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

The solution to expand ((-2x-3x^2)(4-x^2-x^3))/h is (3x^5)/h+(5x^4)/h+(2x^3)/h-(12x^2)/h-(8x)/h

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

expand

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

\frac{3 x ^{5}}{h} + \frac{5 x ^{4}}{h} + \frac{2 x ^{3}}{h} - \frac{12 x ^{2}}{h} - \frac{8 x}{h}

Solution steps

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

Expand

(- 2 x - 3 x^{2}) (4 - x^{2} - x^{3}) : 3 x^{5} + 5 x^{4} + 2 x^{3} - 12 x^{2} - 8 x

= \frac{3 x ^{5} + 5 x ^{4} + 2 x ^{3} - 12 x ^{2} - 8 x}{h}

Apply the fraction rule:

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

\frac{3 x ^{5} + 5 x ^{4} + 2 x ^{3} - 12 x ^{2} - 8 x}{h} = \frac{3 x ^{5}}{h} + \frac{5 x ^{4}}{h} + \frac{2 x ^{3}}{h} - \frac{12 x ^{2}}{h} - \frac{8 x}{h}

= \frac{3 x ^{5}}{h} + \frac{5 x ^{4}}{h} + \frac{2 x ^{3}}{h} - \frac{12 x ^{2}}{h} - \frac{8 x}{h}

3 sin (\frac{( π ( x - 5 ) )}{( 6 )})

\frac{1}{( j x + 4 ) \cdot ( j x + 1 )}

\frac{\partial ^{2}}{\partial u \partial x} ((x) (- 2 x + u (x)))

\frac{( A s ) + B}{( s ^{2} + 1 ) ^{2}} + \frac{( C s ) + D}{s ^{2} + 1}

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

What is expand ((-2x-3x^2)(4-x^2-x^3))/h ?
The solution to expand ((-2x-3x^2)(4-x^2-x^3))/h is (3x^5)/h+(5x^4)/h+(2x^3)/h-(12x^2)/h-(8x)/h