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

I got this as my answer is it correct?
Positive: 3 or 1
Negative: 1
Complex: 2 or 0

Answer 2

I gather you've covered the Descartes Law of Signs already?

Answer 3

yes I'm just checking to see if I got the right answer so I know if im doing it right.

Answer 4

3x^4 – 5x^3 – x^2 – 8x + 4
+ - - - +

Answer 5

and for f(-x) to get the negative roots

3x^4 + 5x^3 – x^2 + 8x + 4
+ + - + +

Answer 6

the degree of the expression is 4 because is the highest term is \(3x^4\)
so that means it'll have 4 roots

Answer 7

These are my choices for answers:

Positive: 2 or 0; Negative: 2 or 0; Complex: 4 or 2 or 0

Positive: 2 or 0; Negative: 2 or 0; Complex: 4 or 2

Positive: 3 or 1; Negative: 1; Complex: 2 or 0

Positive: 4 or 2 or 0; Negative: 2 or 0; Complex: 4 or 2 or 0

Answer 8

so, how many "sign changes" do you see in the f(x) function?

Answer 9

I got 3 sign changes in f(x)

Answer 10

let's see

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

keep in mind from - to -, there's no change, or from + to +
change would be from - to + or + to -

Answer 11

and for f(-x) to get the negative roots

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

Answer 12

so, you have 2 OR 2-2=0 real positive, 2 or 2-2=0 real negative
so, if you say have
2 pos, 2 neg, then no complex ones
0 pos, 2 neg, then 2 complex ones
2 pos, 0 neg, then 2 complex ones
0 pos, 0 neg, then 4 complex ones

Answer 13

in the cases the real positive and negative do not add up to "4", that means, the "rest" are complex ones

Answer 14

oh ok I wrote the problem wrong when solving it I had it writen as -3x when it was suppose to be 3x. so the answer would be :
Positive: 2 or 0; Negative: 2 or 0; Complex: 4 or 2 or 0

is this correct?

Answer 15

yes

Answer 16

ok thank you!!!! :)

Answer 17

yw