How can I solve this Entrance Exam problem? This question is from NEST 2018.
Suppose ${2+ \sqrt{3}}$ and $1-i$  are roots of the equation $(x^2+px+1)(x^2-2x+q)=0$ where $p,q$ are integers and $i=\sqrt{-1}$. Then what is $p+q$ ?

If you substitute ${2+ \sqrt{3}}$ and $1-i$ in $(x^2+px+1)(x^2-2x+q)=0$, I obtain: 
$$\left\{\begin{matrix}
(({2+ \sqrt{3}})^2+p({2+ \sqrt{3}})+1)(({2+ \sqrt{3}})^2-2({2+ \sqrt{3}})+q)=0
\\ ((1-i)^2+p(1-i)+1)((1-i)^2-2(1-i)+q)=0
\end{matrix}\right.$$
From the first equation, I obtain: $(2+\sqrt3)(p+4)(3+\sqrt3+q)=0$, so $p=-4$. Now, plugging in the second equation, I have: $(2i-4)(q-2)=0$ and $q=2$. 
In conclusion $p+q=-4+2=-2$
