Question

Part 1 [3 points]: What are the possible number of positive, negative, and complex zeros of f(x) = –2x3 – 5x2 – 6x + 4 ?

Part 2 [4 points]: Use complete sentences to explain the method used to solve this equation.

i know you have to use descartes rule of signs but idk how to do that

Answer 1

i know you have to change the signs but im not sure what to do

Answer 2

I don't remember decartes rule sorry

Answer 3

its okay :)

Answer 4

the number of positive real zeros equals the number of variations in the sign of the non-zero coefficients of f(x) or else equals that that number less an even integer.

Answer 5

so do i change just the signs inside to find the positives?

Answer 6

I haven't used this rule either, but these might help you:
- http://www.purplemath.com/modules/drofsign.htm
- http://www.youtube.com/watch?v=5YAmwfT3Esc

Answer 7

You just count. The signs of the non-zero coefficients are -, -, -. So, the number of changes is zero.

Answer 8

okay thank you

Answer 9

but dont you have to change them to get postive.

Answer 10

f(x) = –2x3 – 5x2 – 6x + 4 so then you would change
-2x^3+5x^2+6x-4 then you would have 2 positive.2 negative?

Answer 11

The second part is "the number of NEGATIVE real zeros of f either equals the number of variations in the sign of the nonzero coefficients of f(-x) or else equals that number less an even integer. So, replace x with -x, but be careful with the signs, then count.

Answer 12

-2x^3-5x^2-6x+4 so 3 negatvie?

Answer 13

-2(-x)^3 = -2(-x^3) = +2x^3

Answer 14

so then 2 negative

Answer 15

?

Answer 16

So, f(-x) = 2x^3 - 5x^2 + 6x +4. You only count the CHANGES, so it's two or zero.

Answer 17

but y 0?

Answer 18

I graphed it. It looks like this drawing. So, we are correct.