# `3/(x^3 + x - 2)` Write the partial fraction decomposition of the rational expression. Check your result algebraically.

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

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

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

`3/((x-1)(x^2+ x + 2)) = A/(x-1) + (Bx+C)/(x^2+ x + 2)`

You need to bring the fractions to a common denominator:

`3 = Ax^2 + Ax + 2A + Bx^2 - Bx + Cx - C`

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

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

Comparing both sides yields:

`A+B =0 => A = -B`

A - B + C = 0 => 2A + C = 0

`2A - C = 3`

Adding the relations yields:

`4A =3 => A = 3/4 => B = -3/4 => C = -6/4`

**Hence, the partial fraction decomposition is `3/((x-1)(x^2+ x + 2)) = 3/(4x-4) + (-3x-6)/(4x^2+ 4x + 8).` **

Factor

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

Write PFD

`3/((x-1)(x^2+x+2)) = A/(x-1)+(Bx+C)/(x^2+x+2)`

Write as single fraction

`3/((x-1)(x^2+x+2)) = ((x-1)(Bx+C) + (x^2+x+2)A)/((x-1)(x^2+x+2))`

Denominators are equal, so we require equality of numerators

`3=(x−1)(Bx+C)+(x^2+x+2)A`

Expand right-hand side

`3=x^2A+x^2B+xA−xB+xC+2A−C`

Collect up like terms

`3=x^2(A+B)+x(A−B+C)+2A−C`

Coefficients near like terms should be equal, so following system is obtained

`A+B=0`

`A-B+C=0`

`2A-C=3`

Solve for it, getting `A= 3/4. B= -3/4, C= -3/2`

Therefore,

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