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

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

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

simplify

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

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

Solution steps

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

Factor

k^{14} + 4 k^{7} + 3 : (k + 1) (k^{6} - k^{5} + k^{4} - k^{3} + k^{2} - k + 1) (k^{7} + 3)

= \frac{( k + 1 ) ( k ^{6} - k ^{5} + k ^{4} - k ^{3} + k ^{2} - k + 1 ) ( 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)

= \frac{( k + 1 ) ( k ^{6} - k ^{5} + k ^{4} - k ^{3} + k ^{2} - k + 1 ) ( k ^{7} + 3 )}{2 k ^{4} ( k + 1 ) ( k ^{6} - k ^{5} + k ^{4} - k ^{3} + k ^{2} - k + 1 )}

Cancel the common factor:

k + 1

= \frac{( k ^{6} - k ^{5} + k ^{4} - k ^{3} + k ^{2} - k + 1 ) ( k ^{7} + 3 )}{2 k ^{4} ( k ^{6} - k ^{5} + k ^{4} - k ^{3} + k ^{2} - k + 1 )}

Cancel the common factor:

k^{6} - k^{5} + k^{4} - k^{3} + k^{2} - k + 1

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

- x^{3} + 2 x^{2} + 13 x + 10 > 0

- 5 + 6 ∣ 2 x + 2 ∣ = 7

x + 4 = 5

\frac{ln ( \frac{1}{2} )}{1620}

177147 = 54 ln (3) (2) + b

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