# How do you write the partial fraction decomposition of the rational expression  (x^3 - 6x^2 + 11x - 6) / (4x^3 - 28x^2 + 56x - 32) ?

Mar 21, 2016

$\frac{{x}^{3} - 6 {x}^{2} + 11 x - 6}{4 {x}^{3} - 28 {x}^{2} + 56 x - 32} = \frac{1}{4} + \frac{1}{4 \left(x - 4\right)}$

with exclusions $x \ne 1$ and $x \ne 2$

#### Explanation:

Let us factor the numerator and denominator first:

${x}^{3} - 6 {x}^{2} + 11 x - 6$

$= \left(x - 1\right) \left({x}^{2} - 5 x + 6\right)$

$= \left(x - 1\right) \left(x - 2\right) \left(x - 3\right)$

$\textcolor{w h i t e}{}$

$4 {x}^{3} - 28 {x}^{2} + 56 x - 32$

$= 4 \left({x}^{3} - 7 {x}^{2} + 14 x - 8\right)$

$= 4 \left(x - 1\right) \left({x}^{2} - 6 x + 8\right)$

$= 4 \left(x - 1\right) \left(x - 2\right) \left(x - 4\right)$

So:

$\frac{{x}^{3} - 6 {x}^{2} + 11 x - 6}{4 {x}^{3} - 28 {x}^{2} + 56 x - 32}$

$= \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{\left(x - 1\right)}}} \textcolor{red}{\cancel{\textcolor{b l a c k}{\left(x - 2\right)}}} \left(x - 3\right)}{4 \textcolor{red}{\cancel{\textcolor{b l a c k}{\left(x - 1\right)}}} \textcolor{red}{\cancel{\textcolor{b l a c k}{\left(x - 2\right)}}} \left(x - 4\right)}$

$= \frac{x - 3}{4 \left(x - 4\right)}$

$= \frac{x - 4 + 1}{4 \left(x - 4\right)}$

$= \frac{1}{4} + \frac{1}{4 \left(x - 4\right)}$

with exclusions $x \ne 1$ and $x \ne 2$