What is expand ((4z+3)(2z-1))/(72) ?

The solution to expand ((4z+3)(2z-1))/(72) is (z^2)/9+z/(36)-1/24

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expand ((4z+3)(2z-1))/(72)

expand

\frac{( 4 z + 3 ) ( 2 z - 1 )}{72}

\frac{z ^{2}}{9} + \frac{z}{36} - \frac{1}{24}

Solution steps

\frac{( 4 z + 3 ) ( 2 z - 1 )}{72}

Expand

(4 z + 3) (2 z - 1) : 8 z^{2} + 2 z - 3

= \frac{8 z ^{2} + 2 z - 3}{72}

Apply the fraction rule:

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

\frac{8 z ^{2} + 2 z - 3}{72} = \frac{8 z ^{2}}{72} + \frac{2 z}{72} - \frac{3}{72}

= \frac{8 z ^{2}}{72} + \frac{2 z}{72} - \frac{3}{72}

Cancel

\frac{8 z ^{2}}{72} : \frac{z ^{2}}{9}

= \frac{z ^{2}}{9} + \frac{2 z}{72} - \frac{3}{72}

Cancel

\frac{2 z}{72} : \frac{z}{36}

= \frac{z ^{2}}{9} + \frac{z}{36} - \frac{3}{72}

Cancel

\frac{3}{72} : \frac{1}{24}

= \frac{z ^{2}}{9} + \frac{z}{36} - \frac{1}{24}

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

(4 x + 1) e^{5 x}

(x - 0.5 y + 1 = 0,) (2 x - y + 6 = 0)

ln (x) (x^{2} - 1)

(2 + 3 x - 4 x^{2})^{5}

What is expand ((4z+3)(2z-1))/(72) ?
The solution to expand ((4z+3)(2z-1))/(72) is (z^2)/9+z/(36)-1/24