SECTION 11.5 Partial Fraction Decomposition 815 NOTE Again, instead of using subscripts, such as A A , , 1 2 and A ,3 it is common to use A, B, and C to represent the unknown numbers. j Decomposing P Q where Q Has Repeated Linear Factors Find the partial fraction decomposition of + − + x x x x 2 2 . 3 2 Solution EXAMPLE 3 This equation is an identity in x. Equate the coefficients of like powers of x to get = + =− − ⎧ ⎨ ⎪⎪ ⎩⎪⎪ A B A B 1 0 3 2 Equate the coefficients of x: ( ) = + x A B x 1 Equate the constants: =− −A B 0 3 2 This system of two equations containing two unknowns A and B, can be solved using whatever method you wish. The solution is = − = A B 2 3 From equation (2), the partial fraction decomposition is − + = − − + − x x x x x 5 6 2 2 3 3 2 Check: The decomposition can be checked by adding the rational expressions. ( ) ( ) ( )( ) ( )( ) − − + − = − − + − − − = − − = − + x x x x x x x x x x x x 2 2 3 3 2 3 3 2 2 3 2 3 5 6 2 First, factor the denominator, ( ) ( ) − + = − + = − x x x x x x x x 2 2 1 1 3 2 2 2 and notice that the denominator has the nonrepeated linear factor x and the repeated linear factor ( ) −x 1 . 2 By Case 1, the term A x is in the decomposition; and by Case 2, the terms ( ) − + − B x C x 1 1 2 are in the decomposition. The numbers to be found in the partial fraction decomposition can sometimes be found more easily by using suitable choices for x in the identity obtained after fractions have been cleared. In Example 2, the identity after clearing fractions is equation (3): ( ) ( ) = − + − x A x B x 3 2 Let = x 2 in this identity and the term containing B drops out, leaving ( ) = − A 2 1 , or = − A 2. Similarly, let = x 3, and the term containing A drops out, leaving = B 3 . As before, = − A 2 and = B 3. Now Work PROBLEM 17 2 Decompose P Q where Q Has Repeated Linear Factors Case 2: Q has repeated linear factors. If the polynomial Q has a repeated linear factor, say ( ) − ≥ x a n, 2 n an integer, then, in the partial fraction decomposition of P Q , allow for the terms ( ) ( ) − + − + + − A x a A x a A x a n n 1 2 2 where the numbers A A A , , . . . , n 1 2 are to be determined. (continued)
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