Question

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

Answer 1

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

Answer 2

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.

Answer 3

i explained the solution of this question so it doesn't need to be general, @tkhunny

Answer 4

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