Question

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

Answer 1

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)`

Answer 2

`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))`

Answer 3

`(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]