Question

use descartes rule of signs to find how many possible negative real roots this polynomial has:
$p(x) = -7x^4 + 3x^3 - 9x^2 + x + 6$
3 or 1
2 or 0
1

Explanation:

Step1: Substitute $x$ with $-x$

$$P(-x) = -7(-x)^4 + 3(-x)^3 - 9(-x)^2 + (-x) + 6$$

Step2: Simplify the polynomial

$$P(-x) = -7x^4 - 3x^3 - 9x^2 - x + 6$$

Step3: Count sign changes

Count sign changes in $P(-x)$: only 1 change (from $-x$ to $+6$).

Step4: Apply Descartes' Rule

For negative real roots, the number is equal to the number of sign changes in $P(-x)$, or less by an even integer. Since 1 minus an even integer would be negative (invalid), only 1 possible negative real root.

Answer:

1