Let S = {x^{3} + ax^{2} + bx + c : a, b, c in | vedclass

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(D) Let $P(x) = x^{3} + ax^{2} + bx + c$. For $P(x)$ to be divisible by $x^{2} + 2$,we perform polynomial division or equate coefficients.Dividing $x^

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$10$

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(D) Let $P(x) = x^{3} + ax^{2} + bx + c$. For $P(x)$ to be divisible by $x^{2} + 2$,we perform polynomial division or equate coefficients.Dividing $x^{3} + ax^{2} + bx + c$ by $x^{2} + 2$ gives a quotient of $(x + a)$ and a remainder of $(b - 2)x + (c - 2a)$.For the polynomial to be divisible,the remainder must be zero,so $(b - 2)x + (c - 2a) = 0$.This implies $b - 2 = 0$ and $c - 2a = 0$.Thus,$b = 2$ and $c = 2a$.Given $a, b, c \in \mathbb{N}$ and $a, b, c \le 20$,we have $b = 2$ (which is fixed).For $c = 2a$,since $c \le 20$,we have $2a \le 20$,which means $a \le 10$.Since $a \in \mathbb{N}$,$a$ can take values from $\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$.Therefore,there are $10$ such polynomials.

Let $S = \{x^{3} + ax^{2} + bx + c : a, b, c \in \mathbb{N} \text{ and } a, b, c \le 20\}$ be a set of polynomials. Then the number of polynomials in $S$,which are divisible by $x^{2} + 2$,is

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