What is expand ((-1-2x)(3-2x^2))/3 ?

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

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expand ((-1-2x)(3-2x^2))/3

expand

\frac{( - 1 - 2 x ) ( 3 - 2 x ^{2} )}{3}

\frac{4 x ^{3}}{3} + \frac{2 x ^{2}}{3} - 2 x - 1

Solution steps

\frac{( - 1 - 2 x ) ( 3 - 2 x ^{2} )}{3}

Expand

(- 1 - 2 x) (3 - 2 x^{2}) : - 3 + 2 x^{2} - 6 x + 4 x^{3}

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

Apply the fraction rule:

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

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

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

Apply rule

\frac{a}{a} = 1

\frac{3}{3} = 1

= - 1 + \frac{2 x ^{2}}{3} - \frac{6 x}{3} + \frac{4 x ^{3}}{3}

Divide the numbers:

\frac{6}{3} = 2

= - 1 + \frac{2 x ^{2}}{3} - 2 x + \frac{4 x ^{3}}{3}

Rewrite in standard form

= \frac{4 x ^{3}}{3} + \frac{2 x ^{2}}{3} - 2 x - 1

(2, - 7) y (- 4, 11)

e^{y} (y - 1)

(e^{\frac{y}{4}} + 6)^{2}

\frac{12 s + 57}{( s + 2 ) ( s + 4 )}

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

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