How do you express #1/((x+6)(x^2+3))# in partial fractions?

1 Answer

#1/((x+6)(x^2+3))=(1/39)/(x+6)+(-1/39x+2/13)/(x^2+3)#

Explanation:

We start setting up the equation with the unknown constants A, B, C

#1/((x+6)(x^2+3))=A/(x+6)+(Bx+C)/(x^2+3)#

After setting this right like this, convert it to one fraction
using the LCD#=(x+6)(x^2+3)#

so that

#1/((x+6)(x^2+3))=(A(x^2+3)+(Bx+C)(x+6))/((x+6)(x^2+3))#

simplify

#1/((x+6)(x^2+3))=(Ax^2+3A+(Bx^2+Cx+6Bx+6C))/((x+6)(x^2+3))#

#1/((x+6)(x^2+3))=(Ax^2+3A+Bx^2+Cx+6Bx+6C)/((x+6)(x^2+3))#

Rearrange according to degree from highest to lowest

#1/((x+6)(x^2+3))=(Ax^2+Bx^2+6Bx+Cx+3A+6C)/((x+6)(x^2+3))#

Determine the appropriate numerical coefficients

#(0*x^2+0*x+1*x^0)/((x+6)(x^2+3))#

#=((A+B)x^2+(6B+C)x+(3A+6C)x^0)/((x+6)(x^2+3))#

We can now obtain the equations to solve for A, B, C

#A+B=0" " #first equation
#6B+C=0" "#second equation
#3A+6C=1" "#third equation

Simultaneous solution of these three equations result to

#A=1/39# and #B=-1/39# and #C=2/13#

so that our final answer is

#1/((x+6)(x^2+3))=(1/39)/(x+6)+(-1/39x+2/13)/(x^2+3)#

God bless ....I hope the explanation is useful.