`(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...
`(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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