Question

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?Drag the choices to the boxes to correctly complete the table.

Answer 1

According to Descartes’ Rule of Signs, if we let

$f(x)=a_nx^n+a_{n-1}x^{n-1}+\cdot\cdot\cdot+a_1x+a_0$

be a polynomial function with real coefficients:

then,

the number of positive real zeros is either equal to the number of sign changes of f(x) or is less than the number of sign changes by an even integer.

the number of negative real zeros is either equal to the number of sign changes of f(-x) or is less than the number of sign changes by an even integer.

In this case,

$f(x)=-4x^6-3x^5+2x^2-4x+1$

We will now check the number of sign changes of f(x)

There are 3 sign changes.

Therefore, f(x) has 3 or 1 positive real roots

Next we find f(-x),

$\begin{gathered} f(-x)=-4(-x)^6-3(-x)^5+2(-x)^2-4(-x)+1 \\ \text{hence,} \\ f(-x)=-4x^6+3x^5+2x^2+4x+1 \end{gathered}$

We will now check the number of sign changes of f(-x)

There are 1 sign changes.

Therefore, f(x) has 1 negative real root

Number of possible positive roots: 1 or 3

Number of possible negative roots: 1 only