What is expand (y(-y^4+16y^2+8y-32))/8 ?

The solution to expand (y(-y^4+16y^2+8y-32))/8 is -(y^5)/8+2y^3+y^2-4y

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expand (y(-y^4+16y^2+8y-32))/8

expand

\frac{y ( - y ^{4} + 16 y ^{2} + 8 y - 32 )}{8}

- \frac{y ^{5}}{8} + 2 y^{3} + y^{2} - 4 y

Solution steps

\frac{y ( - y ^{4} + 16 y ^{2} + 8 y - 32 )}{8}

Expand

y (- y^{4} + 16 y^{2} + 8 y - 32) : - y^{5} + 16 y^{3} + 8 y^{2} - 32 y

= \frac{- y ^{5} + 16 y ^{3} + 8 y ^{2} - 32 y}{8}

Apply the fraction rule:

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

\frac{- y ^{5} + 16 y ^{3} + 8 y ^{2} - 32 y}{8} = - \frac{y ^{5}}{8} + \frac{16 y ^{3}}{8} + \frac{8 y ^{2}}{8} - \frac{32 y}{8}

= - \frac{y ^{5}}{8} + \frac{16 y ^{3}}{8} + \frac{8 y ^{2}}{8} - \frac{32 y}{8}

Divide the numbers:

\frac{16}{8} = 2

= - \frac{y ^{5}}{8} + 2 y^{3} + \frac{8 y ^{2}}{8} - \frac{32 y}{8}

Divide the numbers:

\frac{8}{8} = 1

= - \frac{y ^{5}}{8} + 2 y^{3} + y^{2} - \frac{32 y}{8}

Divide the numbers:

\frac{32}{8} = 4

= - \frac{y ^{5}}{8} + 2 y^{3} + y^{2} - 4 y

(x^{2} - 9 x \circ x - 9) (- 2)

(b - a) (b + a)

- 2 y y^{'} (x) + (x - y^{2}) y^{'} (x)

(2 x^{2} - \frac{3}{x})^{6}

sin (λ - \frac{π}{3}) + cos (λ + \frac{π}{6})

What is expand (y(-y^4+16y^2+8y-32))/8 ?
The solution to expand (y(-y^4+16y^2+8y-32))/8 is -(y^5)/8+2y^3+y^2-4y