# How do you simplify (x-3)/(-x^3+3x^2)*(x^2+2x+1)?

May 24, 2017

$- 1 - \frac{2 x + 1}{{x}^{2}}$

#### Explanation:

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

$= - \frac{1}{{x}^{2}} \cdot \left({x}^{2} + 2 x + 1\right)$

$= - 1 + \frac{- 2 x - 1}{{x}^{2}}$

May 24, 2017

$\frac{- {x}^{2} - 2 x - 1}{x} ^ 2$

#### Explanation:

$\frac{x - 3}{- {x}^{3} + 3 {x}^{2}} \cdot {x}^{2} + 2 x + 1$

$\therefore = {\cancel{x - 3}}^{1} / \left(- {x}^{2} \cancel{{\left(x - 3\right)}^{1}}\right) \cdot {x}^{2} + 2 x + 1$

$\therefore = \frac{1}{- {x}^{2}} \cdot {x}^{2} + 2 x + 1$

$\therefore = \frac{{x}^{2} + 2 x + 1}{-} {x}^{2}$

$\therefore = \left(\frac{{x}^{2} + 2 x + 1}{-} {x}^{2}\right) \cdot \frac{- {x}^{2}}{-} {x}^{2}$

$\therefore = \frac{- {x}^{2}}{- {x}^{2}} = 1$

$\therefore = \frac{{\cancel{- {x}^{2}}}^{\textcolor{b l u e}{- 1}} \left({x}^{2} + 2 x + 1\right)}{\cancel{{x}^{4}}} ^ \textcolor{b l u e}{{x}^{2}}$

$\therefore = \frac{- 1 \left({x}^{2} + 2 x + 1\right)}{x} ^ 2$

:.=color(blue)((-x^2-2x-1)/x^2