Question

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

Answer 1

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)`

Answer 2

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)`