Question

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

Positive: 4, 2, or 0; negative: 2 or 0; complex: 6, 4, 2, or 0

Positive: 1; negative: 5, 3, or 1; complex: 4, 2, or 0

Positive: 3 or 1; negative: 2 or 0; complex: 4, 2, or 0

Positive: 3 or 1; negative: 3 or 1; complex: 4, 2, or 0

Answer 1

count the sign change in the original equation. Then times (-1) to the each exponent.

Answer 2

?? i DONT UNDERSTAND HOW TO DO IT ALL AT

Answer 3

It's the rules of sign

Answer 4

can you explain

Answer 5

Ok, you have -x^6 -x^5-x^4-4x^3-12x^2 + 12

Answer 6

there is one sign change, do you see the sign change?

Answer 7

yes the +12 right?

Answer 8

yes, now times (-1). -(-1)^6 -(-1)^5-(-1)^4-4(-1)^3-12(-1)^2+12

Answer 9

It'll be x^6-x^+x^4-4x^3+12x^2+12

Answer 10

It has to equal to 6 because of the leading exponent

Answer 11

wait wat has to equal to 6?

Answer 12

yes

Answer 13

it's part of the rule, not this one but another rule that follows this rule

Answer 14

the answer is "C"

Answer 15

thank you

Answer 16

no problem