Question

use descartes rule of signs to find how many possible negative real roots this polynomial has:
$p(x) = -2x^5 + 3x^4 - 9x^3 + x^2 - 7x$
4, 2 or 0
0
3 or 1

Explanation:

Step1: Substitute $x$ with $-x$

$$\begin{align*} P(-x)&=-2(-x)^5 + 3(-x)^4 - 9(-x)^3 + (-x)^2 - 7(-x)\\ &=2x^5 + 3x^4 + 9x^3 + x^2 + 7x \end{align*}$$

Step2: Count sign changes in $P(-x)$

There are 0 sign changes between consecutive terms of $P(-x)$.

Step3: Apply Descartes' Rule

For negative real roots, the number of possible negative real roots is equal to the number of sign changes in $P(-x)$, or less by an even integer. Since there are 0 sign changes, the only possible number of negative real roots is 0.

Answer:

0