What is expand-1/8 (x+2)^2(x-4)^2 ?

The solution to expand-1/8 (x+2)^2(x-4)^2 is -(x^4)/8+(x^3)/2+(3x^2)/2-4x-8

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expand-1/8 (x+2)^2(x-4)^2

expand

- \frac{1}{8} (x + 2)^{2} (x - 4)^{2}

- \frac{x ^{4}}{8} + \frac{x ^{3}}{2} + \frac{3 x ^{2}}{2} - 4 x - 8

Solution steps

- \frac{1}{8} (x + 2)^{2} (x - 4)^{2}

Expand

(x + 2)^{2} : x^{2} + 4 x + 4

= - \frac{1}{8} (x^{2} + 4 x + 4) (x - 4)^{2}

= - \frac{1}{8} (x - 4)^{2} (x^{2} + 4 x + 4)

Expand

(x - 4)^{2} : x^{2} - 8 x + 16

= - \frac{1}{8} (x^{2} - 8 x + 16) (x^{2} + 4 x + 4)

Expand

(x^{2} - 8 x + 16) (x^{2} + 4 x + 4) : x^{4} - 4 x^{3} - 12 x^{2} + 32 x + 64

= - \frac{1}{8} (x^{4} - 4 x^{3} - 12 x^{2} + 32 x + 64)

Distribute parentheses

= - \frac{1}{8} x^{4} - \frac{1}{8} (- 4 x^{3}) - \frac{1}{8} (- 12 x^{2}) - \frac{1}{8} \cdot 32 x - \frac{1}{8} \cdot 64

Simplify

- \frac{1}{8} x^{4} - \frac{1}{8} (- 4 x^{3}) - \frac{1}{8} (- 12 x^{2}) - \frac{1}{8} \cdot 32 x - \frac{1}{8} \cdot 64 : - \frac{x ^{4}}{8} + \frac{x ^{3}}{2} + \frac{3 x ^{2}}{2} - 4 x - 8

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

(1 - \frac{1}{x ^{2}}) \cdot (2 + x)

(2 - 2)^{5}

f (12 x - 6) x^{2} d x

\frac{d}{d t} ((θ) (θ^{2} - t^{2})^{2})

\frac{\partial}{\partial x} ((z) (t x^{2} + yz - 2))

What is expand-1/8 (x+2)^2(x-4)^2 ?
The solution to expand-1/8 (x+2)^2(x-4)^2 is -(x^4)/8+(x^3)/2+(3x^2)/2-4x-8