In general a polynomial of degree n hase n root, some of them possibly equal. The roots may be real or complex, but if the coefficients of the polynomial are real and a complex number  {jatex options:inline}z{/jatex}  is a root of the polynomial, then the complex conjugate of  {jatex options:inline}z{/jatex}  written  {jatex options:inline}z*{/jatex}  is also a root.
For example the polynomial  {jatex options:inline}z^2+2z+2{/jatex}  has a root  {jatex options:inline}z=-1+i{/jatex}  so according to the argument above  {jatex options:inline}z^*=-1-i{/jatex}  is also a root.
We can solve the equation  {jatex options:inline}z^2+2z+2=0{/jatex}  using the quadratic formula  {jatex options:inline}z=\frac{-b \pm \sqrt{b^2-4ac}}{2a}{/jatex}  (with  {jatex options:inline}a=1, \: b=2, \: c=2{/jatex}) to give  {jatex options:inline}z=\frac{-2 \pm \sqrt{2^2-4 \times 1 \times 2}}{2 \times 1}=\frac{-2 \pm \sqrt{=4}}{2}= -1 \pm i{/jatex}.
We can write the polynomal above as  {jatex options:inline}(z-(1+i))(z-(1-i)){/jatex}.
Suppose now that  {jatex options:inline}w, \: w^*{/jatex}  are complex conjugates.
{jatex options:inline}(z-w)(z-w^*)=z^2-zw^*-zw+ww*=z^2-z(w+w^*)+ww^*{/jatex}.
The sum and product of complex conjugates are real (If  {jatex options:inline}w=x+iy{/jatex}  then  {jatex options:inline}w^*=x-iy{/jatex}  sp  {jatex options:inline}w+w^*=2x{/jatex}  and  {jatex options:inline}ww^*=(x+iy)(x-iy)=x^2+ixy-ixy-i^2y^2=x^2+y^2{/jatex}.