What is expand ((k+1)(k+2)(4k-9))/6 ?

The solution to expand ((k+1)(k+2)(4k-9))/6 is (2k^3)/3+(k^2)/2-(19k)/6-3

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expand ((k+1)(k+2)(4k-9))/6

expand

\frac{( k + 1 ) ( k + 2 ) ( 4 k - 9 )}{6}

\frac{2 k ^{3}}{3} + \frac{k ^{2}}{2} - \frac{19 k}{6} - 3

Solution steps

\frac{( k + 1 ) ( k + 2 ) ( 4 k - 9 )}{6}

Expand

(k + 1) (k + 2) (4 k - 9) : 4 k^{3} + 3 k^{2} - 19 k - 18

= \frac{4 k ^{3} + 3 k ^{2} - 19 k - 18}{6}

Apply the fraction rule:

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

\frac{4 k ^{3} + 3 k ^{2} - 19 k - 18}{6} = \frac{4 k ^{3}}{6} + \frac{3 k ^{2}}{6} - \frac{19 k}{6} - \frac{18}{6}

= \frac{4 k ^{3}}{6} + \frac{3 k ^{2}}{6} - \frac{19 k}{6} - \frac{18}{6}

Cancel

\frac{4 k ^{3}}{6} : \frac{2 k ^{3}}{3}

= \frac{2 k ^{3}}{3} + \frac{3 k ^{2}}{6} - \frac{19 k}{6} - \frac{18}{6}

Cancel

\frac{3 k ^{2}}{6} : \frac{k ^{2}}{2}

= \frac{2 k ^{3}}{3} + \frac{k ^{2}}{2} - \frac{19 k}{6} - \frac{18}{6}

Divide the numbers:

\frac{18}{6} = 3

= \frac{2 k ^{3}}{3} + \frac{k ^{2}}{2} - \frac{19 k}{6} - 3

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What is expand ((k+1)(k+2)(4k-9))/6 ?
The solution to expand ((k+1)(k+2)(4k-9))/6 is (2k^3)/3+(k^2)/2-(19k)/6-3