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 sine for this one

Answer 2

\[ f(x) = -7x^4 - 12x^3 + 9x^2 - 17x + 3\]
there is a "change of sign" from -12 to 9, from 9 to -17 and from -17 to 9
three changes all together

Answer 3

then you count down by twos
there are three changes in sign, so there are either 3 or 1 positive real zero

Answer 4

Ok i have answers for this
a.Positive Real: 3 or 1
Negative Real: 1
Complex: 2 or 0
b.Positive Real: 3 or 1
Negative Real: 2 or 0
Complex: 1
c.Positive Real: 1
Negative Real: 3 or 1
Complex: 2 or 0
d.Positive Real: 4, 2 or 0
Negative Real: 1
Complex: 0 or 1 or 3

Answer 5

i can't read all that, it is easier just to solve

Answer 6

a or b?

Answer 7

is it clear that there are three changes in sign?

Answer 8

yes

Answer 9

ok
now for the possible negative real zeros, find the number of changes of sign in \(f(-x)\)

Answer 10

Are there two changes of sign?

Answer 11

\[f(x) = -7x^4 - 12x^3 + 9x^2 - 17x + 3\]
\[f(-x) = -7(-x)^4 - 12(-x)^3 + 9(-x)^2 - 17(-x) + 3\]\[f(-x)=-7x^4+12x^3+9x^2+17x+3\]

Answer 12

i.e. change the sign of the terms with odd exponents, leave the even exponents alone

Answer 13

So there are two?

Answer 14

not the way i count, no

Answer 15

unless i made a mistake, the only change in sign is from -7 to 12

Answer 16

i may have made a mistake, but i think that is right

Answer 17

9 to 17

Answer 18

they are both positive

Answer 19

oh, now i see thank you so much

Answer 20

so now there must be on negative real zero, because you count down by twos, and there can't be minus one negative zero (makes no sense)

Answer 21

there are 4 real zeros, so there can be either 3 positive, 1 negative or
1 positive, 1 negative, and therefore 2 complex