What is expand-((x+1)(x-2)(x-5))/8 ?

The solution to expand-((x+1)(x-2)(x-5))/8 is -(x^3)/8+(3x^2)/4-(3x)/8-5/4

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expand-((x+1)(x-2)(x-5))/8

expand

- \frac{( x + 1 ) ( x - 2 ) ( x - 5 )}{8}

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

Solution steps

- \frac{( x + 1 ) ( x - 2 ) ( x - 5 )}{8}

Expand

(x + 1) (x - 2) (x - 5) : x^{3} - 6 x^{2} + 3 x + 10

= - \frac{x ^{3} - 6 x ^{2} + 3 x + 10}{8}

Apply the fraction rule:

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

\frac{x ^{3} - 6 x ^{2} + 3 x + 10}{8} = - (\frac{x ^{3}}{8}) - (- \frac{6 x ^{2}}{8}) - (\frac{3 x}{8}) - (\frac{10}{8})

= - (\frac{x ^{3}}{8}) - (- \frac{6 x ^{2}}{8}) - (\frac{3 x}{8}) - (\frac{10}{8})

Remove parentheses:

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

= - \frac{x ^{3}}{8} + \frac{6 x ^{2}}{8} - \frac{3 x}{8} - \frac{10}{8}

Cancel

\frac{6 x ^{2}}{8} : \frac{3 x ^{2}}{4}

= - \frac{x ^{3}}{8} + \frac{3 x ^{2}}{4} - \frac{3 x}{8} - \frac{10}{8}

Cancel

\frac{10}{8} : \frac{5}{4}

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

3 x - 2 \cdot (3 x - 2)

(4 + 7) (8 - 7)

x^{2} + 3 x - 28 x^{2} + 3 x - 28

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

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

What is expand-((x+1)(x-2)(x-5))/8 ?
The solution to expand-((x+1)(x-2)(x-5))/8 is -(x^3)/8+(3x^2)/4-(3x)/8-5/4