Question regarding the proof of the Fundamental Theorem of Algebra Every non-constant polynomial in $\mathbb C[x]$ has a complex root
To  prove  this  we  first  consider  a  polynomial  with  complex  coefficients $f(x)=a_{0}+a_{1}x+.......a_{n}x^{n}$  and  its  conjugate  polynomial $\bar f(x)=\bar a_{0}+ \bar a_{1}x+\bar a_{2}x^{2}+....+\bar a_{n}x^{n}$
Then  we  find $$f(x)\bar f(x) \in \mathbb R[x]$$  Then  it  says
$f(x)$ has a complex root iff $f(x)\bar f(x)$ has a complex root
Now  if  $f(x)$  has  a  complex  root  then  definitely  $f(x)\bar f(x)$  has  that  complex  number  as  its  root   but  I  am  stuck with  the  reverse .
Then  there   is  another  problem.  Proving  the  above ensures  that  It's sufficient to prove the statement for real polynomials. 
And  then  it  takes  the  polynomial $$f(x)=p(x)(x^{2}+1)$$  where $p(x)$  is  irreducible  over  $\mathbb R$ and  considers  the  Galois group  of  $f(x)$ .  
I  cannot   understand  why  they  needed  to  take  a  product  of  two   irreducible   polynomials  instead  of  one polynomial  irreducible  in  $\mathbb R[x]$  or  at  what  step  of  this  proof  it  was  put  to  use.  
Please  don't  mind  anybody, I  am  uploading  an  image   of  the  proof.  Please   help  me  understand  these  two  things   properly.

Thanks  for  any  help.
 A: Consider the homomorphism $\varphi\colon\mathbb{C}[x]\to\mathbb{C}[x]$ defined by $\varphi(a)=\bar{a}$ for $a\in\mathbb{C}$ and $\varphi(x)=x$. It is clear that $\varphi\circ\varphi$ is the identity, so $\varphi$ is an automorphism. In particular, since having a root is equivalent to being divisible by a degree one polynomial, $f\in\mathbb{C}[x]$ has a root if and only if $\varphi(f)$ has a root; note that $\varphi(f)=\bar{f}$ according to your notation.
If $f$ has a root, then clearly $f(x)\bar{f}(x)$ has a root. If $f(x)\bar{f}(x)$ has a root, then either $f$ has a root or $\bar{f}$ has a root; in the latter case also $f$ has a root.
Clearly $\overline{f(x)\bar{f}(x)}=f(x)\bar{f}(x)$, so every coefficient of $f(x)\bar{f}(x)$ is real.
Finally, considering the splitting field $E$ of $(x^2+1)p(x)$ ensures that $\mathbb{C}$ is (isomorphic to a) subfield of $E$.
A: For the first question, if $z_0$ is a root of $f(x)\bar{f}(x)$, $f(z_0)\bar{f}(z_0) = 0$ which means $f(z_0)\overline{f(\bar{z_0})} = 0$ . 
thus either $f(z_0) = 0$ or $f(\bar{z_0}) = 0$ (in both case $f$ admits a complex root).
So we can conclude that if $f(x)\bar{f}(x)$ has a complex root, then $f(x)$ has a complex root.
For the second question. $\mathbb{C}$ is the splitting field of $x^2 + 1$ over $\mathbb{R}$ so in the answer $x^2 + 1$ is multiplied to make sure that $E$ "contains" $\mathbb{C}$($\mathbb{C}$ is isomorphic to a subfield of $E$).
