How do you express #(3x+2) / (x^(2)+3x-4)# in partial fractions?

3 Answers
Apr 16, 2018

The answer is #=2/(x+4)+1/(x-1)#

Explanation:

Perform the decomposition into partial fractions

#(3x+2)/(x^2+3x-4)=(3x+2)/((x+4)(x-1))#

#=A/(x+4)+B/(x-1)#

#=(A(x-1)+B(x+4))/((x+4)(x-1))#

The denominators are the same, compare the numerators

#3x+2=A(x-1)+B(x+4)#

Let #x=-4#, #=>#, #-10=-5A#, #=>#, #A=2#

Let #x=1#, #=>#, #5=5B#, #=>#, #B=1#

Therefore,

#(3x+2)/(x^2+3x-4)=2/(x+4)+1/(x-1)#

Apr 16, 2018

#color(blue)(2/(x+4)+1/(x-1))#

Explanation:

#(3x+2)/(x^2+3x-4)#

Factor denominator:

#(3x+2)/((x+4)(x-1))#

For linear factors we expect the partial fraction form to be:

#(3x+2)/((x+4)(x-1))-= A/(x+4)+B/(x-1)#

Adding #RHS#

#(3x+2)/((x+4)(x-1))-= (A(x-1))/(x+4)+(B(x+4))/(x-1)#

Since this is an identity numerators are identical, so:

#3x+2-=A(x-1)+B(x+4)#

We now proceed to find the values of A and B:

Let #x=-4#

#3(-4)+2-=A((-4)-1)+B((-4)+4)#

#-10=-5A#

#A=2#

Let #x=1#

#3(1)+2-=A((1)-1)+B((1)+4)#

#5=5B#

#B=1#

The partial fractions are therefore:

#color(blue)(2/(x+4)+1/(x-1))#

Apr 16, 2018

# (3 x+2)/((x+4)(x-1))= 2/(x+4)+1/(x-1)#

Explanation:

# (3 x+2)/(x^2+3 x-4) = (3 x+2)/((x+4)(x-1))#

Let # (3 x+2)/((x+4)(x-1))= A/(x+4)+B/(x-1)# or

# (3 x+2)/cancel((x+4)(x-1))= (A(x-1)+B(x+4))/cancel((x+4)(x-1))#

or # (3 x+2)= A(x-1)+B(x+4)#

Let #x=1 :. 3*1+2= A(1-1)+B(1+4)# or

#5 B = 5 :. B=1#

Let #x=-4 :. 3*(-4)+2= A(-4-1)+B(-4+4)# or

#-5 A = -10 :. A=2 :. #

# (3 x+2)/((x+4)(x-1))= 2/(x+4)+1/(x-1)# [Ans]