# How do I find the partial-fraction decomposition of (-3x^3 + 8x^2 - 4x + 5)/(-x^4 + 3x^3 - 3x^2 + 3x - 2)?

Sep 14, 2014

Break the polynomial into two groups where each group is factorable on its own.

(-3x^3+8x^2-4x+5)/((-x^4-3x^2-2)+(3x^3+3)

Factor each group.

(-3x^3+8x^2-4x+5)/(1(x^2+2)(x^2+1)+3x(x^2+1)

Factor out the corresponding factor of ${x}^{2} + 1$ from each group.

$\frac{- 3 {x}^{3} + 8 {x}^{2} - 4 x + 5}{\left(- {x}^{2} - 2\right) \left({x}^{2} - 1\right) + \left(3 x\right) \left({x}^{2} + 1\right)}$

Combine the individual factors into a single factored expression.

(-3x^3+8x^2- 4x+5)/((x^2+2)(-x^2+3x-2))

In this problem $- 1 \cdot - 2 = 2$ and $- 1 - 2 = - 3$, so insert -1 as the right hand term of one factor and -2 as the right-hand term of the other factor.

(-3x^3+8x^2-4x+54)/(-(x^2+1)(x-1)(x+2)

Multiply the numerator by -1.

$\frac{3 {x}^{3} - 8 {x}^{2} + 4 x - 5}{\left({x}^{2} + 1\right) \left(x - 1\right) \left(x - 2\right)}$