How do you express # (3x+5)/((3x+2)(2x+1) )# in partial fractions?

2 Answers
maganbhai P.
Jul 22, 2018

Here ,
#7(3x+2)-9(2x+1)=21x+14-18x-9=3x+5#

#(3x+5)/((3x+2)(2x+1))=(7(3x+2)-9(2x+1))/((3x+2)(2x+1))#

#(3x+5)/((3x+2)(2x+1))=7/(2x+1)-9/(3x+2)#

Explanation:

Let ,

#color(red)((3x+5)/((3x+2)(2x+1))=A/(3x+2)+B/(2x+1)to(X)#

#:.3x+5=A(2x+1)+B(3x+2)#

#=>3x+5=2xA+A+3xB+2B#

#=>3x+5=2xA+3xB+A+2B#

#=>3x+5=x(2A+3B)+(A+2B)#

Comparing coefficient of #x and # constant term :

#color(blue)(2A+3B=3 to(1) and A+2B=5 to (2)#

From #(2),# we get #color(blue)( A=5-2B to(3)#

Subst. #A=5-2B# into #(1)#

#2(5-2B)+3B=3#

#:.10-4B+3B=3#

#:.-4B+3B=3-10#

#:.-B=-7#

#=>color(violet)(B=7#

Subst. #B=7# into #(3)#

#A=5-2(7)=5-14#

#:.color(violet)(A=-9#

Subst. #A=-9 and B=7# into #(X)#

#color(red)((3x+5)/((3x+2)(2x+1))=(-9)/(3x+2)+7/(2x+1)#

OR

#(3x+5)/((3x+2)(2x+1))=7/(2x+1)-9/(3x+2)#

Narad T.
Jul 22, 2018

The answer is #=-9/(3x+2)+7/(2x+1)#

Explanation:

Perform the decomposition into partial fractions

#(3x+5)/((3x+2)(2x+1))=A/(3x+2)+B/(2x+1)#

#=(A(2x+1)+B(3x+2))/((3x+2)(2x+1))#

Compare the numerators

#3x+5=A(2x+1)+B(3x+2)#

Let #x=-2/3#

Then,

#3=-1/3A#

#=>#, #A=-9#

Let #x=-1/2#

Then,

#7/2=1/2B#

#=>#, #B=7#

Finally,

#(3x+5)/((3x+2)(2x+1))=-9/(3x+2)+7/(2x+1)#

Related questions
See all questions in Partial Fraction Decomposition (Linear Denominators)
Impact of this question
3768 views around the world
You can reuse this answer
Creative Commons License