What is expand (4(x^{0.5}-3)^2)/x ?

The solution to expand (4(x^{0.5}-3)^2)/x is 4-(24}{x^{1/2}}+\frac{36)/x

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expand (4(x^{0.5}-3)^2)/x

expand

\frac{4 ( x ^{0.5} - 3 ) ^{2}}{x}

4 - \frac{24}{x ^{\frac{1}{2}}} + \frac{36}{x}

Solution steps

\frac{4 ( x ^{\frac{1}{2}} - 3 ) ^{2}}{x}

(x^{\frac{1}{2}} - 3)^{2} = x - 6 x^{\frac{1}{2}} + 9

= \frac{4 ( x - 6 x ^{\frac{1}{2}} + 9 )}{x}

Expand

4 (x - 6 x^{\frac{1}{2}} + 9) : 4 x - 24 x^{\frac{1}{2}} + 36

= \frac{4 x - 24 x ^{\frac{1}{2}} + 36}{x}

Apply the fraction rule:

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

\frac{4 x - 24 x ^{\frac{1}{2}} + 36}{x} = \frac{4 x}{x} - \frac{24 x ^{\frac{1}{2}}}{x} + \frac{36}{x}

= \frac{4 x}{x} - \frac{24 x ^{\frac{1}{2}}}{x} + \frac{36}{x}

Cancel

\frac{4 x}{x} : 4

= 4 - \frac{24 x ^{\frac{1}{2}}}{x} + \frac{36}{x}

Cancel

\frac{24 x ^{\frac{1}{2}}}{x} : \frac{24}{x ^{\frac{1}{2}}}

= 4 - \frac{24}{x ^{\frac{1}{2}}} + \frac{36}{x}

s^{2} - k (1 - e^{s T})

d x (2 x^{3} - 9 x^{2} + 9 x - 2)

(4 x - 7) d x

2 (2 x + 8)^{7} d x

\frac{1}{( 2 k + 1 ) ( 2 k + 3 )}

What is expand (4(x^{0.5}-3)^2)/x ?
The solution to expand (4(x^{0.5}-3)^2)/x is 4-(24}{x^{1/2}}+\frac{36)/x