What is expand ((4x^3y^3-16x^4-2xy))/(-2xy) ?

The solution to expand ((4x^3y^3-16x^4-2xy))/(-2xy) is -2x^2y^2+(8x^3)/y+1

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expand ((4x^3y^3-16x^4-2xy))/(-2xy)

expand

\frac{( 4 x ^{3} y ^{3} - 16 x ^{4} - 2 x y )}{- 2 x y}

- 2 x^{2} y^{2} + \frac{8 x ^{3}}{y} + 1

Solution steps

\frac{4 x ^{3} y ^{3} - 16 x ^{4} - 2 x y}{- 2 x y}

Apply the fraction rule:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{4 x ^{3} y ^{3} - 16 x ^{4} - 2 x y}{2 x y}

Cancel

\frac{4 x ^{3} y ^{3} - 16 x ^{4} - 2 x y}{2 x y} : \frac{- 8 x ^{3} + 2 x ^{2} y ^{3} - y}{y}

= - \frac{- 8 x ^{3} + 2 x ^{2} y ^{3} - y}{y}

Apply the fraction rule:

\frac{a \pm b}{c} = \frac{a}{c} \pm \frac{b}{c}

\frac{2 x ^{2} y ^{3} - 8 x ^{3} - y}{y} = - (\frac{2 x ^{2} y ^{3}}{y}) - (- \frac{8 x ^{3}}{y}) - (- \frac{y}{y})

= - (\frac{2 x ^{2} y ^{3}}{y}) - (- \frac{8 x ^{3}}{y}) - (- \frac{y}{y})

Remove parentheses:

(a) = a, - (- a) = a

= - \frac{2 x ^{2} y ^{3}}{y} + \frac{8 x ^{3}}{y} + \frac{y}{y}

Cancel

\frac{2 x ^{2} y ^{3}}{y} : 2 x^{2} y^{2}

= - 2 x^{2} y^{2} + \frac{8 x ^{3}}{y} + \frac{y}{y}

Apply rule

\frac{a}{a} = 1

= - 2 x^{2} y^{2} + \frac{8 x ^{3}}{y} + 1

\frac{2 ( - 27 t ^{\frac{13}{12}} + 13 t ^{\frac{9}{4}} )}{351}

\frac{1}{2} (x - 10) (30 - x)

(x + \frac{9}{5}) (x + \frac{2}{5})

\frac{2 ( x + 3 )}{3 ( x + 1 )}

\frac{21}{x ( x + 3 )}

What is expand ((4x^3y^3-16x^4-2xy))/(-2xy) ?
The solution to expand ((4x^3y^3-16x^4-2xy))/(-2xy) is -2x^2y^2+(8x^3)/y+1