What is expand ((4x-1)(2x^2+x+2))/(12) ?

The solution to expand ((4x-1)(2x^2+x+2))/(12) is (2x^3)/3+(x^2)/6+(7x)/(12)-1/6

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expand ((4x-1)(2x^2+x+2))/(12)

expand

\frac{( 4 x - 1 ) ( 2 x ^{2} + x + 2 )}{12}

\frac{2 x ^{3}}{3} + \frac{x ^{2}}{6} + \frac{7 x}{12} - \frac{1}{6}

Solution steps

\frac{( 4 x - 1 ) ( 2 x ^{2} + x + 2 )}{12}

Expand

(4 x - 1) (2 x^{2} + x + 2) : 8 x^{3} + 2 x^{2} + 7 x - 2

= \frac{8 x ^{3} + 2 x ^{2} + 7 x - 2}{12}

Apply the fraction rule:

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

\frac{8 x ^{3} + 2 x ^{2} + 7 x - 2}{12} = \frac{8 x ^{3}}{12} + \frac{2 x ^{2}}{12} + \frac{7 x}{12} - \frac{2}{12}

= \frac{8 x ^{3}}{12} + \frac{2 x ^{2}}{12} + \frac{7 x}{12} - \frac{2}{12}

Cancel

\frac{8 x ^{3}}{12} : \frac{2 x ^{3}}{3}

= \frac{2 x ^{3}}{3} + \frac{2 x ^{2}}{12} + \frac{7 x}{12} - \frac{2}{12}

Cancel

\frac{2 x ^{2}}{12} : \frac{x ^{2}}{6}

= \frac{2 x ^{3}}{3} + \frac{x ^{2}}{6} + \frac{7 x}{12} - \frac{2}{12}

Cancel

\frac{2}{12} : \frac{1}{6}

= \frac{2 x ^{3}}{3} + \frac{x ^{2}}{6} + \frac{7 x}{12} - \frac{1}{6}

s (203 sin (θ), 12 + 123 sin (θ))

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(9 x - 1) l e f t (2 x + 3) - 6 x (3 x - 2)

(2 \cdot x - 4)^{2}

What is expand ((4x-1)(2x^2+x+2))/(12) ?
The solution to expand ((4x-1)(2x^2+x+2))/(12) is (2x^3)/3+(x^2)/6+(7x)/(12)-1/6