’s an example to illustrate. A = -4, B = 5, C = 1

20. Repeated quadratic factors We have a similar situation with repeated quadratic factors. A = 5, B = -3, C = 2, D = 0