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
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1If $3+2i$ is one root, what would the other root be, you think? Take your time...:) – imranfat Apr 13 '17 at 15:40
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look up vietas relation – abel Apr 13 '17 at 15:41
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This can be helpful – kingW3 Apr 13 '17 at 15:42
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Then is the other root 3-2i? – K.Enoch Apr 13 '17 at 15:46
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@K.Enoch correct. So you know that $x-(3+2i)$ and $x-(3-2i)$ are factors of your polynomial... – Cameron Williams Apr 13 '17 at 16:06
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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, $5+3(-6)+B=0 \rightarrow B=-13$.
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