Most often in undergraduate mathematics, work is done with polynomials that have real number coefficients.
So, even though the Fundamental Theorem of Algebra deals
with polynomials with complex number coefficients,
you won't often find yourself working with polynomials having coefficients involving the imaginary number $\,i\,.$
By the Fundamental Theorem of Algebra, every nonconstant polynomial with complex coefficients has at least one zero in the set of complex numbers. But—when the polynomial under consideration has all real number coefficients—we can say even more about the zeros!
We'll show that for every quadratic function with real nunber coefficients,
nonreal zeros must occur in complex conjugate pairs.
The proof here easily generalizes to any polynomial with real number coefficients.
Recall that the complex conjugate of $\,x + yi\,$ (for real numbers $\,x\,$ and $\,y\,$) is $\,x  yi\,$;
that is, to find the complex conjugate, the real part remains the same, and you take the opposite of the imaginary part.
The complex conjugate is frequently denoted with an overbar:
$$
\cssId{s39}{\underbrace{\overline{x + yi}}_{\text{the complex conjugate of $\,x+yi\,$}}}
\cssId{s40}{= \underbrace{x  yi}_{\text{same real part; opposite imaginary part}}}
$$
Here are two properties of the complex conjugate that were proved in an earlier lesson:
PROOF:
Let $\,P(x) := ax^2 + bx + c\,,$ where $\,a\,,$ $\,b\,$ and $\,c\,$ are real numbers, and $\,a\ne 0\,.$
Suppose that $\,w\,$ is a zero of $\,P\,.$
Then:
$$
\begin{alignat}{2}
\cssId{s50}{P(\bar w)}\
&\cssId{s51}{= a(\bar w)^2 + b\bar w + c}
&&\qquad \cssId{s52}{\text{REASON: definition of $P$, function evaluation}}\cr
&\cssId{s53}{= a(\overline{w^2}) + b\bar w + c}
&&\qquad \cssId{s54}{\text{REASON: } \overline{\vphantom{b}a}\cdot\overline{b} = \overline{ab}}\cr
&\cssId{s55}{= \bar a(\overline{w^2}) + \bar b\bar w + \bar c}
&&\qquad \cssId{s56}{\text{REASON: for real numbers $x$, $\bar x = x$}}\cr
&\cssId{s57}{= \overline{a\,w^2} + \overline{bw} + \bar c}
&&\qquad \cssId{s58}{\text{REASON: } \overline{\vphantom{b}a}\cdot\overline{b} = \overline{ab}}\cr
&\cssId{s59}{= \overline{aw^2 + bw + c}}
&&\qquad \cssId{s60}{\text{REASON: } \overline{\vphantom{b}a} + \overline{b} = \overline{a + b}}\cr
&\cssId{s61}{= \overline{P(w)}}
&&\qquad \cssId{s62}{\text{REASON: definition of $P$, function evaluation}}\cr
&\cssId{s63}{= \bar 0}
&&\qquad \cssId{s64}{\text{REASON: definition of zero}}\cr
&\cssId{s65}{= 0}
&&\qquad \cssId{s66}{\text{REASON: for real numbers $x$, $\bar x = x$}}\cr
\end{alignat}
$$
Q.E.D.
Recall that there is a beautiful relationship between the zeros of a polynomial and its factors:
if $\,c\,$ is a zero, then $\,xc\,$ is a factor.
In particular:
Now, put everything together:
On this exercise, you will not key in your answer. However, you can check to see if your answer is correct. 
PROBLEM TYPES:
