Determine the number of possible positive and negative real zeros using Descartes' Rule of Signs. $p ( x ) = x ^ { 5 } - 4 x ^ { 4 } - 13 x ^ { 3 } + 70 x ^ { 2 } - 90 x + 36$
A) Positive real: exactly one; Negative real: exactly one
B) Positive real: 2 or 0; Negative real: exactly 2
C) Positive real: 4 or 0; Negative real: exactly one
D) Positive real: exactly 4; Negative real: exactly one
E) Positive real: 4, 2, or 0; Negative real: exactly one

Question

Quizplus · Accepted Answer

For positive real zeros, the sign changes in $p(x)$ occur four times, so there could be 4, 2, or 0 positive real zeros. For negative real zeros, substituting $x$ with $-x$ gives $p(-x) = -x^5 - 4x^4 + 13x^3 + 70x^2 + 90x + 36$, which has one sign change, indicating exactly one negative real zero.

Determine the Number of Possible Positive and Negative Real Zeros p(x)=x5−4x4−13x3+70x2−90x+36p ( x ) = x ^ { 5 } - 4 x ^ { 4 } - 13 x ^ { 3 } + 70 x ^ { 2 } - 90 x + 36p(x)=x5−4x4−13x3+70x2−90x+36

Determine the Number of Possible Positive and Negative Real Zeros $p ( x ) = x ^ { 5 } - 4 x ^ { 4 } - 13 x ^ { 3 } + 70 x ^ { 2 } - 90 x + 36$