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konrad509
[tex]5x^3+4x-2=0\\\\ (5x^3+4x-2)'=15x^2+4\\\\ 15x^2+4=0\\ 15x^2=-4\\ x^2=-\dfrac{4}{15}\\ x\in\emptyset [/tex]

The derivative of [tex]5x^3+4x-2[/tex] is positive for all real numbers, which means the function is increasing in its all domain and therefore there is only one intersection with the x-axis ⇒ one solution.

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