Thread: 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 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

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 ...



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



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

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 ...

... 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?

Originally Posted by skeeter

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!

if is a complex root, then so is its conjugate ... therefore and are factors of the quartic.












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

Thank you! This makes so much sense!