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 $D$?



Use Vieta's formula


where $t$ is the fifth root

Can you complete the solution now?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.