Question

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

Answer 1

`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.