Find polynomial when given a complex root

[jkristia]

Hi there,

I have over the last couple of month worked my way through Algebra Demystified and College Algebra Demystified (well almost), and I just completed the chapter on polynomial functions and I’m stuck at one of the questions in the chapter review where I’m given a complex root and the asked to match it to one of four functions.

The complex root given is 2-3i, so I know the polynomial for the complex root pair is x^2-4x+13 (I hope I got that right – if not that might be emy problem).
I have then tried to do a long division of this to all of the functions as I expected the remainder to be zero for the correct function, but ..hmm .. that didn’t happen.

Isn’t it correct that I should end up with a zero remainder when I do a long division of x^2-4x+14| polynomial from answer ?

I refuse to look at the answer until I at least have come up with an answer, so I appreciate if someone can point me at what I might be doing wrong.

Thanks
Jesper

 

[rock.freak667]

Yes you should end up with no remainder. What were the four functions?

 

[fzero]

Isn’t it correct that I should end up with a zero remainder when I do a long division of x^2-4x+14| polynomial from answer ?

Assuming that you mean x^2-4x+13, yes it should. A simpler way to check would be to take the function and compute f(2-3i), which will be zero if 2-3i is a root. When you find the right function, you can go back and check your long division for errors.

 

[jkristia]

ha - found it.. a bit embarrassing, but I had flipped the sign in one of the divisions and just didn’t notice it.
The solution is f(x) = x^3 - 5x^2 + 17x – 13.

 

[Mentallic]

Since the quadratic has all real coefficients, then if one of its roots is complex, there must be another root that is the complex conjugate. So if you have one root of a+ib then there must be another root of a-ib IFF the coefficients of the polynomial are real.

So another way to check would be to expand (x-(2-3i))(x-(2+3i)) and check to see if it is equivalent to x²-4x+13

 

[jkristia]

>>So another way to check would be to expand (x-(2-3i))(x-(2+3i)) and check to see if it is equivalent to x2-4x+13

That is kind of what I did I think, I found x^2-4x+13 and then used that with a long division of the answer polynomial until I found the one with a remainder of zero.

Thankts to all of you for your help. I will most likely come back with more questions :)