Need some help with finding zeros of a polynomial given one complex zero

Hi all! having some trouble with this question and have an exam in 3 days so hoping to get some help! :) leaving it late, I know.

Anyway, so $\displaystyle f(x)= x^4 - 3x^3+6x^2+2x-60$ is the polynomial and 1+3i is the complex zero.

So I used synthetic division to obtain this 1 4+3i 1+15i - 42+18i remainder -36-108i

Then I used 1-3i, the conjugate which should also be a zero, yes? to obtain: 1 5 6 -36 and a remainder of -72 for my co-efficients. Which if the sign had been reversed I would have had an nice neat 0 remainder and I could have gone on from there but ive been over my working twice and cant find any mistakes I made- I am hoping that I have, as that means I don't have to learn how to deal with these remainders.

Hoping someone can help me out here, thanks so much!

Rhys

Re: Need some help with finding zeros of a polynomial given one complex zero

I find that synthetic division with complex zeros are a bit too tedious (for me) ...

since the given complex zero and its conjugate are roots ...

$\displaystyle [x - (1+3i)][x - (1-3i)] = x^2 - 2x + 10$

... the above quadratic is a factor of the original polynomial.

$\displaystyle \frac{x^4-3x^3+6x^2+2x-60}{x^2-2x+10} = x^2-x-6 = (x-3)(x+2)$

from which we can find the other two (real, in this case) roots.

Re: Need some help with finding zeros of a polynomial given one complex zero

Quote:

Originally Posted by

**skeeter** I find that synthetic division with complex zeros are a bit too tedious (for me) ...

since the given complex zero and its conjugate are roots ...

$\displaystyle [x - (1+3i)][x - (1-3i)] = x^2 - 2x + 10$

... the above quadratic is a factor of the original polynomial.

Ok thank you but I am unsure as to how you obtained the factor there, I can see how that quadratic is a factor but I dont know you obtained the complex factor from the original polynomial, if you might be able to explain the steps involved?

Quote:

Originally Posted by

**skeeter** $\displaystyle \frac{x^4-3x^3+6x^2+2x-60}{x^2-2x+10} = x^2-x-6 = (x-3)(x+2)$

So did you use log division here to obtain the result or some other method? Sorry I only really know the synthetic division method

Thanks again!

Re: Need some help with finding zeros of a polynomial given one complex zero

if $\displaystyle z$ is a complex root, then so is its conjugate $\displaystyle \bar{z}$ ... therefore $\displaystyle (x - z)$ and $\displaystyle (x-\bar{z})$ are factors of the quartic.

$\displaystyle (x - z)(x-\bar{z}) =$

$\displaystyle x^2 - zx - \bar{z} x + z \bar{z} = $

$\displaystyle x^2 - (1+3i)x - (1-3i)x + (1+3i)(1-3i) =$

$\displaystyle x^2 - x - 3ix - x + 3ix + (1 - 9i^2) =$

$\displaystyle x^2 - 2x + 10$

... and, yes, I performed long division.

Re: Need some help with finding zeros of a polynomial given one complex zero

Thank you! This makes so much sense! :)