Let P(x) = x^3 - ax^2 + bx + c where a, b, c | vedclass

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(B) Given $P(x) = (x - \alpha)(x - \beta)(x - \gamma)$ where $\alpha, \beta, \gamma$ are integral roots.Substituting $x = 6$ into the equation,we get:

Vedclass · Accepted Answer

$15$

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(B) Given $P(x) = (x - \alpha)(x - \beta)(x - \gamma)$ where $\alpha, \beta, \gamma$ are integral roots.Substituting $x = 6$ into the equation,we get:$P(6) = (6 - \alpha)(6 - \beta)(6 - \gamma) = 3$.Since $\alpha, \beta, \gamma$ are integers,$(6 - \alpha), (6 - \beta),$ and $(6 - \gamma)$ must be integer factors of $3$. The factors of $3$ are $\{1, -1, 3, -3\}$.Let $x_1 = 6 - \alpha, x_2 = 6 - \beta, x_3 = 6 - \gamma$. Then $x_1 x_2 x_3 = 3$.Possible sets for $(x_1, x_2, x_3)$ are:$1) \{1, 1, 3\} \implies \alpha = 5, \beta = 5, \gamma = 3$. Then $a = \alpha + \beta + \gamma = 5 + 5 + 3 = 13$.$2) \{1, -1, -3\} \implies \alpha = 5, \beta = 7, \gamma = 9$. Then $a = \alpha + \beta + \gamma = 5 + 7 + 9 = 21$.$3) \{-1, -1, 3\} \implies \alpha = 7, \beta = 7, \gamma = 3$. Then $a = \alpha + \beta + \gamma = 7 + 7 + 3 = 17$.Comparing these values with the options,$a$ can be $13, 17,$ or $21$. Thus,$a$ cannot be $15$.

Let $P(x) = x^3 - ax^2 + bx + c$ where $a, b, c \in \mathbb{R}$ have integral roots such that $P(6) = 3$. Then $a$ cannot be equal to:

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