# Find the value of D.

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 $$D$$?

$$6=1+i+1-i+1-2i+1+2i+t$$
where $$t$$ is the fifth root