What is expand (4(x+1)(2x+1))/(3x^2) ?

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

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expand (4(x+1)(2x+1))/(3x^2)

expand

\frac{4 ( x + 1 ) ( 2 x + 1 )}{3 x ^{2}}

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

Solution steps

\frac{4 ( x + 1 ) ( 2 x + 1 )}{3 x ^{2}}

Expand

4 (x + 1) (2 x + 1) : 8 x^{2} + 12 x + 4

= \frac{8 x ^{2} + 12 x + 4}{3 x ^{2}}

Apply the fraction rule:

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

\frac{8 x ^{2} + 12 x + 4}{3 x ^{2}} = \frac{8 x ^{2}}{3 x ^{2}} + \frac{12 x}{3 x ^{2}} + \frac{4}{3 x ^{2}}

= \frac{8 x ^{2}}{3 x ^{2}} + \frac{12 x}{3 x ^{2}} + \frac{4}{3 x ^{2}}

Cancel

\frac{8 x ^{2}}{3 x ^{2}} : \frac{8}{3}

= \frac{8}{3} + \frac{12 x}{3 x ^{2}} + \frac{4}{3 x ^{2}}

Cancel

\frac{12 x}{3 x ^{2}} : \frac{4}{x}

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

\frac{- s ^{2} + 1}{s ^{2} ( s + 2 )}

\frac{s}{( s ^{2} + 2 s + 10 )}

x (- 1 - 18 y^{3} + ln (3) \cdot 3^{y} x y)

(1 - 5 x + 4 x^{2})^{5}

\frac{d}{d θ} ((t) (2 - cos (θ)))

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