What is expand x(2x-1/3)^5 ?

The solution to expand x(2x-1/3)^5 is 32x^6-(80x^5)/3+(80x^4)/9-(40x^3)/(27)+(10x^2)/(81)-x/(243)

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expand x(2x-1/3)^5

expand

x (2 x - \frac{1}{3})^{5}

32 x^{6} - \frac{80 x ^{5}}{3} + \frac{80 x ^{4}}{9} - \frac{40 x ^{3}}{27} + \frac{10 x ^{2}}{81} - \frac{x}{243}

Solution steps

x (2 x - \frac{1}{3})^{5}

Expand

(2 x - \frac{1}{3})^{5} : 32 x^{5} - \frac{80 x ^{4}}{3} + \frac{80 x ^{3}}{9} - \frac{40 x ^{2}}{27} + \frac{10 x}{81} - \frac{1}{243}

= x (32 x^{5} - \frac{80 x ^{4}}{3} + \frac{80 x ^{3}}{9} - \frac{40 x ^{2}}{27} + \frac{10 x}{81} - \frac{1}{243})

Distribute parentheses

= x \cdot 32 x^{5} + x (- \frac{80 x ^{4}}{3}) + x \frac{80 x ^{3}}{9} + x (- \frac{40 x ^{2}}{27}) + x \frac{10 x}{81} + x (- \frac{1}{243})

Simplify

x \cdot 32 x^{5} + x (- \frac{80 x ^{4}}{3}) + x \frac{80 x ^{3}}{9} + x (- \frac{40 x ^{2}}{27}) + x \frac{10 x}{81} + x (- \frac{1}{243}) : 32 x^{6} - \frac{80 x ^{5}}{3} + \frac{80 x ^{4}}{9} - \frac{40 x ^{3}}{27} + \frac{10 x ^{2}}{81} - \frac{x}{243}

= 32 x^{6} - \frac{80 x ^{5}}{3} + \frac{80 x ^{4}}{9} - \frac{40 x ^{3}}{27} + \frac{10 x ^{2}}{81} - \frac{x}{243}

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What is expand x(2x-1/3)^5 ?
The solution to expand x(2x-1/3)^5 is 32x^6-(80x^5)/3+(80x^4)/9-(40x^3)/(27)+(10x^2)/(81)-x/(243)