What is expand ((3c+4)(3c-2))/(144) ?

The solution to expand ((3c+4)(3c-2))/(144) is (c^2)/(16)+c/(24)-1/18

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expand ((3c+4)(3c-2))/(144)

expand

\frac{( 3 c + 4 ) ( 3 c - 2 )}{144}

\frac{c ^{2}}{16} + \frac{c}{24} - \frac{1}{18}

Solution steps

\frac{( 3 c + 4 ) ( 3 c - 2 )}{144}

Expand

(3 c + 4) (3 c - 2) : 9 c^{2} + 6 c - 8

= \frac{9 c ^{2} + 6 c - 8}{144}

Apply the fraction rule:

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

\frac{9 c ^{2} + 6 c - 8}{144} = \frac{9 c ^{2}}{144} + \frac{6 c}{144} - \frac{8}{144}

= \frac{9 c ^{2}}{144} + \frac{6 c}{144} - \frac{8}{144}

Cancel

\frac{9 c ^{2}}{144} : \frac{c ^{2}}{16}

= \frac{c ^{2}}{16} + \frac{6 c}{144} - \frac{8}{144}

Cancel

\frac{6 c}{144} : \frac{c}{24}

= \frac{c ^{2}}{16} + \frac{c}{24} - \frac{8}{144}

Cancel

\frac{8}{144} : \frac{1}{18}

= \frac{c ^{2}}{16} + \frac{c}{24} - \frac{1}{18}

(ln (x) + 1) \cdot (2 ln (x) - 1)

(d x - 4) 3

(- 6, - 10) y (- 2, - 2)

3 (x - 1 = 2 x + 9)

(a - 2) (a + 3)^{''}

What is expand ((3c+4)(3c-2))/(144) ?
The solution to expand ((3c+4)(3c-2))/(144) is (c^2)/(16)+c/(24)-1/18