# Thread: Help with Finding Zeros

1. ## Help with Finding Zeros

Hi all I'm having a little trouble with a math problem,
I'm trying to find the the zeros of:

$\displaystyle f(x)=x^6-4x^5+5x^4-x^2-5$

so far I have:

degree 6 so this function has 6 zeros

5, 3, or 1 positive real zeros
1 negative real zero

possible rational zeros: $\displaystyle -5,-1,1,5$

using synthetic devision I found that -1 and 1 are real zeros and eliminated -5 and 5

and am left with: $\displaystyle f(x)=(x+1)(x-1)(x^4-4x^3+6x^2-4x+5)$

I looked at the graph of this function and it only crosses the x-axis twice so I'm guessing that there are no irrational zeros and that there are two pairs of complex conjugates.

Am I right? If so how do I go about finding the complex conjugates of $\displaystyle x^4-4x^3+6x^2-4x+5$. If not, how do I proceed?

-thanks a bunch

2. Originally Posted by boblee
...
Am I right? If so how do I go about finding the complex conjugates of $\displaystyle x^4-4x^3+6x^2-4x+5$. If not, how do I proceed?

-thanks a bunch
$\displaystyle x^4-4x^3+6x^2-4x+5 =$

$\displaystyle x^4+x^2-(4x^3+4x) + 5x^2+5 =$

$\displaystyle x^2(x^2+1)-4x(x^2+1)+5(x^2+1)$

Can you take it from here?

3. thanks, but, correct me if I'm wrong, after looking it over wouldn't it be the following?

$\displaystyle x^4-4x^3+6x^2-4x-5$

$\displaystyle = x^4+x^2+(-4x^3-4x)+5x^2+5$

$\displaystyle = x^2(x^2+1)-4x(x^2+1)+5(x^2+1)$

from there I get:

$\displaystyle = (x^2-4x+5)(x^2+1)$

$\displaystyle x^2+1=0$

$\displaystyle x^2=-1$

$\displaystyle x=-1^{1/2}$

$\displaystyle x=+/-i$

$\displaystyle x^2-4x+5=0$

$\displaystyle x=(4+/-(-4^2-4*1*5)^{1/2})/2$

$\displaystyle = (4+/-(16-20)^{1/2})/2$

$\displaystyle = (4+/-(-4)^{1/2})/2$

$\displaystyle = 1+/-i$
Did I do that right?

4. Originally Posted by boblee
thanks, but, correct me if I'm wrong, after looking it over wouldn't it be the following?

$\displaystyle x^4-4x^3+6x^2-4x-5$

$\displaystyle = x^4+x^2+(-4x^3-4x)+5x^2+5$

$\displaystyle = x^2(x^2+1)-4x(x^2+1)+5(x^2+1)$

from there I get:

$\displaystyle = (x^2-4x+5)(x^2+1)$ ....... OK

$\displaystyle x^2+1=0$

$\displaystyle x^2=-1$

$\displaystyle x=-1^{1/2}$

$\displaystyle x=+/-i$ ....... OK

$\displaystyle x^2-4x+5=0$

$\displaystyle x=(4+/-(-4^2-4*1*5)^{1/2})/2$

$\displaystyle = (4+/-(16-20)^{1/2})/2$

$\displaystyle = (4+/-(-4)^{1/2})/2$ ....... OK

$\displaystyle = {\color{red}\bold{2}}+/-i$ ... !!!
Did I do that right?
Btw \pm will yield $\displaystyle \pm$