What is expand ((x^2+1)(x^3-1))/(2x) ?

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

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expand ((x^2+1)(x^3-1))/(2x)

expand

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

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

Solution steps

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

Expand

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

= \frac{x ^{5} - x ^{2} + x ^{3} - 1}{2 x}

Apply the fraction rule:

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

\frac{x ^{5} - x ^{2} + x ^{3} - 1}{2 x} = \frac{x ^{5}}{2 x} - \frac{x ^{2}}{2 x} + \frac{x ^{3}}{2 x} - \frac{1}{2 x}

= \frac{x ^{5}}{2 x} - \frac{x ^{2}}{2 x} + \frac{x ^{3}}{2 x} - \frac{1}{2 x}

Cancel

\frac{x ^{5}}{2 x} : \frac{x ^{4}}{2}

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

Cancel

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

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

Cancel

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

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

- 3 x - (7 x + 4 = - 4 (3 x + 1))

\frac{( 25 + 5 s ) ( 100 + 300 s )}{s}

\frac{3 x}{x}

\frac{( x + 7 ) 2 x + 6}{2}

s (x + sy + s^{2} z)^{3}

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