What is expand-((2y+1)(2y^2-y-2))/y ?

The solution to expand-((2y+1)(2y^2-y-2))/y is -4y^2+5+2/y

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expand-((2y+1)(2y^2-y-2))/y

expand

- \frac{( 2 y + 1 ) ( 2 y ^{2} - y - 2 )}{y}

- 4 y^{2} + 5 + \frac{2}{y}

Solution steps

- \frac{( 2 y + 1 ) ( 2 y ^{2} - y - 2 )}{y}

Expand

(2 y + 1) (2 y^{2} - y - 2) : 4 y^{3} - 5 y - 2

= - \frac{4 y ^{3} - 5 y - 2}{y}

Apply the fraction rule:

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

\frac{4 y ^{3} - 5 y - 2}{y} = - (\frac{4 y ^{3}}{y}) - (- \frac{5 y}{y}) - (- \frac{2}{y})

= - (\frac{4 y ^{3}}{y}) - (- \frac{5 y}{y}) - (- \frac{2}{y})

Remove parentheses:

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

= - \frac{4 y ^{3}}{y} + \frac{5 y}{y} + \frac{2}{y}

Cancel

\frac{4 y ^{3}}{y} : 4 y^{2}

= - 4 y^{2} + \frac{5 y}{y} + \frac{2}{y}

Cancel

\frac{5 y}{y} : 5

= - 4 y^{2} + 5 + \frac{2}{y}

32

\frac{d}{d t} ((x) (t^{2} - 4 t))

\frac{2 π x ^{2} - 44 x ^{2} + 1440 x}{9}

(0, - 2) y (0, - 2)

\frac{5 ( 9 - 21 )}{3}

What is expand-((2y+1)(2y^2-y-2))/y ?
The solution to expand-((2y+1)(2y^2-y-2))/y is -4y^2+5+2/y