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

Hint: To find the number of solutions we will go with the concept of differentiation. So we will differentiate the given function. The number of zeroes the first derivative will give, then according to concept, is the number of real solutions the given function consists of. And if its first derivative does not give zero for any value, and it is either always increasing or decreasing in all real numbers then the given function will have only one root.

Complete step-by-step solution:
Moving ahead with the question in step wise manner, we have ${{x}^{7}}+14{{x}^{5}}+16{{x}^{3}}+30x-560=0$.
So by using the concept of differentiation, which says that the number of zeroes the first derivative will give, then according to the concept than the number of real solutions the given function consists. And if its first derivative does not give zero for any value, and it is either always increasing or decreasing in all real numbers then the given function will have only one root.
So moving with the concept let us first differentiate the given function, so we will get;
\[\Rightarrow \dfrac{dy}{dx}=7{{x}^{6}}+70{{x}^{4}}+48{{x}^{2}}+30\]
Now to find out the zeroes of the above first derivative, is also somewhat difficult as it has a power ‘6’ function. But we can use the logic to find out the zeroes. That is, in the first derivative the power of ‘x’ is even, and all terms are in the addition, which means whether we put any value of ‘x’ it will always give positive value, and then they all are in addition, which means the value will never decrease always it will come greater than zero.
So we can say that whatever the value we will keep in the first derivative it will always give greater than zero, which means the first derivative will always be greater than zero, it will never give zero. Means the first derivative will not have any zero. In fact it will always be increasing in all real numbers.
So by the concept we can say that the main function will only have one root, as the first derivative is always increasing in all real numbers.
Hence the answer is one root, i.e. the function will have only one root.

Note: It can happen that the first derivative can be continuously increasing or decreasing in some interval rather than in all real numbers, then we cannot say that it will have only one root. Then we have to find out the zeroes.