Help with Finding Zeros

**boblee** · October 11th 2008, 01:48 AM

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

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

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: $-5,-1,1,5$

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

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

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 $x^4-4x^3+6x^2-4x+5$ . If not, how do I proceed?

-thanks a bunch

**earboth** · October 11th 2008, 02:00 AM

Originally Posted by boblee

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

-thanks a bunch

$x^4-4x^3+6x^2-4x+5 =$

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

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

Can you take it from here?

**boblee** · October 11th 2008, 02:55 PM

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

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

$<br /> = x^4+x^2+(-4x^3-4x)+5x^2+5$

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

from there I get:

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

$x^2+1=0$

$x^2=-1$

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

$x=+/-i$

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

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

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

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

$= 1+/-i$
Did I do that right?
Thanks for your help.

**earboth** · October 12th 2008, 12:58 AM

Originally Posted by boblee

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

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

$<br /> = x^4+x^2+(-4x^3-4x)+5x^2+5$

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

from there I get:

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

$x^2+1=0$

$x^2=-1$

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

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

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

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

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

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

$= {\color{red}\bold{2}}+/-i$ ... !!!
Did I do that right?
Thanks for your help.

Btw \pm will yield $\pm$

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