Question

What are the possible number of positive, negative, and complex zeros of f(x) = 3x^4 – 5x^3 – x^2 – 8x + 4 ?

Answer 1

Descartes Law of Signs

let's find the positive real ones using f(x)

+ 3x^4 – 5x^3 – x^2 – 8x + 4
yes no no yes

it changed signs twice, from + to -, and later one from - to +
that means it has 2 or (2-2) = 0, real positive zeros/roots

Answer 2

let's find the negative real ones using f( -x )

+ 3x^4 + 5x^3 – x^2 + 8x + 4
no yes yes no

so it changed signs twice again
that means it has 2 or (2-2) =0, real negative zeros/roots

Answer 3

so is a 4th degree polynomial, it'll have 4 roots/zeros
and they'll be

2 positive real, and 2 negative real, that leaves room for no complex
2 positive real, and 0 negative real, we have room for 2 complex ones
or
0 positive real, and 2 negative real, that also leaves room for 2 complex ones

Answer 4

keep in mind that when using f( -x )
all odd exponent terms will change sign, why? well

2x^3 => 2(-x)^3 => 2(-x * -x * -x) => 2 * -x^3 => -2x^3