What is expand (2(x^4-3x^3+1))/(x^2) ?

The solution to expand (2(x^4-3x^3+1))/(x^2) is 2x^2-6x+2/(x^2)

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expand (2(x^4-3x^3+1))/(x^2)

expand

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

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

Solution steps

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

Expand

2 (x^{4} - 3 x^{3} + 1) : 2 x^{4} - 6 x^{3} + 2

= \frac{2 x ^{4} - 6 x ^{3} + 2}{x ^{2}}

Apply the fraction rule:

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

\frac{2 x ^{4} - 6 x ^{3} + 2}{x ^{2}} = \frac{2 x ^{4}}{x ^{2}} - \frac{6 x ^{3}}{x ^{2}} + \frac{2}{x ^{2}}

= \frac{2 x ^{4}}{x ^{2}} - \frac{6 x ^{3}}{x ^{2}} + \frac{2}{x ^{2}}

Cancel

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

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

Cancel

\frac{6 x ^{3}}{x ^{2}} : 6 x

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

sin (\frac{( 2 k + 1 ) π}{2})

\frac{( 32 y + 1 ) ^{2}}{16}

(x^{3} + 2 x) e^{5} x - 1

\frac{1 - e ^{- 7 s}}{s ( s ^{2} + 2 s + 5 )}

\frac{1}{x ( 1 + b x )}

What is expand (2(x^4-3x^3+1))/(x^2) ?
The solution to expand (2(x^4-3x^3+1))/(x^2) is 2x^2-6x+2/(x^2)