# Thread: Complex Zero's of a Function

1. ## 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.

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

2. 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.

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

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

Aha!

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

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

Aha!
Thanks for your help Mr. Fantastic (haha ) 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?

4. Originally Posted by Anandymous
Thanks for your help Mr. Fantastic (haha ) 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 $\displaystyle (x^2 + 8) (x^2 + 10x - 39)$.

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 ....)

5. 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)

6. 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 $\displaystyle i$ in an appropriate way (or $\displaystyle j$ if you're an electrical engineer in the making).