# Determine the number of real solutions of an equation

I need to calculate the number of real solutions in a polynomial equation. From the Fundamental Theorem of Algebra, I know that an equation of degree n has exactly n roots (but, here the complex ones are also included). And I just need to determine how many real ones.

For example, I am asked to the following equation:

$$3x^5+15x-8=0$$

From the Fundamental Algebra Theorem, it is known to have 5 roots (real and complex). Next, I have made a sketch of the function and it comes out that it only cuts the X-axis once between zero and one. Then, this equation would have a real solution and 4 complex solutions. But is there any way to solve it analytically without having to graph the function?

• Hint: will have at least one real root because of odd degree of the polynomial. Then consider the derivative $15(x^4+1) >0$, so the function is increasing. – Anurag A Dec 25 '18 at 8:58
• According to en.wikipedia.org/wiki/Descartes%27_rule_of_signs there's exactly one and that zero is positive. – Michael Hoppe Dec 25 '18 at 19:00

Let $$P(x)=3x^5+15x-8$$. Since $$\lim_{x\to\infty}P(x)=\infty$$ and $$\lim_{x\to-\infty}P(x)=-\infty$$, it follows from the intermediate value theorem that $$P(x)$$ has at least one real root. Now, imagine that it has two distinct roots $$x_1$$ and $$x_2$$. Then, by Rolle's theorem, there is some $$y$$ between $$x_1$$ and $$x_2$$ such that $$P'(y)=0$$. But $$P'(y)=15y^4+15$$, and therefore it can't be $$0$$.