Question

use descartes rule of signs to find how many possible negative real roots this polynomial has:
$p(x) = 5x^5 - 8x^4 + 3x^3 - 7x^2 + 2x$
3 or 1
4, 2 or 0
0

Explanation:

Step1: Substitute $x$ with $-x$

$P(-x) = 5(-x)^5 - 8(-x)^4 + 3(-x)^3 - 7(-x)^2 + 2(-x)$

Step2: Simplify each term

$P(-x) = -5x^5 - 8x^4 - 3x^3 - 7x^2 - 2x$

Step3: Count sign changes

No sign changes between consecutive terms.

Step4: Apply Descartes' Rule

Number of negative real roots = number of sign changes (or subtract even integers). Since there are 0 sign changes, there are 0 possible negative real roots.

Answer:

0