What is expand (z^2(1+2z)^3)/6 ?

The solution to expand (z^2(1+2z)^3)/6 is (4z^5)/3+2z^4+z^3+(z^2)/6

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expand (z^2(1+2z)^3)/6

expand

\frac{z ^{2} ( 1 + 2 z ) ^{3}}{6}

\frac{4 z ^{5}}{3} + 2 z^{4} + z^{3} + \frac{z ^{2}}{6}

Solution steps

\frac{z ^{2} ( 1 + 2 z ) ^{3}}{6}

(1 + 2 z)^{3} = 1 + 6 z + 12 z^{2} + 8 z^{3}

= \frac{z ^{2} ( 8 z ^{3} + 12 z ^{2} + 6 z + 1 )}{6}

Expand

z^{2} (1 + 6 z + 12 z^{2} + 8 z^{3}) : z^{2} + 6 z^{3} + 12 z^{4} + 8 z^{5}

= \frac{z ^{2} + 6 z ^{3} + 12 z ^{4} + 8 z ^{5}}{6}

Apply the fraction rule:

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

\frac{z ^{2} + 6 z ^{3} + 12 z ^{4} + 8 z ^{5}}{6} = \frac{z ^{2}}{6} + \frac{6 z ^{3}}{6} + \frac{12 z ^{4}}{6} + \frac{8 z ^{5}}{6}

= \frac{z ^{2}}{6} + \frac{6 z ^{3}}{6} + \frac{12 z ^{4}}{6} + \frac{8 z ^{5}}{6}

Divide the numbers:

\frac{6}{6} = 1

= \frac{z ^{2}}{6} + z^{3} + \frac{12 z ^{4}}{6} + \frac{8 z ^{5}}{6}

Divide the numbers:

\frac{12}{6} = 2

= \frac{z ^{2}}{6} + z^{3} + 2 z^{4} + \frac{8 z ^{5}}{6}

Cancel

\frac{8 z ^{5}}{6} : \frac{4 z ^{5}}{3}

= \frac{z ^{2}}{6} + z^{3} + 2 z^{4} + \frac{4 z ^{5}}{3}

Rewrite in standard form

= \frac{4 z ^{5}}{3} + 2 z^{4} + z^{3} + \frac{z ^{2}}{6}

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What is expand (z^2(1+2z)^3)/6 ?
The solution to expand (z^2(1+2z)^3)/6 is (4z^5)/3+2z^4+z^3+(z^2)/6