Question

According to Descartes's rule of sign, how many possible positive and negative roots are there for the equation

Answer 1

There are 3 or 1 postive root, and 1 negative root.

Answer 2

Descartes Rule of Sign

Answer:

There are $3$ Real Roots for the Equation

Step-by-step explanation:

According to Descartes Rule of Sign, The maximum number of real roots a polynomial equation has is the number of times the sign of term changes from a higher power term to a lower power term.

In the given equation, $0=-4x^6-3x^5+2x^2-4x+1$, we can already see that from $\red -4x^6$ to $\red -3x^5$ we can see that the sign didn't change so there might be no real roots. From $\red -3x^5$ to $\blue +2x^2$, we can see that the signs change so we now know that there might be $1$ real root for equation. From $\blue +2x^2$ to $\red -4x$, the signs changed so there might be $2$ real roots. From $\red -4x$ to $\blue +1$, it changed again so there might be $3$ real roots. There are no other term with lesser power to compare for $\blue +1$ so we are left with $3$ real roots.