# Complex Zero's of a Function

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• Dec 7th 2008, 03:02 PM
Anandymous
Complex Zero's of a Function
Hi everyone. First time here and I was wondering if someone could assist me with this math problem that I'm having trouble with.

The problem reads:

find all complex zeros of the polynomial function, give exact values, list multiple zeros as necessary

f(x)= x^4+10x^3-31x^2+80x-312

I know I'm supposed to do something with rational zero principle or maybe synthetic division but I wasn't sure.

Any help would be appreciated thanks (Happy)
• Dec 7th 2008, 03:23 PM
mr fantastic
Quote:

Originally Posted by Anandymous
Hi everyone. First time here and I was wondering if someone could assist me with this math problem that I'm having trouble with.

The problem reads:

find all complex zeros of the polynomial function, give exact values, list multiple zeros as necessary

f(x)= x^4+10x^3-31x^2+80x-312

I know I'm supposed to do something with rational zero principle or maybe synthetic division but I wasn't sure.

Any help would be appreciated thanks (Happy)

Something worth checking since $312 = 39 \cdot 8$:

$x^4 + 10x^3 - 31x^2 + 80x - 312 = (x^4 + 10x^3 - 39x^2) + (8x^2 + 80x - 312)$ $= x^2 (x^2 + 10x - 39) + 8(x^2 + 10x - 39) = \, ....$

Aha!
• Dec 7th 2008, 03:39 PM
Anandymous
Quote:

Originally Posted by mr fantastic
Something worth checking since $312 = 39 \cdot 8$:

$x^4 + 10x^3 - 31x^2 + 80x - 312 = (x^4 + 10x^3 - 39x^2) + (8x^2 + 80x - 312)$ $= x^2 (x^2 + 10x - 39) + 8(x^2 + 10x - 39) = \, ....$

Aha!

Thanks for your help Mr. Fantastic (haha (Bow)) but perhaps you could shed some light on your response. I think I somewhat understand where you are going with this, but I am not familiar with this method of solution. Should I be looking at the factors in front of the parenthesis (x^2 and 8) in order to form a solution?
• Dec 7th 2008, 03:43 PM
mr fantastic
Quote:

Originally Posted by Anandymous
Thanks for your help Mr. Fantastic (haha (Bow)) but perhaps you could shed some light on your response. I think I somewhat understand where you are going with this, but I am not familiar with this method of solution. Should I be looking at the factors in front of the parenthesis (x^2 and 8) in order to form a solution?

Take out the common factor and you get $(x^2 + 8) (x^2 + 10x - 39)$.

Now factorise each quadratic factor over C.

(Once you've done this you'll see that your alternative was to spot x = 3 as a zero and go from there ....)
• Dec 7th 2008, 04:47 PM
Anandymous
Doh, I'm dumb. So basically, X^2 + 8 solves into X^2 = -8 and then I simply have to solve for X, which is +- sqrt(-8)
• Dec 7th 2008, 06:50 PM
mr fantastic
Quote:

Originally Posted by Anandymous
Doh, I'm dumb. So basically, X^2 + 8 solves into X^2 = -8 and then I simply have to solve for X, which is +- sqrt(-8)

Yes. And feel free to use the symbol $i$ in an appropriate way (or $j$ if you're an electrical engineer in the making).