Question

What are the possible number of positive, negative, and complex zeros of f(x) = -x6 + x5- x4 + 4x3 - 12x2 + 12

Answer 1

a Positive: 4, 2, or 0; Negative: 2 or 0; Complex: 6, 4, 2, or 0
b Positive: 3 or 1; Negative: 3 or 1; Complex: 4, 2, or 0
c Positive: 5, 3, or 1; Negative: 1; Complex: 4, 2, or 0
d Positive: 2 or 0; Negative: 2 or 0; Complex: 6, 4, or 2

Answer 2

oik here we go with descartes rule of sign
do you know how to use it?

Answer 3

i think.. would it be option c?

Answer 4

\[f(x)=x^5- x^4 + 4x^3 - 12x^2 + 12\] has 5 changes in sign, so either 5, 3, or 1 positive zero
yes that part is right

Answer 5

now we need \(f(-x)=-x^5-x^4-4x^3-12x^2+12\) which has only one change in sign

Answer 6

since you count down by 2's there must be one negative zero

Answer 7

option.. d?

Answer 8

ooh damn damn damn!

Answer 9

i didn't see the first term, have to start again

Answer 10

\(f(x)= f(x) = -x^6 + x^5- x^4 + 4x^3 - 12x^2 + 12\)

Answer 11

ok yes 5 changes in sign right?

Answer 12

yes

Answer 13

so 5,3,1 for positive zeros

Answer 14

ooohh, so C right?

Answer 15

\[ f(-x) = -x^6 - x^5- x^4 - 4x^3 - 12x^2 + 12\]

Answer 16

yes it is C

Answer 17

thank you!

Answer 18

yw, but i think you did all the work