Thursday, March 7, 2013

`(x^3 + 2x^2 - x + 1)/(x^2 + 3x - 4)` Write the partial fraction decomposition of the improper rational expression.

`(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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