`(x^3+2x^2-x+1)/(x^2+3x-4)`

Since the rational expression is an improper expression , we have to express the expression as a sum of simpler fractions with the degree of the polynomial in the numerator less than the degree of the polynomial in the denominator.

Dividing using the long division yields,

`(x^3+2x^2-x+1)/(x^2+3x-4)=x-1+(6x-3)/(x^2+3x-4)`

Polynomials do not completely divide , so we have to continue with the partial fractions of the remainder expression,

Let's factorize the denominator of the remainder fraction,

`x^2+3x-4=x^2+4x-x-4`

`=x(x+4)-1(x+4)`

`=(x-1)(x+4)`

Let `(6x-3)/(x^2+3x-4)=A/(x-1)+B/(x+4)`

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

`=(Ax+4A+Bx-B)/((x-1)(x+4))`

`:.(6x-3)=Ax+4A+Bx-B`

`6x-3=x(A+B)+4A-B`

equating the coefficients of the like terms,

`A+B=6` ----- equation 1

`4A-B=-3` ------ equation 2

Now we have to solve the above equations to get the solutions of A and B,

Adding the equation 1 and 2 yields,

`A+4A=6+(-3)`

`5A=3`

`A=3/5`

Plug the value of A in equation 1 ,

`3/5+B=6`

`B=6-3/5`

`B=27/5`

`(6x-3)/(x^2+3x-4)=3/(5(x-1))+27/(5(x+4))`

`:.(x^3+2x^2-x+1)/(x^2+3x-4)=x-1+3/(5(x-1))+27/(5(x+4))`

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