(4-2x)/(2x+1)(x+1)(x+3) = A/(2x+1)+B/(x+1)+C(x+3) Find the values of the constants A, B and C

First, multiply throughout by the denominator of the main function to give as follows: 4-2x = A(x+1)(x+3) + B(2x+1)(x+3) + C(2x+1)(x+1) Then, choose values of x which will cause two of the constants to vanish. If x = -3, then the bracket (x+3) will equal 0, eliminating A and B, giving 4-2(-3) = C(2(-3)+1)(-3+1) Solving this gives as follows: (-5)(-2)C = 10 C = 10/10 = 1 If x = -1, then A and C will vanish, giving B(-1)(2) = 6 => B = -3 If x = -1/2, then B and C will vanish, giving A(1/2)(5/2) = 5 => A = 4 This means that the fraction (4-2x)/(2x+1)(x+1)(x+3) = 4/(2x+1) -3/(x+1) + 1/(x+3)