Let 1, omega,and omega^2 be the cube roots of | vedclass

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(B) Given that $1, \omega, \omega^2$ are the cube roots of unity,we know that $1+\omega+\omega^2=0$. The roots of the polynomial are given as: $z_1 =

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$5$

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(B) Given that $1, \omega, \omega^2$ are the cube roots of unity,we know that $1+\omega+\omega^2=0$. The roots of the polynomial are given as: $z_1 = 2\omega$ $z_2 = 2\omega^2$ $z_3 = 3+4\omega$ $z_4 = 3+4\omega^2$ $z_5 = 5-(\omega+\omega^2) = 5-(-1) = 6$ Since the polynomial has real coefficients,if a complex number $z$ is a root,its conjugate $\bar{z}$ must also be a root. $1$. For $z_1 = 2\omega$,the conjugate is $\bar{z_1} = 2\bar{\omega} = 2\omega^2$,which is $z_2$. $2$. For $z_3 = 3+4\omega$,the conjugate is $\bar{z_3} = 3+4\bar{\omega} = 3+4\omega^2$,which is $z_4$. $3$. For $z_5 = 6$,the conjugate is $\bar{z_5} = 6$,which is $z_5$ itself. All these roots are already present in the given set. Thus,the set of roots is $\{2\omega, 2\omega^2, 3+4\omega, 3+4\omega^2, 6\}$. There are $5$ distinct roots. Therefore,the minimum degree of the polynomial is $5$.

Let $1, \omega$,and $\omega^2$ be the cube roots of unity. The least possible degree of a polynomial with real coefficients,having $2\omega, 2\omega^2, 3+4\omega, 3+4\omega^2$,and $5-\omega-\omega^2$ as roots is

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