What is expand ((k^{14}+4k^7+3))/(2k^{11)+2k^4} ?

The solution to expand ((k^{14}+4k^7+3))/(2k^{11)+2k^4} is

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expand ((k^{14}+4k^7+3))/(2k^{11)+2k^4}

expand

\frac{( k ^{14} + 4 k ^{7} + 3 )}{2 k ^{11} + 2 k ^{4}}

Solution steps

\frac{k ^{14} + 4 k ^{7} + 3}{2 k ^{11} + 2 k ^{4}}

Factor

2 k^{11} + 2 k^{4} : 2 k^{4} (k + 1) (k^{6} - k^{5} + k^{4} - k^{3} + k^{2} - k + 1)

Cancel

Expand

Apply the fraction rule:

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

\frac{k ^{7}}{2 k ^{4}} = \frac{k ^{3}}{2}

\frac{3 ^{\frac{3}{7}} k ^{4}}{2 k ^{4}} = \frac{3 ^{\frac{3}{7}}}{2}

\frac{3 ^{\frac{4}{7}} k ^{3}}{2 k ^{4}} = \frac{3 ^{\frac{4}{7}}}{2 k}

\frac{3 ^{\frac{5}{7}} k ^{2}}{2 k ^{4}} = \frac{3 ^{\frac{5}{7}}}{2 k ^{2}}

\frac{3 ^{\frac{6}{7}} k}{2 k ^{4}} = \frac{3 ^{\frac{6}{7}}}{2 k ^{3}}

2 x e^{x^{2} - y^{2}} + 2 x (x^{2} + y^{2}) e^{x^{2} - y^{2}}

\frac{\partial}{\partial x} ((z) (z^{3} - z - x y))

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

\frac{d}{d x} \frac{( φ ( x ^{2} + 2 y ( φ , x ) ) + ψ ( φ , x , y ( φ , x ) ) ( x - 2 y ( φ , x ) ^{2} ) )}{x}

\frac{e ^{- 2 π s}}{( s ^{2} + 1 ) \cdot ( s ^{2} - s + 4 )}

What is expand ((k^{14}+4k^7+3))/(2k^{11)+2k^4} ?
The solution to expand ((k^{14}+4k^7+3))/(2k^{11)+2k^4} is