The sum of non-real roots of the polynomial e | vedclass

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(C) Given the equation $x^3+3x^2+3x+3=0$.Let $f(x) = x^3+3x^2+3x+3$.Then $f'(x) = 3x^2+6x+3 = 3(x+1)^2$.Since $f'(x) \geq 0$ for all $x \in \mathbb{R}

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lies between $-1$ and $0$

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(C) Given the equation $x^3+3x^2+3x+3=0$.Let $f(x) = x^3+3x^2+3x+3$.Then $f'(x) = 3x^2+6x+3 = 3(x+1)^2$.Since $f'(x) \geq 0$ for all $x \in \mathbb{R}$,the function $f(x)$ is monotonically increasing.Thus,the equation $f(x)=0$ has exactly one real root,say $\alpha$.Since $f(-3) = -27+27-9+3 = -6  0$,the real root $\alpha$ lies in the interval $(-3, -2)$.Let the roots of the cubic equation be $\alpha, \beta, \gamma$,where $\beta$ and $\gamma$ are non-real complex conjugate roots.From Vieta's formulas,the sum of the roots is $\alpha + \beta + \gamma = -3$.Therefore,$\beta + \gamma = -3 - \alpha$.Since $-3 Adding $-3$ to all parts: $2-3 Thus,the sum of the non-real roots lies between $-1$ and $0$.

The sum of non-real roots of the polynomial equation $x^3+3x^2+3x+3=0$ is

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