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

Let $f(x)=x^7+14x^5+16x^3+30x-560$.Find the derivative:$f^{\prime }(x)=7x^6+70x^4+48x^2+30$.Note that all terms in $f^{\prime }(x)$ contain even powers of $x$ with positive coefficients, and the constant term is positive.Thus, $f^{\prime }(x)>0$ for all real $x$.This means $f(x)$ is a strictly increasing function.A strictly increasing function can cross the $x$ -axis at most once.As $x\to -\infty ,f(x)\to -\infty$ and as $x\to \infty ,f(x)\to \infty$.By intermediate value theorem, there is exactly one real solution.