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

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

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expand-(x(2x-3)(2x^2+3x+2))/2

expand

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

- 2 x^{4} + \frac{5 x ^{2}}{2} + 3 x

Solution steps

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

Expand

x (2 x - 3) (2 x^{2} + 3 x + 2) : 4 x^{4} - 5 x^{2} - 6 x

= - \frac{4 x ^{4} - 5 x ^{2} - 6 x}{2}

Apply the fraction rule:

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

\frac{4 x ^{4} - 5 x ^{2} - 6 x}{2} = - (\frac{4 x ^{4}}{2}) - (- \frac{5 x ^{2}}{2}) - (- \frac{6 x}{2})

= - (\frac{4 x ^{4}}{2}) - (- \frac{5 x ^{2}}{2}) - (- \frac{6 x}{2})

Remove parentheses:

(a) = a, - (- a) = a

= - \frac{4 x ^{4}}{2} + \frac{5 x ^{2}}{2} + \frac{6 x}{2}

Divide the numbers:

\frac{4}{2} = 2

= - 2 x^{4} + \frac{5 x ^{2}}{2} + \frac{6 x}{2}

Divide the numbers:

\frac{6}{2} = 3

= - 2 x^{4} + \frac{5 x ^{2}}{2} + 3 x

(3 x - 8 y - 18 = 0,) (3 x - 8 y + 1 = 0)

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

(- 2 + 6 i)^{4}

(1 - \frac{( x ^{2} )}{2}) \cdot (x - \frac{( x ^{3} )}{6})

l im x \Rightarrow 0 - (cos (2 x) + x^{2}) cot (3 x)

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