What is expand 4/((s-3)(s^2+9)) ?

The solution to expand 4/((s-3)(s^2+9)) is 4/(s^3+9s-3s^2-27)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

expand 4/((s-3)(s^2+9))

expand

\frac{4}{( s - 3 ) ( s ^{2} + 9 )}

\frac{4}{s ^{3} + 9 s - 3 s ^{2} - 27}

Solution steps

\frac{4}{( s - 3 ) ( s ^{2} + 9 )}

Expand

(s - 3) (s^{2} + 9) : s^{3} + 9 s - 3 s^{2} - 27

= \frac{4}{s ^{3} + 9 s - 3 s ^{2} - 27}

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

\frac{7 ( y - 4 )}{6 ( y + 4 )}

(6 x y) d x + (4 y + 9 x^{2}) d y

(y^{3} - y^{2} sin (x) - x) d x

(x^{2} + 4 x + 3 \circ x + 3) (3)

What is expand 4/((s-3)(s^2+9)) ?
The solution to expand 4/((s-3)(s^2+9)) is 4/(s^3+9s-3s^2-27)