What is expand ((x^2+2)(3x^3-2))/(4x^3) ?

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

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expand ((x^2+2)(3x^3-2))/(4x^3)

expand

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

\frac{3 x ^{2}}{4} - \frac{1}{2 x} + \frac{3}{2} - \frac{1}{x ^{3}}

Solution steps

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

Expand

(x^{2} + 2) (3 x^{3} - 2) : 3 x^{5} - 2 x^{2} + 6 x^{3} - 4

= \frac{3 x ^{5} - 2 x ^{2} + 6 x ^{3} - 4}{4 x ^{3}}

Apply the fraction rule:

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

\frac{3 x ^{5} - 2 x ^{2} + 6 x ^{3} - 4}{4 x ^{3}} = \frac{3 x ^{5}}{4 x ^{3}} - \frac{2 x ^{2}}{4 x ^{3}} + \frac{6 x ^{3}}{4 x ^{3}} - \frac{4}{4 x ^{3}}

= \frac{3 x ^{5}}{4 x ^{3}} - \frac{2 x ^{2}}{4 x ^{3}} + \frac{6 x ^{3}}{4 x ^{3}} - \frac{4}{4 x ^{3}}

\frac{3 x ^{5}}{4 x ^{3}} = \frac{3 x ^{2}}{4}

\frac{2 x ^{2}}{4 x ^{3}} = \frac{1}{2 x}

\frac{6 x ^{3}}{4 x ^{3}} = \frac{3}{2}

\frac{4}{4 x ^{3}} = \frac{1}{x ^{3}}

= \frac{3 x ^{2}}{4} - \frac{1}{2 x} + \frac{3}{2} - \frac{1}{x ^{3}}

\frac{- 1}{4} (- x + 2)^{2} (- x - 1)^{2}

(1 - 3 x) (x + 2)^{'}

2 t (2 t - 2)

[(x + 1) (x - 1)] 2

(- t^{3} + \frac{9}{10} t + \frac{1}{10}) (1 + t + t^{2})

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