`(2x^3-x^2+x+5)/(x^2+3x+2)`

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

Dividing using long division yields,

`(2x^3-x^2+x+5)/(x^2+3x+2)=2x-7+(18x+19)/(x^2+3x+2)`

Since the polynomial don not completely, so we have to continue with the partial fractions of the remainder expression,

Let's factorize the denominator of the remainder expression,

`x^2+3x+2=x^2+x+2x+2`

`=x(x+1)+2(x+1)`

`=(x+1)(x+2)`

Let`(18x+19)/(x^2+3x+2)=A/(x+1)+B/(x+2)`

`=(A(x+2)+B(x+1))/((x+1)(x+2))`

`=(Ax+2A+Bx+B)/((x+1)(x+2))`

`=(x(A+B)+2A+B)/((x+1)(x+2))`

`:.(18x+19)=x(A+B)+2A+B`

Equating the coefficients of the like terms,

`A+B=18` --------- equation 1

`2A+B=19` -------- equation 2

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

Subtract equation 1 from equation 2

`2A-A=19-18`

`A=1`

plug the value of A in equation 1,

`1+B=18`

`B=18-1`

`B=17`

`(18x+19)/(x^2+3x+2)=1/(x+1)+17/(x+2)`

`:.(2x^3-x^2+x+5)/(x^2+3x+2)=2x-7+1/(x+1)+17/(x+2)`

You need to decompose the fraction in simple irreducible fractions, such that:

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

`(2x^3 - x^2 + x + 5)/(x^2 + 3x + 2) = A/(x+1) + B/(x+2) `

You need to bring to the same denominator all fractions, such that:

`2x^3 - x^2 + x + 5 = A(x+2) +B(x+1) `

You need to group the terms having the same power of x:

`2x^3 - x^2 + x + 5 = x(A+B) + 2A+B`

Comparing the expressions both sides yields:

`A+B = 1 => B = 1-A`

`2A+B=5 => 2A + 1 - A = 5 => A = 4 => B = -3`

**Hence, the partial fraction decomposition of the improper rational expression is `(2x^3-x^2+x+5)/((x+1)(x+2)) = 4/(x+1) - 3/(x+2)`**

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