What is expand ((2x^4-3x^2+6x^5-x^3))/((2x^2)) ?

The solution to expand ((2x^4-3x^2+6x^5-x^3))/((2x^2)) is 3x^3+x^2-x/2-3/2

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expand ((2x^4-3x^2+6x^5-x^3))/((2x^2))

expand

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

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

Solution steps

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

Factor

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

= \frac{x ^{2} ( 2 x ^{2} - 3 + 6 x ^{3} - x )}{2 x ^{2}}

Cancel the common factor:

x^{2}

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

Apply the fraction rule:

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

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

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

Divide the numbers:

\frac{2}{2} = 1

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

Divide the numbers:

\frac{6}{2} = 3

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

Rewrite in standard form

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

\frac{s ^{2} + 5}{( s - 1 ) ( s ^{2} - 2 s - 3 )}

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

\frac{d ^{3}}{d x ^{3}} ((y) (- sin (x) + 3 x^{2} - 8 x + 3))

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

3 x^{3} (- ln (x) + c)

What is expand ((2x^4-3x^2+6x^5-x^3))/((2x^2)) ?
The solution to expand ((2x^4-3x^2+6x^5-x^3))/((2x^2)) is 3x^3+x^2-x/2-3/2