Use Descartesʹ Rule of Signs to determine the possible number of positive real zeros and the possible negative real zeros for the function.
-$$2 x ^ { 5 } - 2 x ^ { 4 } + 6 x ^ { 3 } - 3 = 0$$
A) Positive (1), negative $( 3,1 )$
B) Positive $( 4,2,0 )$, negative ( $0 )$
C) Positive $( 3,1 )$, negative $( 3,1 )$
D) Positive $( 3,1 )$, negative $( 0 )$

Question

Quizplus · Accepted Answer

Descartes' Rule of Signs for positive real zeros counts the sign changes in the original polynomial: $2x^5 - 2x^4 + 6x^3 - 3$, which has 1 sign change, indicating 1 or 0 positive real zeros (since it's impossible to have a negative number of zeros, the only option is 1). For negative real zeros, we substitute $x$ with $-x$ and get $2(-x)^5 - 2(-x)^4 + 6(-x)^3 - 3$, simplifying to $-2x^5 - 2x^4 - 6x^3 - 3$, which has no sign changes, indicating 0 negative real zeros.

Use Descartesʹ Rule of Signs to Determine the Possible Number 2x5−2x4+6x3−3=02 x ^ { 5 } - 2 x ^ { 4 } + 6 x ^ { 3 } - 3 = 02x5−2x4+6x3−3=0

Use Descartesʹ Rule of Signs to Determine the Possible Number $2 x ^ { 5 } - 2 x ^ { 4 } + 6 x ^ { 3 } - 3 = 0$