According to Descartes's rule of sign, how many possible positive and negative roots are there for the equation
0= -4x6-3x5+2x2-4x+1?

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Happil

Descartes Rule of Sign

Answer:

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

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