What is expand ((x+1)(x-3)(x-5))/9 ?

The solution to expand ((x+1)(x-3)(x-5))/9 is (x^3)/9-(7x^2)/9+(7x)/9+5/3

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expand ((x+1)(x-3)(x-5))/9

expand

\frac{( x + 1 ) ( x - 3 ) ( x - 5 )}{9}

\frac{x ^{3}}{9} - \frac{7 x ^{2}}{9} + \frac{7 x}{9} + \frac{5}{3}

Solution steps

\frac{( x + 1 ) ( x - 3 ) ( x - 5 )}{9}

Expand

(x + 1) (x - 3) (x - 5) : x^{3} - 7 x^{2} + 7 x + 15

= \frac{x ^{3} - 7 x ^{2} + 7 x + 15}{9}

Apply the fraction rule:

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

\frac{x ^{3} - 7 x ^{2} + 7 x + 15}{9} = \frac{x ^{3}}{9} - \frac{7 x ^{2}}{9} + \frac{7 x}{9} + \frac{15}{9}

= \frac{x ^{3}}{9} - \frac{7 x ^{2}}{9} + \frac{7 x}{9} + \frac{15}{9}

Cancel

\frac{15}{9} : \frac{5}{3}

= \frac{x ^{3}}{9} - \frac{7 x ^{2}}{9} + \frac{7 x}{9} + \frac{5}{3}

A (- 5, 1) B (1, - 3)

(x - \frac{b}{2 a})^{2} - b (x - \frac{b}{2 a}) + c

\frac{6 ^{7}}{6 ^{4}}

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

(- 3)^{4}

What is expand ((x+1)(x-3)(x-5))/9 ?
The solution to expand ((x+1)(x-3)(x-5))/9 is (x^3)/9-(7x^2)/9+(7x)/9+5/3