What is expand-(x(-2x+12))/9 ?

The solution to expand-(x(-2x+12))/9 is (2x^2)/9-(4x)/3

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expand-(x(-2x+12))/9

expand

- \frac{x ( - 2 x + 12 )}{9}

\frac{2 x ^{2}}{9} - \frac{4 x}{3}

Solution steps

- \frac{x ( - 2 x + 12 )}{9}

Expand

x (- 2 x + 12) : - 2 x^{2} + 12 x

= - \frac{- 2 x ^{2} + 12 x}{9}

Apply the fraction rule:

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

\frac{- 2 x ^{2} + 12 x}{9} = - (- \frac{2 x ^{2}}{9}) - (\frac{12 x}{9})

= - (- \frac{2 x ^{2}}{9}) - (\frac{12 x}{9})

Remove parentheses:

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

= \frac{2 x ^{2}}{9} - \frac{12 x}{9}

Cancel

\frac{12 x}{9} : \frac{4 x}{3}

= \frac{2 x ^{2}}{9} - \frac{4 x}{3}

r sin (θ) (e^{r^{2} c o s (θ)} + r sin (θ))

(e^{t} + e^{- t})^{2}

\frac{2 ( - 3 x ^{2} + 7 )}{( 3 x ^{2} + 7 ) ^{2}}

\frac{8}{( e ^{2 x} + e ^{- 2 x} ) ^{2}}

p o in t s (42 x - 6 y^{2} - ln ∣ x + y ∣)

What is expand-(x(-2x+12))/9 ?
The solution to expand-(x(-2x+12))/9 is (2x^2)/9-(4x)/3