How many Real solution does the equation x^7 + 14x^5 + 16x^3 + 30x = 0 have ?

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How many Real solution does the equation x^7 + 14x^5 + 16x^3 + 30x = 0 have ?

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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x(x^6+14x^4+16x^2+30)=0 so x=0 or x^6+14x^4+16x^2+30=0 but if x is a real number x^6+14x^4+16x^2+30>0 so it has one solution: x=0
One at x = 0. Factor this out and move on. If we define the LHS (with x factored out) as f(x), all coefficient signs are positive. There cannot be a positive Real Solution. No sign change if f(-x), either. We are done. x = 0 is the only Real Solution. marsss identical conclusion is much simpler, but less general.
i explained the solution of this question so it doesn't need to be general, @tkhunny

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@marss Absolutely agree with your comment. I just wanted the rest along for the ride. Not a complaint or criticism at all. :-)

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