What are the possible number of positive, negative, and complex zeros of f(x)= -2x^3+5x^2-6x+4

Question

anonymous · Answer

descartes rule of sign for this one right?

anonymous · Answer

$$f(x)= -2x^3+5x^2-6x+4 $$ count the changes in sign of the coefficents
from $-2$ to $+5$ is one
from $+5$ to $-6$ is two
from $-6$ to $+4$ is three
three changes in sign

anonymous · Answer

that means there are 3 positive zeros or 1 positive zero (you count down by twos)

anonymous · Answer

for the possible negative zeros, count the changes in sign of $f(-x)$
in this case 
$$ f(x)= -2x^3+5x^2-6x+4$$ so $$f(-x)=-2(-x)^3+5(-x)^2-6(-x)+4$$ or 
$$f(-x)=2x^3+5x^2+6x+4$$

anonymous · Answer

no changes in sign, all the coefficients are positive
so no negative zeros

whpalmer4 · Answer

Complex zeros come in conjugate pairs $(a\pm bi)$ in polynomials with only real coefficients such as this one.  We know there are 3 total zeros because the highest power of $x$ present is $x^3$.  What are the different ways that we can come up with 3 zeros with what you know about the zeros?