What is expand (3(x-1)(x-4)(x-5))/4 ?

The solution to expand (3(x-1)(x-4)(x-5))/4 is (3x^3)/4-(15x^2)/2+(87x)/4-15

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expand (3(x-1)(x-4)(x-5))/4

expand

\frac{3 ( x - 1 ) ( x - 4 ) ( x - 5 )}{4}

\frac{3 x ^{3}}{4} - \frac{15 x ^{2}}{2} + \frac{87 x}{4} - 15

Solution steps

\frac{3 ( x - 1 ) ( x - 4 ) ( x - 5 )}{4}

Expand

3 (x - 1) (x - 4) (x - 5) : 3 x^{3} - 30 x^{2} + 87 x - 60

= \frac{3 x ^{3} - 30 x ^{2} + 87 x - 60}{4}

Apply the fraction rule:

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

\frac{3 x ^{3} - 30 x ^{2} + 87 x - 60}{4} = \frac{3 x ^{3}}{4} - \frac{30 x ^{2}}{4} + \frac{87 x}{4} - \frac{60}{4}

= \frac{3 x ^{3}}{4} - \frac{30 x ^{2}}{4} + \frac{87 x}{4} - \frac{60}{4}

Cancel

\frac{30 x ^{2}}{4} : \frac{15 x ^{2}}{2}

= \frac{3 x ^{3}}{4} - \frac{15 x ^{2}}{2} + \frac{87 x}{4} - \frac{60}{4}

Divide the numbers:

\frac{60}{4} = 15

= \frac{3 x ^{3}}{4} - \frac{15 x ^{2}}{2} + \frac{87 x}{4} - 15

\frac{y ^{2} + 1}{y \cdot ( 2 - x )}

\frac{2 ^{n}}{n ( n + 1 )}

\frac{( 1 + 2 ) π}{4}

((2 y - x^{3}) d x + x d g (y, x) = 0) x

16 π (\frac{16}{3} - 8 ln (2))

What is expand (3(x-1)(x-4)(x-5))/4 ?
The solution to expand (3(x-1)(x-4)(x-5))/4 is (3x^3)/4-(15x^2)/2+(87x)/4-15