Decomposing Rational Expressions Into Partial Fractions

Transcription

1 Decomposing Rational Expressions Into Partial Fractions Say we are ked to add x to 4. The first step would be to write the two fractions in equivalent forms with the same denominators. Thus we write: x 4 () (x ) () 4(x ) (x ) () Now we can add the two fractions because they have the same denominator. expression x x 8 (x ) () = 7x + 1 (x ) () The result is the rational Decomposing a rational expression into partial fractions is doing the exact opposite of what we did above. Thus given the rational expression (x 1) (x 5) then splitting it up unto partial fractions means writing it a sum of simpler fractions. To proceed, we would argue that the fraction is most probably the sum of a fraction with denominator (x 1) and a fraction with denominator (x 4). Therefore, (x 1) (x 5) = A (x 1) + B (x 5) (1) where A and B are numbers which have to be determined. To clear the fractions, multiply each term in equation (1) by (x 1) (x 5). The result is = A (x 5) + B (x 1) () If we substitute x = 5 in equation (), A drops out and we eily solve for B. In other words, 8 (5) = A (0) + B (5 1) = 4B Solve to get B = 10. If we substitute x = 1 in the same equation, B drops out and we eily solve for A: Now solve to get A =. It follows that 8 (1) = A (1 5) + B (0) = 4A (x 1) (x 5) = (x 1) + 10 (x 5) The two fractions (x 1) and 10 are called the partial fractions of the given rational expression (x 5) (x 1) (x 5). The numbers x = 5 and x = 1 which we substituted into () to determine A and B are the zeros of the denominator (x 1) (x 5). If the denominator of the given rational expression h more than two distinct factors, we would expect each one to contribute a partial fraction. For example, to split x + 5 x (x 1) () 1

2 into partial fractions, we would argue that the rational expression is most probably a sum of three fractions, one with denominator x, another one with denominator (x 1) and a third one with denominator (x 5). Thus x + 5 x (x 1) () = A x + B (x 1) + C () where A, B and C are constants to be determined. We would then multiply every term in the equation by x (x 1) () to clear all the fractions. We get x + 5 = A (x 1) () + Bx () + Cx (x 1) Substituting x = 0 gives 5 = A( ) therefore A = 5. Substituting x = 1 gives 8 = B (4) therefore B =. Finally, substituting x = gives therefore C = 1. It follows that 4 = C( ) ( 4) x + 5 x (x 1) () = 5 x + (x 1) + 1 () = 5 x + (x 1) 1 () In general, given a rational expression with a denominator that is a product of distinct linear factors and a numerator that h a degree that is lower than the degree of the denominator, we sume that it is a sum of fractions contributed by the different linear factors in the denominator and proceed to find them described in the above examples. Example 1 To split x + x(x + 1) (x ) into partial fractions We expect the rational expression to be a sum of three fractions with denominators x, (x + 1) and (x ) respectively. Thus x + x(x + 1) (x ) = A x + B (x + 1) + C x To clear fractions, we multiply each by x(x + 1) (x ). The result is x + = A(x + 1) (x ) + Bx (x ) + Cx (x + 1) substituting x = 0 gives = A( ), therefore A = 1. Substituting x = 1 gives ( + = B 1 ) ( ) which simplifies to 1 = 7B 4 Therefore B = 7. Substituting x = gives ) ( ) ( ( + = C ( ) ) + 1 which simplifies to 4 = 14C 9 Therefore C = 18 7 and x + x(x + 1) (x ) = 1 18 x + 7 (x + 1) + 7 (x ) = 1 x + 7 (x + 1) (x ) Exercise Split each rational expression into partial fractions x (x + ) (x + 1) 4x (x + 6) (x ) x 5 (x + 4) (x 5) d) x () (x 1)

3 A Rational Expression With Repeated Factors The following are examples of rational expressions with repeated factors in their denominators: x (x ) 5x 4 (x + 4) (x 1) x + x (x 1) (x 5) 4 d) x (x 5) (x + 1) In, x is the repeated factor, since it may be viewed xx. In, (x + 4) is the repeated factor because it may be viewed () () (). In (x 1) and (x 5) 4 are repeated factors. Both factors in the denominator of d) are repeated factors. A sum of fractions involving repeated factors yields a clue to splitting a rational expression with repeated factors For an example, consider adding the three fractions (x 1), x (x 1) and 6. The first step is to look for a common denominator. The simplest one is (x 1) (). Now we write (x 1) (x 1) () (x 1) () x (x 1) x () (x 1) () 6 Therefore the sum of the three fractions is 6 (x 1) (x 1) () (x 1) + x (x 1) 6 = (x 1) () + x () 6 (x 1) = x + 5x 1 (x 1) () (x 1) () Note that the denominator of x + 5x 1 (x 1) contains ONLY THE HIGHEST POWER of (x 1). In () other words, adding fractions with denominators that include (x 1) and (x 1) yields a fraction, in its lowest form, with a denominator that includes (x 1), NOT (x 1) (x 1). This suggests that, given a 5x 1 rational expression like (x 1) to split into partial fractions, we should expect the factor (x 1) (x + 4) to contribute two fractions, one with denominator (x 1) and another one with denominator (x 1). In general, if a rational expression h a repeated factor (x in the denominator, then that factor is expected to contribute two partial fractions, one with denominator (x and another one with denominator (x. Example To split x + 1 (x + 1) (x 4) into partial fractions x + 1 Solution: We expect (x + 1) to be a sum of three fractions, one with denominator (x + 1), (x 4) another one with denominator (x + 1) and a third one with denominator (x 4). Thus x + 1 (x + 1) (x 4) = A (x + 1) + B (x + 1) + C (x 4) To clear fractions, we multiply every term by (x + 1) (x 4). The result is x + 1 = A (x + 1) (x 4) + B (x 4) + C (x + 1) ()

4 Substituting x = 1 gives +1 = B( 5), therefore B =. Substituting x = 4 gives 5 = C(5), therefore C = 1. We have run out of zeros of the denominator (x + 1) (x 4) and we still have to determine A. One way to do this is to expand the right hand side of () and compare terms. Since B = and c = 1, the result is This simplifies to x + 1 = Ax Ax 4A (x 4) + 1(x + x + 1). x + 1 = (A + 1)x + Ax + (9 4A) (4) Now we may compare the different terms in (4). For example, there is no x term in the left hand side, (which is x + 1). Therefore there must be no x term in the right hand side, (which is (A + 1)x + Ax + (9 4A)). In other words, A + 1 = 0 This implies that A = 1. Of course we may compare the "x terms" instead. side and it is Ax in the right hand side. This implies that It is x in the left hand = A, which gives the same value A = 1. If you choose to compare the constant terms, you would get the equation 1 = 9 4A and solving also gives A = 1. Therefore you are free to choose what terms to compare. You get the same result A = 1. We conclude that x + 1 (x + 1) (x 4) = 1 (x + 1) (x + 1) + 1 (x 4). A Rational Expression With Quadratic Prime Factors An example of a quadratic prime factor is ( x + 1 ). It is a quadratic expression ( and you cannot factor it into linear factors, therefore it is a prime quadratic factor. More examples: x + 5 ) and x +. Examples of rational expressions that have prime quadratic factors in the denominator are x + 1 (x ) (x + 14), 1 x (x + 1), x 7 (x 1) (x + x + 4) It turns out that a quadratic prime factor contributes a fraction with a "linear numerator". For example, ( x + 5 ) contributes a fraction of the form Example 4 To split 7x 1 () (x + ) Ax + B x + 5 Solution: The given rational expression is a sum of a fraction of the form the form Bx + C x +. In other words 7x 1 () (x + ) = A + Bx + C x + A and another one of To clear fractions, we multiply every term in the above equation by () ( x + ). The result is 7x 1 = A ( x + ) + (Bx + C) () (5) 4

5 There is only one zero of the denominator, (because we cannot factor x + ), and it is x =. Substituting it into (5) gives 1 1 = A(9 + ) which simplifies to = 11A Therefore A =. Now we expand (5) but with the knowledge that A = This simplifies to 7x 1 = ( x + ) + Bx + Bx + CC 7x 1 = (B ) x + (B + C) C 4 (6) We now examine like terms. The constant term in the right hand side of (6) is C 4 and it is 1 in the left hand side. The conclusion is that C 4 = 1 which we eily solve for C to get C = 1. The x term in the right hand side is (B ) x and there is none in the left hand side. It follows that (B ) = 0, which implies that B =. Of course one may also obtain B by comparing the x terms. It is 7x in the left hand side and (B + 1) x in the right hand side, (we have already determined that C = 1). Solving (B + 1) = 7 also gives B =. Therefore 7x 1 () (x + ) = + x + 1 x + When The Degree Of The Numerator Is Not Less Than The Degree Of The Denominator Examples of such rational expressions are x (x + 1) (x ) x + 4 (x + 1) (x 1) x x (x + 4) (x ) In, the degree of the numerator, (which is ), is higher than the degree of the denominator, (which is ). In and, the degree of the numerator is equal to the degree of the denominator. Given a rational expression N(x) G(x) with a numerator N(x) whose degree is NOT LESS than the degree of the denominator G(x), divide N(x) by G(x), (the usual long division), to get a quotient Q(x) and a remainder R(x) whose degree is smaller than the degree of G(x). After that, decompose R (x) G (x) Example 5 To decompose x + x () into partial fractions: Solution: Since the degree of the numerator is not lower than the degree of the denominator, we first divide x + by x () = x + x. The result is We now split 9x + x () into partial fractions: x + x () = x + 9x + x () 9x + x () = A x + B 5

6 Clearing fractions gives 9x + = A () + Bx Substituting x = 0 gives = A, therefore A =. Substituting x = gives 9 + = B, therefore B = 7. Combining the different expressions gives x 7 + x () = x + We may write this x + x () = x + x + 7 () Exercise 6 1. In each of the following rational expressions, the denominator h distinct linear factors. Split each expression x + 11 (x + 1) () x 5 (x + 4) (x 1) 1 x (x 1) (x ) e) x + 5 () (x + 4) f) x + 1 (x 4) (x 1) g) 1 x (x ) (x + 1) h) 9x + 1 (x + 5) (x ) i) x + 1 (x ) (x ) j) 4x 1 (x + ) (x 1) k) x + 14 (x 4) (x + ) l) 6 x (x + ) (x 4) m) x + 14 (x 4) (x + ) n) x (x ) (x ) o) 9x + 5 (x + 5) (x ) p) 5x + 6 x (x + ) (x ) q) 9x + 5 (x + 5) (x ). In each of the following rational expressions, the denominator h a repeated factor. Split each expression 7x + 6 x (x + ) x + x (x + 1) 8 5x (x + ) (x 1) d) 10x x (x ) e) x (x 1) (x ) f) x + 1 x (x + 1) g) x 1 (x 1) (x ) h) 1x (x ) () i) x + 1 (x + 1) (x 4) j) x + 5 (x + 1) (x ) k) x 8 x (x ) l) 4x (x 1) (x + 1). Split x + (x 1) (x + 1) 4. Split x x + x (x ) 5. Split 4 x (x + 4) 6