Finding the number of real roots of a polynomial

Question

I want to find the number of real roots of $f(t)=t^4-2t^2+4t+1$. As $f(0)=1>0, f(-1)=-4 <0$ and $f(-2)=1>0$, I can say that there are two real roots since the polynomial is continuous. For the rest of the roots (which are complex), I think Rolle's theorem may be used but I could not find a way to show it. How can I proceed?

Answer 1

The derivative is given by $f'(t)=4(t^3-t+1)$ so roots arise from the cubic polynomial $g(t):=t^3-t+1$. Such a polynomial has a nice formula for the discriminant. You should be able to analyze the number of real roots of the derivative. You have figured out that there are at least two real roots of $f$ due to a local minimum dipping below the $x$-axis. The information about the derivative should allow you to conclude more about the shape of $f$.

Answer 2

Well, there are no non-negative roots as $f(t)=(t^2-1)^2+4t$, and Descartes rule of signs says there are either $2$ or no negative roots. As you observed, there are indeed two negative roots as $f(-1)=-4<0$, so there are no more real roots.