How can I solve this Entrance Exam problem?

Question

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 $p+q$ ?

Answer 1

As $p,q$ are integers, the other two roots are $2-\sqrt{3}$ and $1+i$ by examining the quadratic equation. Hence our equation would be equivalent to $$(x-(2+\sqrt{3}))(x-(2-\sqrt{3}))(x-(1-i))(x-(1+i))=0$$ $$(x^2-4x+1)(x^2-2x+2)=0$$ $$\therefore p+q=(-4)+(2)=-2$$

Answer 2

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 $p+q=-4+2=-2$