Question

Use Descartes' Rule of Signs to find the possible number of negative zeros of p(x)=2x5+x4+x3−4x2−x−6

Answer 1

$$P(x)=2x^5+x^4+x^3−4x^2−x−6$$

The coefficients are $2,1,1,−4,−1,−6.$

There is one change. This means that there is one positive real root.

To find the number of negative real roots, substitute $x$  with $-x$ in the given polynomial: $2x^5+x^4+x^3−4x^2−x−6$ becomes $-2x^5+x^4-x^3−4x^2+x−6$

The coefficients are $-2,1,-1,−4,1,−6.$

There are four changes. This means that there are 4 or 2 or 0 negative real roots.

1 positive real root.

4 or 2 or 0 negative real roots.