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Answer 1

polinomul

$f = X^{4} + mX^{3} + mX^{2} + 1$

a)

$f(-m) = (-m)^{4} + m(-m)^{3} + m(-m)^{2} + 1 = m^{4} - m^{4} + m^{3} + 1 = m^{3} + 1$

$f(-m) = 2 \iff m^{3} + 1 = 2$

$m^{3} = 1 \implies m = \sqrt[3]{1} \implies m = 1$

b)

$f = X^{4} + 2X^{3} + 2X^{2} + 1$

$\dfrac{ {x^{3}_{1}} - 1}{x_{1}} + \dfrac{ {x^{3}_{2}} - 1}{x_{2}} + \dfrac{ {x^{3}_{3}} - 1}{x_{3}} + \dfrac{ {x^{3}_{4}} - 1}{x_{4}} =$

$= x^{2}_{1} - \dfrac{1}{x_{1}} + x^{2}_{2} - \dfrac{1}{x_{2}} + x^{2}_{3} - \dfrac{1}{x_{3}} + x^{2}_{4} - \dfrac{1}{x_{4}}$

$= (x^{2}_{1} + x^{2}_{2} + x^{2}_{3} + x^{2}_{4}) - \bigg(\dfrac{1}{x_{1}} + \dfrac{1}{x_{2}} + \dfrac{1}{x_{3}} + \dfrac{1}{x_{4}} \bigg)$

se folosesc Relațiile lui Viete:

$\left\{\begin{matrix} S_{1} = x_{1}+x_{2}+x_{3}+x_{4} = - 1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ S_{2} = x_{1}x_{2}+x_{1}x_{3}+x_{1}x_{4}+x_{2}x_{3}+x_{2}x_{4}+x_{3}x_{4} = +2 \\ S_{3} = x_{1}x_{2}x_{3}+x_{1}x_{2}x_{4}+x_{1}x_{3}x_{4}+x_{2}x_{3}x_{4} = -2 \ \ \ \ \ \ \ \ \ \ \\ S_{4} = x_{1}x_{2}x_{3}x_{4} = +1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \end{matrix} \right.$

$(x_{1}+x_{2}+x_{3}+x_{4})^{2} =$

$= x^{2}_{1}+x^{2}_{2}+x^{2}_{3}+x^{2}_{4} - 2(x_{1}x_{4}+x_{2}x_{3}+x_{2}x_{4}+x_{3}x_{4})$

$= \bf S^{2}_{1} - 2S_{2}$

$\dfrac{1}{x_{1}} + \dfrac{1}{x_{2}} + \dfrac{1}{x_{3}} + \dfrac{1}{x_{4}} =$

$= \dfrac{x_{1}x_{2}x_{3} + x_{1}x_{2}x_{4} + x_{1}x_{3}x_{4} + x_{2}x_{3}x_{4}}{x_{1}x_{2}x_{3}x_{4}}$

$= \bf \dfrac{S_{3}}{S_{4}}$

=>

$= S^{2}_{1} - 2S_{2} - \dfrac{S_{3}}{S_{4}} = ( - 1)^{2} - 2 \cdot 2 - \dfrac{ - 2}{1} = 1 - 4 + 2 = \bf - 1$

c)

$m = 0 \implies f = X^{4} + 1$

$X^{4} + 1 = X^{4} - 1 + 2 = (X^{2} - 1)(X^{2} + 1) + 2$

=>

$(X^{4} + 1) : (X^{2} - 1) = (X^{2} + 1) + 2$