What is expand (x(-8+3x)(16x-3x^2-16)(1+2x))/8 ?

The solution to expand (x(-8+3x)(16x-3x^2-16)(1+2x))/8 is -(9x^5)/4+(135x^4)/8-35x^3+10x^2+16x

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expand (x(-8+3x)(16x-3x^2-16)(1+2x))/8

expand

\frac{x ( - 8 + 3 x ) ( 16 x - 3 x ^{2} - 16 ) ( 1 + 2 x )}{8}

- \frac{9 x ^{5}}{4} + \frac{135 x ^{4}}{8} - 35 x^{3} + 10 x^{2} + 16 x

Solution steps

\frac{x ( - 8 + 3 x ) ( 16 x - 3 x ^{2} - 16 ) ( 1 + 2 x )}{8}

Expand

x (- 8 + 3 x) (16 x - 3 x^{2} - 16) (1 + 2 x) : - 18 x^{5} + 135 x^{4} - 280 x^{3} + 80 x^{2} + 128 x

= \frac{- 18 x ^{5} + 135 x ^{4} - 280 x ^{3} + 80 x ^{2} + 128 x}{8}

Apply the fraction rule:

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

\frac{- 18 x ^{5} + 135 x ^{4} - 280 x ^{3} + 80 x ^{2} + 128 x}{8} = - \frac{18 x ^{5}}{8} + \frac{135 x ^{4}}{8} - \frac{280 x ^{3}}{8} + \frac{80 x ^{2}}{8} + \frac{128 x}{8}

= - \frac{18 x ^{5}}{8} + \frac{135 x ^{4}}{8} - \frac{280 x ^{3}}{8} + \frac{80 x ^{2}}{8} + \frac{128 x}{8}

\frac{18 x ^{5}}{8} = \frac{9 x ^{5}}{4}

\frac{280 x ^{3}}{8} = 35 x^{3}

\frac{80 x ^{2}}{8} = 10 x^{2}

\frac{128 x}{8} = 16 x

= - \frac{9 x ^{5}}{4} + \frac{135 x ^{4}}{8} - 35 x^{3} + 10 x^{2} + 16 x

- 20 (θ sin^{2} (θ) + sin (θ) - cos^{2} (θ))

π (x + y)

ln (z) (z - 1)^{9}

ln (\frac{x - 65}{x})

[P o w er [4, n]] l e f t (402 n + 1)

What is expand (x(-8+3x)(16x-3x^2-16)(1+2x))/8 ?
The solution to expand (x(-8+3x)(16x-3x^2-16)(1+2x))/8 is -(9x^5)/4+(135x^4)/8-35x^3+10x^2+16x