SOLUTION: I am having extreme difficulty with this section of my homework. I have to find a polynomial function of least degree having only real coefficients with zeros. The book tells me

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Question 272272: I am having extreme difficulty with this section of my homework. I have to find a polynomial function of least degree having only real coefficients with zeros.
The book tells me that f(x)=x-(1+i) has 1+i as a zero, but the conjugate 1-i is not a zero. Why? If I use the synthetic division (as the book says) to figure out if f(k)=0, I get don't understand why 1-i is not a zero. All I keep getting is the same equation by using synthetic division.
The problem I am having to do it:
2++sqrt%28+3+%29+, 2-+sqrt%28+3+%29+, and 2+3i.
By not understanding why why 1-i is not a zero, I am confused as to what is a zero in the above equation.
I THINK that 2++sqrt%28+3+%29+ and 2+3i are going to be zeros for the problem, giving me an equation that looks something like:
f(x)= [x-(2++sqrt%28+3+%29+)][x-(2-+sqrt%28+3+%29+)][x-(2+3i)][x-(2-3i)]
If someone could please help me with this I would GREATLY appreciate it.
Answer by edjones(8007) About Me  (Show Source):
You can put this solution on YOUR website!
x-(1+i)=0
let 1+i=a so you will understand better
x-a=0
x=a
x=1+i piece of cake. you are over-thinking this one.
A first degree equation has only one zero.
.
Ed