Use Descartes' Rule of Signs to determine the possible number of positive and negative zeros of $f ( x ) = 2 x ^ { 3 } - 2 x ^ { 2 } + 3 x - 4$ .
A)3 positive reals or 1 positive real; 3 negative reals or 1 negative real
B)3 positive reals or 1 positive real; no negative reals
C)1 positive real; 3 negative reals or 1 negative real
D)3 positive reals or 1 positive real; 1 negative real
E)no positive reals; no negative reals

Question

Quizplus · Accepted Answer

For positive zeros, the sign changes in $f(x)$ occur twice: from $2x^3$ to $-2x^2$ and from $3x$ to $-4$, indicating up to 2 positive real zeros or none due to Descartes' rule (even reductions). For negative zeros, $f(-x) = -2x^3 - 2x^2 - 3x - 4$ has no sign changes, indicating no negative real zeros.

Use Descartes' Rule of Signs to Determine the Possible Number f(x)=2x3−2x2+3x−4f ( x ) = 2 x ^ { 3 } - 2 x ^ { 2 } + 3 x - 4f(x)=2x3−2x2+3x−4

Use Descartes' Rule of Signs to Determine the Possible Number $f ( x ) = 2 x ^ { 3 } - 2 x ^ { 2 } + 3 x - 4$