If 2 + i is a root of the equation {x^3} - 5{ | vedclass

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(A) Since the coefficients of the polynomial ${x^3} - 5{x^2} + 9x - 5 = 0$ are real,complex roots must occur in conjugate pairs.Given one root is $z_1

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$1$ and $2 - i$

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(A) Since the coefficients of the polynomial ${x^3} - 5{x^2} + 9x - 5 = 0$ are real,complex roots must occur in conjugate pairs.Given one root is $z_1 = 2 + i$,the second root must be $z_2 = 2 - i$.Let the third root be $\alpha$.According to Vieta's formulas,the sum of the roots is equal to the negative of the coefficient of ${x^2}$ divided by the coefficient of ${x^3}$.Sum of roots $= z_1 + z_2 + \alpha = -(-5)/1 = 5$.Substituting the known roots: $(2 + i) + (2 - i) + \alpha = 5$.$4 + \alpha = 5 \Rightarrow \alpha = 1$.Thus,the other roots are $1$ and $2 - i$.

If $2 + i$ is a root of the equation ${x^3} - 5{x^2} + 9x - 5 = 0$,then the other roots are

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