Question

What are the possible number of positive real, negative real, and complex zeros of
f(x) = –7x4 – 12x3 + 9x2 – 17x + 3?
Answer:

A)Positive Real: 1
Negative Real: 3 or 1
Complex: 2 or 0

B)Positive Real: 3 or 1
Negative Real: 2 or 0
Complex: 1

C)Positive Real: 3 or 1
Negative Real: 1
Complex: 2 or 0

D)Positive Real: 4, 2 or 0
Negative Real: 1
Complex: 0 or 1 or 3
This the last question like this, help me please:)

Answer 1

descartes rule of sign for this one
http://www.purplemath.com/modules/drofsign.htm

Answer 2

three changes of sign in your example,
\[–7x^4 – 12x^3 + 9x^2 – 17x + 3\]
\[–7x^4 – 12x^3_{\text{ here }} + 9x^2_{\text{ here }} – 17x_{\text{ here }} + 3\]

Answer 3

so possible positive roots: 3 or 1

Answer 4

\[f(-x) = –7(-x)^4 – 12(-x)^3 + 9(-x)^2 – 17(-x) + 3\]
\[f(-x) = –7x^4 + 12x^3 + 9x^2 + 17x + 3\] this has one change in sign

Answer 5

so one possible negative root. degree is 4, so there are 4 roots in total
Positive Real: 3 or 1
Negative Real: 1
Complex: 2 or 0

Answer 6

Okay thank you very much!:))

Answer 7

yw
btw the link i sent has a brief clear explanation with examples, if you have to take a test on this it is easy enough with a little practice

Answer 8

okay thank you, i am understanding bits and pieces of this lesson:)