Find the value of the polynomial $y^{3} - 5 y - 2 y^{2} + 3$ when, $y = - b$

- b^{3} - 2 b^{2} + 5 b + 3

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b^{2} - 2 b^{3} + b + 1

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- b^{3} - 2 b^{2} - b + 3

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b^{3} - 2 b^{2} + b - 3

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Solution

The correct option is A $- b^{3} - 2 b^{2} + 5 b + 3$
Let $f (y) = y^{3} - 5 y - 2 y^{2} + 3$
When $y = - b$
$f (- b) = (- b)^{3} - 5 (- b) - 2 (- b)^{2} + 3$
$= - b^{3} + 5 b - 2 b^{2} + 3$
$= - b^{3} - 2 b^{2} + 5 b + 3$

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\frac{2 b^{2} + x^{2}}{b^{3} - x^{3}} - \frac{2 x}{b x + b^{2} + x^{2}} + \frac{1}{x - b} = 0

f (x) = ⎧ ⎨ ⎩ \begin{matrix} \frac{b^{3} + b - 2 b^{2} - 2}{b^{2} + 5 b + 6} - x^{2}; & 0 \leq x < 1 3 x - 4; & 1 \leq x \leq 3 \end{matrix}

b \in R

f (x)

x = 1,

b

\frac{2 b^{2} + x^{2}}{b^{3} - x^{3}} - \frac{2 x}{b x + b^{2} + x^{2}} + \frac{1}{x - b} = 0

\frac{a}{b} = \frac{5}{3},

\frac{a^{3} - b^{3}}{b^{3}} .

B

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b_{1} > b_{2} > b_{3}

λ

b_{1}, b_{2}, (b_{3} λ)

G . P .

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