What is expand-(5pi(x+2)(x-2)(3x^2-20))/(16) ?

The solution to expand-(5pi(x+2)(x-2)(3x^2-20))/(16) is -(15pix^4)/(16)+10pix^2-25pi

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

expand-(5pi(x+2)(x-2)(3x^2-20))/(16)

expand

- \frac{5 π ( x + 2 ) ( x - 2 ) ( 3 x ^{2} - 20 )}{16}

- \frac{15 π x ^{4}}{16} + 10 π x^{2} - 25 π

Solution steps

- \frac{5 π ( x + 2 ) ( x - 2 ) ( 3 x ^{2} - 20 )}{16}

Expand

5 π (x + 2) (x - 2) (3 x^{2} - 20) : 15 π x^{4} - 160 π x^{2} + 400 π

= - \frac{15 π x ^{4} - 160 π x ^{2} + 400 π}{16}

Apply the fraction rule:

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

\frac{15 π x ^{4} - 160 π x ^{2} + 400 π}{16} = - (\frac{15 π x ^{4}}{16}) - (- \frac{160 π x ^{2}}{16}) - (\frac{400 π}{16})

= - (\frac{15 π x ^{4}}{16}) - (- \frac{160 π x ^{2}}{16}) - (\frac{400 π}{16})

Remove parentheses:

(a) = a, - (- a) = a

= - \frac{15 π x ^{4}}{16} + \frac{160 π x ^{2}}{16} - \frac{400 π}{16}

Divide the numbers:

\frac{160}{16} = 10

= - \frac{15 π x ^{4}}{16} + 10 π x^{2} - \frac{400 π}{16}

Divide the numbers:

\frac{400}{16} = 25

= - \frac{15 π x ^{4}}{16} + 10 π x^{2} - 25 π

(x + L \cdot cos (T H))^{2}

2 (e^{3 y x} + 3 y e^{3 y x} x) + 2 y x - y^{3}

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

(1 + u^{\frac{1}{2}})^{3}

cos^{2} (t) (1 - t) (7 - t^{3})

What is expand-(5pi(x+2)(x-2)(3x^2-20))/(16) ?
The solution to expand-(5pi(x+2)(x-2)(3x^2-20))/(16) is -(15pix^4)/(16)+10pix^2-25pi