$$\frac{k^2+2k-8}{3k^3} = \frac{1}{3k^2}+\frac{1}{k^2}$$
Answer
$$ \begin{matrix}k_1 = -0.73459 & k_2 = 2.12755 & k_3 = -3.94925 \\[1 em] k_4 = 0.27815+0.63929i & k_5 = 0.27815-0.63929i \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} \frac{k^2+2k-8}{3k^3} &= \frac{1}{3k^2}+\frac{1}{k^2}&& \text{multiply ALL terms by } \color{blue}{ 3k^3k^2 }. \\[1 em]3k^3k^2\frac{k^2+2k-8}{3k^3} &= 3k^3k^2\cdot\frac{1}{3k^2}+3k^3k^2\cdot\frac{1}{k^2}&& \text{cancel out the denominators} \\[1 em]k^4+2k^3-8k^2 &= k+\frac{3}{k^1}&& \text{multiply ALL terms by } \color{blue}{ k^1 }. \\[1 em]k^1\cdot1k^4+k^1\cdot2k^3-k^1\cdot8k^2 &= k^1\cdot1k+k^1\cdot\frac{3}{k^1}&& \text{cancel out the denominators} \\[1 em]k^5+2k^4-8k^3 &= k^2+3&& \text{move all terms to the left hand side } \\[1 em]k^5+2k^4-8k^3-k^2-3 &= 0&& \\[1 em] \end{aligned} $$
This polynomial has no rational roots that can be found using Rational Root Test.
Roots were found using Newton method.
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