Question

What are the possible number of positive, negative, and complex zeros of f(x) = –2x3 + 5x2 + 6x – 4 ?

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

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

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

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

Answer 1

have you covered Descartes Law of Signs yet?

Answer 2

yes but i dont know very well how to use it

Answer 3

i know there is only 3 sign changes

Answer 4

well, let's find the POSITIVE real roots first, using f( x )

– 2x^3 + 5x^2 + 6x – 4
- + + -
yes no yes

it changed signs twice, from - to + and later from + to -
that means it has 2 or (2-2) real POSITIVE roots, so 2 or 0

Answer 5

okay thanks 2 or 0 positive and how about negative and complex thats where i get confused alot

Answer 6

let's find the NEGATIVE real roots first, using f( -x ) <--- notice negative "x"

+ 2x^3 + 5x^2 - 6x – 4
+ + - -
no yes no

so it changed signs once only, that means it has only 1 NEGATIVE root

Answer 7

so the polynomial has a degree of 3, that means it will have 3 roots
so our combination for real ones is

2 positive ones, and 1 negative, that leaves no room for any complex ones
or
0 positive ones, and 1 negative, we now have room for 2 complex ones

Answer 8

so the answer is c

Answer 9

indeed

Answer 10

positive 2 or 0 negative 1 2 or 0 complex. got it thanks so much

Answer 11

yw