Partial Fractions Adding fractions To add fractions we rewrite the fractions with a common denominator then add the numerators. Example Find the sum of 3 2 + 3x + 4 x − 5 The common denominator of b b g b gb g Rewrite each fraction with a denominator of 3x + 4 x − 5 b gb g 3 2 and is 3x + 4 x − 5 3x + 4 x −5 3 x −5 3 = 3x + 4 3x + 4 x − 5 gb g and b g 2 3x + 4 2 = x − 5 3x + 4 x − 5 b gb g b g b g b g b g b gb g b gb g b gb g Add the numerators, then expand all brackets and simplify 3 x −5 2 3x + 4 3 x − 5 + 2 3x + 4 + = 3x + 4 x − 5 3x + 4 x − 5 3x + 4 x − 5 = 9x − 7 3x − 11x − 20 2 3 2 9x − 7 + = 2 3x + 4 x − 5 3x − 11x − 20 Therefore The reverse of this process is to split a fraction into partial fractions. In the above example 9x − 7 3 2 = + 3x − 11x − 20 3x + 4 x − 5 2 Algebraic fraction Partial fractions If the degree of the numerator of the algebraic fraction is greater than that of the denominator, divide the denominator into the numerator then express the remaining fractional part as partial fractions. H:\Projects docs\maths downloads\Partial Fractions.doc, Created by Sue Thomas/LSU/APS/FELCS;Created on 22/07/2002 10:10 Page 1 of 8 The form of the numerator in the partial fractions depends only on the type of the factors in the denominator of the original fraction, as indicated below. Each distinct linear factor eg. (x− a) has a corresponding partial fraction of the form A ( x − a) Each repeated linear factor eg. (x− a)2 has a corresponding partial fractions of the form A B + A and B constants. ( x − a ) ( x − a )2 where A is a constant. . 2 Each quadratic factor eg. ax + bx + c has a corresponding partial fraction of the form. Ax + B ax + bx + c A and B constants. 2 Examples The following fractions have been written in partial fraction form (without evaluating the constants in the numerators). (1) (2) 3x + 1 A B = + x −1 x + 2 x −1 x+2 b gb g b g b g 3x + 1 linear factors only in the denominator A B C + + x +1 x+3 x+3 b x + 1gb x + 3g b g b g b g 2 = 2 repeated linear factor in the denominator gives rise to a partial fraction for both b x − 3g and b x − 3g (3) d 3x + 1 i x x2 − 2x + 3 = A Bx + C + 2 x x − 2x + 3 d i 2 quadratic factor in the denominator In general the numerator of a partial fraction is a polynomial of degree one less than the factor in the denominator. Note the special case of repeated factors, example (2) above. Exercise 1 Write the following in partial fraction form, but do not calculate the numerical values for the constants in the numerator. If possible, factorise the quadratic factor first. x+6 5 (a) (b) 2x − 3 x + 4 x x + 4 x 2 − 3x + 2 b (c) gb g 2x d x + 3ib x + 1g 2 b gd (d) dx i x2 − x + 3 2 ib g − 3x + 7 x + 2 2 H:\Projects docs\maths downloads\Partial Fractions.doc, Created by Sue Thomas/LSU/APS/FELCS;Created on 22/07/2002 10:10 Page 2 of 8 Linear factors Example The next task is to find values for the constants in the numerator. The first example will contain linear factors in the denominator only. Write x −5 as the sum of partial fractions. x + 2x − 3 2 factorise the denominator write the general expression for the partial fractions add the fractions on the right side the equation holds for all values of x (except x = 1 and x =− 3) so we can equate the numerators expand brackets and collect like terms equate coefficients of powers of x on both sides solve these simultaneous equations to give substitute for A and B the partial fractions are therefore x −5 x −5 = x −1 x + 3 x + 2x − 3 b gb g 2 x −5 A B = + x −1 x + 3 x −1 x+3 b gb g b g b g b g b g b g b g b gb g A x + 3 + B x −1 A B + = x −1 x+3 x −1 x + 3 b g b g b gb g b gb g x − 5 = Ab x + 3g + Bb x − 1g A x + 3 + B x −1 x −5 = x −1 x + 3 x −1 x + 3 b g b x − 5 = A + B x + 3A − B g A+ B =1 (coefficients of x) 3 A − B = −5 (constant terms) A = −1 and B=2 −1 2 and x −1 x+3 x −5 2 1 = − x + 2x − 3 x + 3 x − 1 2 H:\Projects docs\maths downloads\Partial Fractions.doc, Created by Sue Thomas/LSU/APS/FELCS;Created on 22/07/2002 10:10 Page 3 of 8 Solving the simultaneous equations that result from equating coefficients can sometimes be quite lengthy. An alternative method is to equate the numerators and, before expanding the brackets, substitute a value of x into both sides of the equation so that only one variable remains. Repeat this to find other variables. This method will not necessarily find all variables, but will often make calculations easier. Example Using the previous example, after equating numerators we had b g b g x − 5 = A x + 3 + B x −1 substituting x = 1 will eliminate B and substituting x = −3 will eliminate A. if x = 1 then if x = −3 then 1 − 5 = A(1 + 3) + B(1− 1) −3 −5 = A(−3 + 3) + B(−3 − 1) solving gives A=−1 solving gives B=2 Repeated linear factors In this example the denominator contains a repeated linear factor. Example Express 5x 2 + 3 x + 1 in terms of partial fractions. x 3 − 3x − 2 factorise the denominator write the general expression for the partial fractions add the fractions on the right side 5x 2 + 3 x + 1 5x 2 + 3 x + 1 = 2 x 3 − 3x − 2 x +1 x − 2 b gb g 5x 2 + 3 x + 1 A B C = + + 3 2 x−2 x +1 x − 3x − 2 x +1 b g b g b g b g b gb g b g b g b g b g b gb g 5x + 3x + 1 Ab x + 1g + Bb x + 1gb x − 2g + Cb x − 2g ∴ = x − 3x − 2 b x − 2gb x + 1g 2 A x +1 + B x +1 x − 2 + C x − 2 A B C + + = 2 2 x−2 x +1 x +1 x − 2 x +1 2 2 3 equate numerators 2 b g 2 b gb g b g 5x 2 + 3x + 1 = A x + 1 + B x + 1 x − 2 + C x − 2 H:\Projects docs\maths downloads\Partial Fractions.doc, Created by Sue Thomas/LSU/APS/FELCS;Created on 22/07/2002 10:10 Page 4 of 8 b g b gb g b g b g b gb g b g b g b gb g b g to find A substitute x = 2 5x 2 + 3x + 1 = A x + 1 2 + B x + 1 x − 2 + C x − 2 in the equation (the terms involving B and 2 2 C will equal zero and we 5 × 2 + 3 × 2 + 1 = A 2 + 1 + B 2 + 1 2 − 2 + C 2 − 2 2 can solve for A) = A 3 + B 3 0 +C 0 27 = 9 A A=3 b g b gb g b g b g b g b gb g b g 2 2 to find C substitute x = −1 5x + 3x + 1 = A x + 1 + B x + 1 x − 2 + C x − 2 ( the terms involving A and B will equal zero) 5 × −1 2 + 3 × −1 + 1 = A 0 2 + B 0 −3 + C −3 3 = −3C b g C = −1 b g b gb g b g b g b gb g b g to find B substitute the 2 2 values already found (A=1 5x + 3x + 1 = A x + 1 + B x + 1 x − 2 + C x − 2 and C=−1) and a value 2 for x (substituting x=0 0 + 0 + 1 = 3 1 + B 1 −2 − 1 −2 keeps the arithmetic 1 = −2 B + 5 simple) B=2 therefore A = 3, substitute for A, B, and C the partial fractions are the solution is 3 , ( x − 2) B = 2, and C = − 1 2 1 and 2 ( x + 1) ( x + 1) 5x 2 + 3 x + 1 3 2 1 = + − 3 2 x−2 x +1 x − 3x − 2 x +1 b g b g b g H:\Projects docs\maths downloads\Partial Fractions.doc, Created by Sue Thomas/LSU/APS/FELCS;Created on 22/07/2002 10:10 Page 5 of 8 Quadratic or higher factor Example The numerator for a quadratic factor has the form Ax+B. In general if the denominator is of degree n then the numerator of the partial fraction is a polynomial of degree n−1. Express 5x 2 − 6 x + 5 in terms of partial fractions. x 3 − 2 x 2 + 3x − 2 factorise the denominator 5x 2 − 6 x + 5 5x 2 − 6 x + 5 = x 3 − 2 x 2 + 3x − 2 x − 1 x2 − x + 2 b gd write the general expression for the partial fractions add the fractions on the right side i 5x 2 − 6 x + 5 A Bx + C = + 2 3 2 x −1 x − 2 x + 3x − 2 x − x+2 b g d i d i b b gd gb g i d i b b gd gb g i A x 2 − x + 2 + Bx + C x − 1 A Bx + C + 2 = x −1 x − x+2 x − 1 x2 − x + 2 b g d i A x 2 − x + 2 + Bx + C x − 1 5x 2 − 6 x + 5 = x 3 − 2 x 2 + 3x − 2 x − 1 x2 − x + 2 equate numerators expand the brackets on the right-hand side of the equation equate coefficients of powers of x d i b gb g 5x 2 − 6 x + 5 = A x 2 − x + 2 + Bx + C x − 1 5 x 2 − 6 x + 5 = Ax 2 − Ax + 2 A + Bx 2 − Bx + Cx − C = ( A + B ) x2 + ( C − A − B ) x + ( 2 A − C ) A+ B = 5 C − A − B = −6 2A − C = 5 solve the simultaneous equations to give A = 2, B = 3 , and C =−1 substitute for A, B and C the solution is 2 3x − 1 5x2 − 6 x + 5 = + 2 3 2 x − 2 x + 3 x − 2 ( x − 1) x − x + 2 H:\Projects docs\maths downloads\Partial Fractions.doc, Created by Sue Thomas/LSU/APS/FELCS;Created on 22/07/2002 10:10 Page 6 of 8 Exercise 2 Express the following as partial fractions. (a) (c) (e) 2x x−3 x+2 b gb g x2 + 2x + 4 b x + 1gd x 2 (b) i + 3x + 3 (d) x+2 x − 5x + 6 2 3x b1 − xg 2 x2 − 2x + 2 x3 + x2 + x H:\Projects docs\maths downloads\Partial Fractions.doc, Created by Sue Thomas/LSU/APS/FELCS;Created on 22/07/2002 10:10 Page 7 of 8 Answers Exercise 1 A B (a) + 2x − 3 x + 4 (b) A B C D + + + x x + 4 x −1 x − 2 A Bx + C + 2 x +1 x + 3 (d) Exercise 2 6 4 (a) + 5 ( x − 3) 5 ( x + 2 ) (b) 5 4 − x−3 x−2 (e) 2 x+4 − 2 x x + x +1 (c) (d) 3 (1 − x ) 2 − 3 (1 − x ) A B Cx + D + + 2 2 ( x + 2 ) ( x + 2 ) x − 3x + 7 (c) 3 2x + 5 − 2 ( x + 1) x + 3x + 3 H:\Projects docs\maths downloads\Partial Fractions.doc, Created by Sue Thomas/LSU/APS/FELCS;Created on 22/07/2002 10:10 Page 8 of 8