What is expand ((1-3t)(-27te^{-t}-11t-8))/9 ?

The solution to expand ((1-3t)(-27te^{-t}-11t-8))/9 is 9e^{-t}t^2+(11t^2)/3-3e^{-t}t+(13t)/9-8/9

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expand ((1-3t)(-27te^{-t}-11t-8))/9

expand

\frac{( 1 - 3 t ) ( - 27 t e ^{- t} - 11 t - 8 )}{9}

9 e^{- t} t^{2} + \frac{11 t ^{2}}{3} - 3 e^{- t} t + \frac{13 t}{9} - \frac{8}{9}

Solution steps

\frac{( 1 - 3 t ) ( - 27 t e ^{- t} - 11 t - 8 )}{9}

Expand

(1 - 3 t) (- 27 t e^{- t} - 11 t - 8) : 81 e^{- t} t^{2} + 33 t^{2} - 27 e^{- t} t + 13 t - 8

= \frac{81 e ^{- t} t ^{2} + 33 t ^{2} - 27 e ^{- t} t + 13 t - 8}{9}

Apply the fraction rule:

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

\frac{81 e ^{- t} t ^{2} + 33 t ^{2} - 27 e ^{- t} t + 13 t - 8}{9} = \frac{81 e ^{- t} t ^{2}}{9} + \frac{33 t ^{2}}{9} - \frac{27 e ^{- t} t}{9} + \frac{13 t}{9} - \frac{8}{9}

= \frac{81 e ^{- t} t ^{2}}{9} + \frac{33 t ^{2}}{9} - \frac{27 e ^{- t} t}{9} + \frac{13 t}{9} - \frac{8}{9}

Divide the numbers:

\frac{81}{9} = 9

= 9 e^{- t} t^{2} + \frac{33 t ^{2}}{9} - \frac{27 e ^{- t} t}{9} + \frac{13 t}{9} - \frac{8}{9}

Cancel

\frac{33 t ^{2}}{9} : \frac{11 t ^{2}}{3}

= 9 e^{- t} t^{2} + \frac{11 t ^{2}}{3} - \frac{27 e ^{- t} t}{9} + \frac{13 t}{9} - \frac{8}{9}

Divide the numbers:

\frac{27}{9} = 3

= 9 e^{- t} t^{2} + \frac{11 t ^{2}}{3} - 3 e^{- t} t + \frac{13 t}{9} - \frac{8}{9}

(\frac{1}{24} sin (10 x) - \frac{5}{24} (sin (2 x))) cos (6 x)

4 x^{2} y^{''} (x) - 4 x^{2} y^{'} (x) + (1 - 2 x) y

(2 x - 10) d x

\frac{7 x + 4}{( s + 10 ) ^{2}}

\frac{2}{d ( 1 - r ) ^{2}}

What is expand ((1-3t)(-27te^{-t}-11t-8))/9 ?
The solution to expand ((1-3t)(-27te^{-t}-11t-8))/9 is 9e^{-t}t^2+(11t^2)/3-3e^{-t}t+(13t)/9-8/9