Use Descartes's Rule of Signs to determine the possible number of positive and negative real zeros for the given
function.
-$$f ( x ) = 2 x ^ { 3 } - 3 x ^ { 2 } + x + 6$$
A) 2 or 0 positive zeros, 1 negative zero
B) 3 or 1 positive zeros, 3 or 1 negative zeros
C) 2 or 0 positive zeros, 1 or 0 negative zeros
D) 2 or 0 positive zeros, 2 or 0 negative zeros

Question

Quizplus · Accepted Answer

Descartes's Rule of Signs states that the number of positive real zeros is equal to the number of sign changes in $ f(x) $ or less by an even number. For $ f(x) = 2x^3 - 3x^2 + x + 6 $, there are 2 sign changes, so there can be 2 or 0 positive zeros. For negative zeros, consider $ f(-x) = -2x^3 - 3x^2 - x + 6 $, which has 1 sign change, indicating 1 negative zero.

Use Descartes's Rule of Signs to Determine the Possible Number f(x)=2x3−3x2+x+6f ( x ) = 2 x ^ { 3 } - 3 x ^ { 2 } + x + 6f(x)=2x3−3x2+x+6

Use Descartes's Rule of Signs to Determine the Possible Number $f ( x ) = 2 x ^ { 3 } - 3 x ^ { 2 } + x + 6$