Question

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


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


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


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


Positive Real: 4, 2 or 0
Negative Real: 1
Complex: 0 or 1 or 3

Answer 1

have you covered Descartes Rule of Signs yet?

Answer 2

nope, whats that?

Answer 3

@Squirrels

Answer 4

http://www.youtube.com/watch?v=5YAmwfT3Esc <--- Descarte rule of signs

Answer 5

thank you :) but youtube is blocked on my computer ahaha oops but i apreciate your help thanks again

Answer 6

if someone could help me i'd be really thankful!

Answer 7

f(x) = -7x^4 - 12x^3 + 9x^2 - 17x + 3
For Descartes Rule of Signs, assume x is positive.
An easy thing to do would be to set x = +1
f(1) = -7 - 12 + 9 - 17 + 3
Count the number of times the terms change signs.
one sign change from -12 to +9
then +9 to -17
then -17 to +3
Three sign changes. Therefore, there is a maximum of 3 positive roots.

Try the same but assuming x is negative or set x = -1.
Write down each term and see how many times the signs change. There will be a maximum of that many negative roots.

Answer 8

Thank you :)

Answer 9

You are welcome.

With positive x we got 3 sign changes. That means maximum is 3 positive. But it could also be just1 positive. You always step down 2 at a time.

Answer 10

Okay, so the positive is 3 &1. What about the negative & complex

Answer 11

f(x) = -7x^4 - 12x^3 + 9x^2 - 17x + 3
Let us try x = -1
f(-1) = -7(-1)^4 - 12(-1)^3 + 9(-1)^2 - 17(-1) + 3 =
-7 + 12 + 9 + 17 + 3
There is just ONE sign change from -7 to +12. The rest are all + so no more sign changes.

Here it is definite. There is one negative root.
(had there been say 2 sign changes here we will say, it can have 2 negative roots or 0 negative roots (always step down by 2 from the maximum possibilities)).

Answer 12

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

this is what i got.

Answer 13

This is a fourth degree polynomial. Therefore it will have a total of 4 roots.

We know for sure that it has 1 negative root. That leaves with 3 roots.
If positive is 3, complex must be 0.
If positive is 1, complex must be 2.

Yes, your answer is correct.