How many real solutions does the equation `x^7+14 x^5+16 x^3+30 x-560=0` have?

Explanation

Here,
`f(x) = x^(7) + 14x^(5) + 16x^(3) + 30x - 560`
Hence the degree is odd, so at least one solution is real.
Now `f'(x) = 7x^(6) + 70x^(4)+ 48x^(2) + 30 ge 30 AA x in R`
Thus, y = f(x) has no turning point. Also, when x `to infty`, f(x) `toinfty` and when x `to-infty,f(x) to -infty`.
So, graph corsses only once. Therefore, f(x) = 0 has only one root.