What are the possible number of positive, negative, and complex zeros of f(x) = 2x^3 – 5x^2 – 6x + 4 ? Can someone help me, I am not sure if I am doing this properly.

Question

What are the possible number of positive, negative, and complex zeros of f(x) = 2x^3 – 5x^2 – 6x + 4 ?  Can someone help me, I am not sure if I am doing this properly.

anonymous · Answer

here is a great explanation with example we can work through it in a minute or two if you still need help http://www.purplemath.com/modules/drofsign.htm

liizzyliizz · Answer

Thank you and sure I have some time. I do have some work already done and I am familiar with the steps I am supposed to take, but when I get to the complex zeros, I am getting strange results.

liizzyliizz · Answer

The first thing you need to do is write the equation and look at where the sign changes. 

 2x^3 – 5x^2 – 6x + 4

(Sign changes twice therefore, there are 2 positive real zeros in the function)

 2x^3 – 5x^2 – 6x + 4

The next thing that you do is make the x variables negative. 
-(2x^3) -(-5x^2) -(-6x) - 4

Next,simplify each of these terms by simplifying the variables. 
-2x^3 + 5x^2 + 6x - 4 

(Sign changes twice, therefore there are 2 negative real zeros in the function) 

Using Descartes' rule of signs there's an extra step in which you subtract 2 from the number you have for the possible negative real zeros. Therefore 2-2 = 0, so there's a possibility that there are 2 or 0 negative real zeros.

anonymous · Answer

two changes in sign says there are either 2 or no positive zeros

liizzyliizz · Answer

Whoops lol I knew I was forgetting something, I just wasn't quite sure what it was.

anonymous · Answer

but i think $f(-x)$ has only one change in sign
correct me if i am wrong

anonymous · Answer

this line $$-2x^3 + 5x^2 + 6x - 4 $$ is wrong

anonymous · Answer

$$f(x)= 2x^3 – 5x^2 – 6x + 4 $$
$$f(-x)=-2x^3-5x^2+6x+4$$

liizzyliizz · Answer

hmm perhaps that explains why my work for the complex zeros was coming out strange. O.o You'd think basic math wouldn't be an issue lol.

anonymous · Answer

no i would never think that
basic mistakes at the beginning cause all kinds of problems
you got this now right?

anonymous · Answer

1 negative zero and either 2 or no positive ones

liizzyliizz · Answer

okay, and for the complex zeros would it be 1 or 0 then? Since you'd subtract 3-2-1 and 3-2-0 ?

anonymous · Answer

no

liizzyliizz · Answer

Wait, youre right, I am missing something o.o lol
Trying to figure out what it is exactly.

anonymous · Answer

you count down by twos
$f(-x)$ has one change in sign, so there must be one negative real zero

anonymous · Answer

because there cannot be minus one negative real zero, that doesn't make any sense

anonymous · Answer

and since $f(x)$ has two changes in sign, there are either 2 or no real positive zeros

liizzyliizz · Answer

yeah that's what I was getting before and I knew I was doing something wrong since -1 doesn't make sense. err

anonymous · Answer

there are three zeros all together, counting multiplicity and complex zeros
so there are only two options

$i$ one negative real zero and 2 positive ones 
$ii$ one negative real zero and 2 complex ones

liizzyliizz · Answer

Ok I am kind of grasping what you are saying, can you explain how you got to that conclusion?
Sorry, I am trying to reteach this to myself and I am a bit lost lol.

anonymous · Answer

no problem

anonymous · Answer

once you use descartes rule of sign, you count down by twos

anonymous · Answer

so for example if $f(x)$  has 4 changes in sign, then there are either 4, 2 or 0 positive zeros

anonymous · Answer

and if $f(x)$ has 3 changes in sign, there are either 3 or 1 positive zero

anonymous · Answer

is this more or less clear?   you count down by twos for each one

liizzyliizz · Answer

Yeah this makes sense :)

anonymous · Answer

in this case you have 2 changes in sign for $f(x)$ so there are either 2 or 0 positive zeros

anonymous · Answer

since there is only one change of sign for $f(-x)$ there must be one negative zero

anonymous · Answer

because you can't count down by twos starting at 1
there must be exactly 1

anonymous · Answer

no choices there, but for the positive zeros there are two choices: 2 or 0

anonymous · Answer

and therefore for the complex zeros there are also two choices, 2 or 0, because there are 3 in total for sure (degree is 3)

anonymous · Answer

btw there are always an even number of complex zeros, because if $a+bi$ is a zero, then so is its conjugate $a-bi$

liizzyliizz · Answer

That last fact, I did not know. Thank you so much, it makes a lot more sense now! :)