Find number of real roots of the polynomial $x^3+7x^2+6x+5$. I want to find the number of real roots of the polynomial $x^3+7x^2+6x+5$.
Using Descartes rule, this polynomial has either only one real root or 3 real roots (all are negetive). How will we conclude one answer without doing some long process?
I note that this is "close" to $$(x+5)(x+1)(x+1)=x^3+7x^2+11x+5$$ which has a repeated root at $-1$, and another root at $-5$. The repeated root at $-1$ is a local minimum, considering the general shape of a cubic with positive leading coefficient.

So you have $$(x+5)(x+1)(x+1)-5x$$

Adding that $-5x$ is going to push the local minimum upward, since $-5x$ is positive near $-1$. The doubled root will be perturbed into two non-real complex conjugate roots. And only the perturbed root near $-5$ will still be real.
If you actually want the roots of a cubic, you can use the rational root test to find one of the roots. After you have one of the roots you can easily do a long division to find the resulting second degree polynomial from there you just apply the quadratic formula. This will yield the exact roots, but if you want to just know if you have three real or one real and two complex that is a different scenario.