How do you express #3/((1 + x)(1 - 2x)) # in partial fractions? Precalculus Matrix Row Operations Partial Fraction Decomposition (Linear Denominators) 1 Answer Bdub Mar 21, 2016 #3/((1+x)(1-2x)) = 1/(1+x) +2/(1-2x)# Explanation: #3/((1+x)(1-2x)) = A/(1+x) +B/(1-2x)# #3=A(1-2x)+B((1+x)# #3=A-2Ax+B+Bx# #A+B=3, -2A+B=0# #B=3-A->-2A+3-A=0->-3A+3=0# #-3A=-3,A=1,B=2# #3/((1+x)(1-2x)) = 1/(1+x) +2/(1-2x)# Answer link Related questions What does partial-fraction decomposition mean? What is the partial-fraction decomposition of #(5x+7)/(x^2+4x-5)#? What is the partial-fraction decomposition of #(x+11)/((x+3)(x-5))#? What is the partial-fraction decomposition of #(x^2+2x+7)/(x(x-1)^2)#? How do you write #2/(x^3-x^2) # as a partial fraction decomposition? How do you write #x^4/(x-1)^3# as a partial fraction decomposition? How do you write #(3x)/((x + 2)(x - 1))# as a partial fraction decomposition? How do you write the partial fraction decomposition of the rational expression #x^2/ (x^2+x+4)#? How do you write the partial fraction decomposition of the rational expression # (3x^2 + 12x -... How do you write the partial fraction decomposition of the rational expression # 1/((x+6)(x^2+3))#? See all questions in Partial Fraction Decomposition (Linear Denominators) Impact of this question 1175 views around the world You can reuse this answer Creative Commons License