What is expand-((2+sqrt(x+4))sqrt(x^2-x-2))/x ?

The solution to expand-((2+sqrt(x+4))sqrt(x^2-x-2))/x is -(2sqrt(x^2-x-2))/x-(sqrt(x^3+3x^2-6x-8))/x

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expand-((2+sqrt(x+4))sqrt(x^2-x-2))/x

expand

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

- \frac{2 x ^{2} - x - 2}{x} - \frac{x ^{3} + 3 x ^{2} - 6 x - 8}{x}

Solution steps

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

Expand

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

= - \frac{2 x ^{2} - x - 2 + x ^{3} + 3 x ^{2} - 6 x - 8}{x}

Apply the fraction rule:

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

\frac{2 x ^{2} - x - 2 + x ^{3} + 3 x ^{2} - 6 x - 8}{x} = - (\frac{2 x ^{2} - x - 2}{x}) - (\frac{x ^{3} + 3 x ^{2} - 6 x - 8}{x})

= - (\frac{2 x ^{2} - x - 2}{x}) - (\frac{x ^{3} + 3 x ^{2} - 6 x - 8}{x})

Remove parentheses:

(a) = a

= - \frac{2 x ^{2} - x - 2}{x} - \frac{x ^{3} + 3 x ^{2} - 6 x - 8}{x}

\frac{i ( - 1 ) ^{n}}{2 nπ} + \frac{( - 1 ) ^{n} - 1}{2 n ^{2} π ^{2}}

1537.6 \cdot ((s + 0.6) \cdot (s + 0.8))

\frac{( S ^{3} - 3 S ^{2} + 6 S - 4 )}{( S ^{2} - 2 S + 2 ) ^{2}}

\frac{x ( 2 x - 1 )}{x + 1}

(- 7, 1, - 5) t + (3, - 6, - 8) (1 - t)

What is expand-((2+sqrt(x+4))sqrt(x^2-x-2))/x ?
The solution to expand-((2+sqrt(x+4))sqrt(x^2-x-2))/x is -(2sqrt(x^2-x-2))/x-(sqrt(x^3+3x^2-6x-8))/x