Descartes' Rule of Signs

Question

Shadow · Answer

$$f(x) = 4x^3 -2x^2 + 3x + 7$$

2 sign changes, so 2 positive real zeros

$$f(-x) = 4(-x)^3 -2(-x)^2 + 3(-x) + 7$$
$$f(-x) = -4x^3 -2x^2 -3x + 7$$

1 sign changes, so 1 negative real zero

@Vocaloid

Vocaloid · Answer

iirc for descartes rule of signs the number of real zeros = number of sign changes - 2n 
(so, the number of sign changes minus an even number)

for the first case, for 2 sign changes it can be either 2 ~or~ 0, and in this case, there are actually 0 positive zeros

Vocaloid · Answer

I haven't taken pre-calc in years so I've kind of forgotten the rest ;;

Shadow · Answer

Hmm, knew I was forgetting something. And for this my teacher is fine with 2 or 0.

Shadow · Answer

Haha it is probably the most simple concept in Precal, unless there is some stuff beyond what we have been taught. Was just making sure I got the gist of it.

Shadow · Answer

A faithful constant as always, thank you @Vocaloid