$3+2i$ is one root of a quadratic function $f(x)=x^2+Ax+B$, where $A,B\in\Bbb R$. Compute the ordered pair $(A,B)$

Question

In this question, I'm not exactly sure on how to solve this problem. Also, in $3+2i$, isn't there supposed to be a $x$. I understand that $i=\sqrt{-1}$. Please explain to me in a easy understandable way. I don't have much time, so please answer quickly

Answer 1

Since we know $3+2i$ is one of the roots, this means $x=3+2i$, and $$(3+2i)^2+A(3+2i)+B=0 \\ (5+12i)+(3A+2Ai)+B=0$$ From this, we know $$(5+3A+B)=0 \text{ and } (12i+2Ai)=0$$ Therefore, $2Ai=-12i \rightarrow A=-6$ Finally, $5+3(-6)+B=0 \rightarrow B=-13$.