Question

use descartes rule of signs to find how many possible negative real roots this polynomial has:
$p(x)=4x^{5}-9x^{4}-x^{3}+8x^{2}+x$
4, 2 or 0
1
2 or 0

Explanation:

Step1: Substitute $x$ with $-x$

$$P(-x) = 4(-x)^5 - 9(-x)^4 - (-x)^3 + 8(-x)^2 + (-x)$$

Step2: Simplify $P(-x)$

$$P(-x) = -4x^5 - 9x^4 + x^3 + 8x^2 - x$$

Step3: Count sign changes

Sign sequence: $-, -, +, +, -$. There are 2 sign changes.

Step4: Apply Descartes' Rule

The number of negative real roots is equal to the number of sign changes or that number minus an even integer. So $2$ or $2-2=0$.

Answer:

2 or 0