Question

What are the possible numbers of positive, negative, and complex zeros of f(x) = x^6 - x^5- x^4 + 4x^3 - 12x^2 + 12 ?

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

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

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

Answer 1

@paki
@aaronq

Answer 2

@amistre64 @phi can you guys help me out real quick?

Answer 3

yup. wait pls

Answer 4

when you add columnly, you should get total degree=6

Answer 5

So the answer would be C? I don't understand your explanation...

Answer 6

this is Descarte's Rule of counting + - roots

Answer 7

Yeah I get it now.
Do you agree with my answer? C?

Answer 8

yup C

Answer 9

positive roots: 4 2 0
negative roots: 2 0 0
complex roots: 0 4 6
----------------------
total is 6

Answer 10

Alright thanks

Answer 11

adding columns, nice touch

Answer 12

that's how my teacher taught me :p

Answer 13

x^6 - x^5- x^4 + 4x^3 - 12x^2 + 12

spose x=1, for a positive root test

1 - 1- 1 + 4 - 12 + 12

count the sign changes

1 - 1- 1 + 4 - 12 + 12
^ ^ ^ ^
0 1 2 3

we have at best 3 real positive roots
or 1 real and 2 complex using descartes rule of sign

Answer 14

ugh, i cant count :)

Answer 15

1 - 1- 1 + 4 - 12 + 12
^ ^ ^ ^ ^
0 1 2 3 4

there we go

Answer 16

Sometimes there is no substitute for a plot.