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The complex numbers $1+i$ and $1+2i$ are both roots of the equation $x^5-6x^4+Ax^3+Bx^2+Cx+D=0$, where $A, B, C, D \in R$ Find the value of D.

My attempt: The given equation will have 5 roots (distinct or undistinct) since it is a polynomial equation of degree 5. The coefficients are all real. Since $1+i$ and $1+2i$ are two roots of the given equation, their conjugates are also the roots of this equation. Therefore $1-i$ and $1-2i$ are also the roots of this equation. Therefore the L.H.S can be factorized as $x^5-6x^4+Ax^3+Bx^2+Cx+D=\{x-(1+i)\}\{x-(1-i)\}\{x-(1+2i)\}\{x-(1-2i)\}Q(x)=(x^4-4x^3+11x^2-14x+10)Q(x)$

,where $Q(x)$ is a polynomial of degree 1 are it gives the unknown root of the given equation. We find $Q(x)$ is $x-2$ by long division. Multiplying the other factor by $(x-2)$ we get the constant term is -20. Now $D$ is equal to the constant term. Therefore $D$ is equal to -20.

Am I correct? Is there any other way to find $D$?

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Hint:

Use Vieta's formula

$6=1+i+1-i+1-2i+1+2i+t$

where $t$ is the fifth root

Can you complete the solution now?

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