Question

What are the possible number of positive real, negative real, and complex zeros of f(x) = -7x4 - 12x3 + 9x2 - 17x + 3?

Answer 1

Descartes rule of signs

Answer 2

hmmm, let's take a peek at your equation

Answer 3

what descartes rule of signs?

Answer 4

as ganeshie8 already pointed out, Descartes Rule of signs
meaning
how many times does jumping from one term to the next, changes signs
let's see

-7x4 - 12x3 + 9x2 - 17x + 3
- - + - +

Answer 5

look at the signs of the coefficients, how many times did they CHANGE, that is from negative to positive or positive to negative
if you have negative to negative, there's no CHANGE

Answer 6

so they changed 3 times ?

Answer 7

-7x4 - 12x3 + 9x2 - 17x + 3
- - + - +

^ ^ ^ ^
no yes yes yes

Answer 8

yes, they CHANGED 3 times, that means 3 OR 3-2 = 1 or 3 or 1 positive real roots

Answer 9

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

Answer 10

so this is my answer?

Answer 11

now let's check how many negative ones, by using f(-x)

Answer 12

-7x^4 + 12x3 + 9x2 + 17x + 3
- + + + +

Answer 13

so it CHANGES only between the 1st term and the 2nd one, ONCE
so that means, 1 negative real root

Answer 14

so, the degree of the polynomial is 4, since that's the leading exponential \("7x^4"\)
that means that there'll be 4 roots

3 positive and 1 negative, 3+1 = 4, so 0 complex
if
1 positive and 1 negative, 1+1 = 2, so 2 complex

Answer 15

so, your answer is right

Answer 16

okay! got it! thank you!

Answer 17

yw