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

The solution to expand ((3x+2)(x+2)(-2x+3))/6 is -x^3-(7x^2)/6+(8x)/3+2

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expand ((3x+2)(x+2)(-2x+3))/6

expand

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

- x^{3} - \frac{7 x ^{2}}{6} + \frac{8 x}{3} + 2

Solution steps

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

Expand

(3 x + 2) (x + 2) (- 2 x + 3) : - 6 x^{3} - 7 x^{2} + 16 x + 12

= \frac{- 6 x ^{3} - 7 x ^{2} + 16 x + 12}{6}

Apply the fraction rule:

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

\frac{- 6 x ^{3} - 7 x ^{2} + 16 x + 12}{6} = - \frac{6 x ^{3}}{6} - \frac{7 x ^{2}}{6} + \frac{16 x}{6} + \frac{12}{6}

= - \frac{6 x ^{3}}{6} - \frac{7 x ^{2}}{6} + \frac{16 x}{6} + \frac{12}{6}

Divide the numbers:

\frac{6}{6} = 1

= - x^{3} - \frac{7 x ^{2}}{6} + \frac{16 x}{6} + \frac{12}{6}

Cancel

\frac{16 x}{6} : \frac{8 x}{3}

= - x^{3} - \frac{7 x ^{2}}{6} + \frac{8 x}{3} + \frac{12}{6}

Divide the numbers:

\frac{12}{6} = 2

= - x^{3} - \frac{7 x ^{2}}{6} + \frac{8 x}{3} + 2

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

[(1 - x) (- e^{- x})] - 1

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

\frac{1}{( y - 1 ) ^{3}}

z (u, v, w (z, u, v)) (2 u + v - w (z, u, v) + 3 z (u, v, w (z, u, v)) = 0)

What is expand ((3x+2)(x+2)(-2x+3))/6 ?
The solution to expand ((3x+2)(x+2)(-2x+3))/6 is -x^3-(7x^2)/6+(8x)/3+2