1. Partial Fractions Lesson 7.3, page 737 Objective: To decompose rational expressions into partial fractions.

2. Review: Rational Expressions rational function – a quotient of two polynomials where p(x) and q(x) are polynomials and where q(x) is not the zero polynomial.

3. Review AddingSubtracting Rationals

4. Definition In the problem above, the fractions on the right are called “partial fractions”. In calculus, it is often helpful to write a rational expression as the sum of two or more simpler rational expressions.

5. What is decomposition of partial fractions? Writing a more complex fraction as the sum or difference of simpler fractions. Examples: Why would you ever want to do this? It’s EXTREMELY helpful in calculus!

6. Example 1 Decompose into partial fractions. Begin by factoring the denominator: (x + 2)(2x  3). We know that there are constants A and B such that To determine A and B, we add the expressions on the right…giving us…

7. Equate the numerators: 4x  13 = A(2x  3) + B(x + 2) Since the above equation containing A and B is true for all x, we can substitute any value of x and still have a true equation. If we choose x = 3/2, then 2x  3 = 0 and A will be eliminated when we make the substitution. This gives us 4(3/2)  13 = A[2(3/2)  3] + B(3/2 + 2)  7 = 0 + (7/2)B. B =  2.

8. If we choose x =  2, then x + 2 = 0 and B will be eliminated when we make the substitution. So, 4(  2)  13 = A[2(  2)  3] + B(  2 + 2)  21 =  7A. A = 3. The decomposition is as follows: Example 1 continued

9. Check Point 1, page 740 Find the partial fraction decomposition.

10. What if one on the denominators is a linear term squared? This is accounted for by having the nonsquared term as one denominator and having the squared term as another denominator. What if one denominator is a linear term cubed? There would be 3 denominators in the decomposition:

11. Example 2: Decompose into partial fractions.

12. Example 2 continued Next, we add the expression on the right: Then, we equate the numerators. This gives us Since the equation containing A, B, and C is true for all of x, we can substitute any value of x and still have a true equation. In order to have 2x – 1 = 0, we let x = ½. This gives us Solving, we obtain A = 5.

13. Example 2 continued In order to have x  2 = 0, we let x = 2. Substituting gives us To find B, we choose any value for x except ½ or 2 and replace A with 5 and C with  2. We let x = 1:

14. Example 2 continued The decomposition is as follows:

15. Check Point 2, page 741 Find the partial fraction decomposition.

16. Check Point 3, page 743 Find the partial fraction decomposition.