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Supported in part by funding from a VCCS LearningWare Grant Practice Problems for Partial Fractions by Richard Gill Supported in part by funding from a VCCS LearningWare Grant

Answers and a couple of complete solutions appear on the following slides. Find each partial fraction decomposition.

Answers for the practice problems 1 - 4.

Answers for practice problems 5 – 8.

Complete solution for #3. Begin by factoring the denominator. Multiply both sides by the LCD.

Comparing coefficients…. -2 = A + B 8 = -6A + C 6 = 9A – 9B + 3C -24 = 18A – 3C -18 = 27A – 9B -18 = 9A + 9B -36 = 36A A = -1 (-3) (9) -2 = A + B -2 = -1 + B -1 = B 8 = -6A + C 8 = -6(-1) + C 8 = 6 + C 2 = C Finally….

Complete solution for #7 Complete solution for #7. Since the order of the numerator is larger than the order of the denominator, we start with long division. Now we do partial fraction decomposition on the remainder.

Usually at this point we set up two equations in two unknowns by comparing coefficients. But in this problem we will use an alternative tactic called “strategic substitution”. Sometimes it works nicely, sometimes it needs systems of equations to close out the problem. The basic idea is that if the left side of the equation is equal to the right for all values of x, you can choose values of x that are strategic. In this case those values would be x = 3 and x = -3. From the previous slide… -x –9 = A(x + 3) + B(x – 3) so if we choose x = 3, we eliminate B. If x = 3 -3 –9 = A(3 + 3) + B(3 – 3) -12 = 6A A = -2 -x –9 = A(x + 3) + B(x – 3) so if we choose x = -3, we eliminate A. If x = -3 then –(-3) – 9 = A(-3 + 3) + B(-3 –3) 3–9 = -6B B=1 In conclusion then…..

Complete solution for #8. Factor, then multiply both sides by LCD.

From the previous slide we can compare coefficients….. A = 3 and B = -1 9A + C = 19 9(3) + C = 19 27 + C = 19 C = -8 9B + D = -9 9(-1) + D = -9 D = 0 Finally…..