What is expand ((s-2)(3s+1)^2)/9 ?

The solution to expand ((s-2)(3s+1)^2)/9 is s^3-(4s^2)/3-(11s)/9-2/9

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expand ((s-2)(3s+1)^2)/9

expand

\frac{( s - 2 ) ( 3 s + 1 ) ^{2}}{9}

s^{3} - \frac{4 s ^{2}}{3} - \frac{11 s}{9} - \frac{2}{9}

Solution steps

\frac{( s - 2 ) ( 3 s + 1 ) ^{2}}{9}

(3 s + 1)^{2} = 9 s^{2} + 6 s + 1

= \frac{( s - 2 ) ( 9 s ^{2} + 6 s + 1 )}{9}

Expand

(s - 2) (9 s^{2} + 6 s + 1) : 9 s^{3} - 12 s^{2} - 11 s - 2

= \frac{9 s ^{3} - 12 s ^{2} - 11 s - 2}{9}

Apply the fraction rule:

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

\frac{9 s ^{3} - 12 s ^{2} - 11 s - 2}{9} = \frac{9 s ^{3}}{9} - \frac{12 s ^{2}}{9} - \frac{11 s}{9} - \frac{2}{9}

= \frac{9 s ^{3}}{9} - \frac{12 s ^{2}}{9} - \frac{11 s}{9} - \frac{2}{9}

Divide the numbers:

\frac{9}{9} = 1

= s^{3} - \frac{12 s ^{2}}{9} - \frac{11 s}{9} - \frac{2}{9}

Cancel

\frac{12 s ^{2}}{9} : \frac{4 s ^{2}}{3}

= s^{3} - \frac{4 s ^{2}}{3} - \frac{11 s}{9} - \frac{2}{9}

- \frac{4 s}{( s + 6 ) ( s + 2 )}

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

\frac{( x - 4 ) ( x + 1 )}{( x + 3 ) ( x + 2 )}

lo g_{10} (1 s)

(9 i, - 2 j, - 3 k) (i, - 6 j - 8 k)

What is expand ((s-2)(3s+1)^2)/9 ?
The solution to expand ((s-2)(3s+1)^2)/9 is s^3-(4s^2)/3-(11s)/9-2/9