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

### Core Logic For a polynomial P(x) = x^3 + ax^2 + bx + c$P(x) = x^3 + ax^2 + bx + c$ to be divisible by x^2 + 2$x^2 + 2$, we can perform polynomial long division or use synthetic substitution (roots of x^2+2=0$x^2+2=0$). Alternatively, factorize P(x)$P(x)$: Since the degree is 3 and the leading coefficient is 1, we must have: x^3 + ax^2 + bx + c = (x^2 + 2)(x + k)$$x^3 + ax^2 + bx + c = (x^2 + 2)(x + k)$$ where k$k$ is a constant. ### Step 1: Equating Coefficients Expand the right side: (x^2 + 2)(x + k) = x^3 + kx^2 + 2x + 2k$$(x^2 + 2)(x + k) = x^3 + kx^2 + 2x + 2k$$ Comparing coefficients with x^3 + ax^2 + bx + c$x^3 + ax^2 + bx + c$: x^2$x^2$ coefficient: a = k$a = k$ x$x$ coefficient: b = 2$b = 2$ Constant term: c = 2k$c = 2k$ ### Step 2: Apply Constraints From the coefficient matching, we have a = fracc2$a = \frac{c}{2}$ and b = 2$b = 2$. Since a, b, c in mathbbN$a, b, c \in \mathbb{N}$ and a, b, c leq 20$a, b, c \leq 20$: b = 2$b = 2$ (this is fixed, always valid). c$c$ must be an even natural number such that c leq 20$c \leq 20$. The possible values for c$c$ are \2, 4, 6, 8, 10, 12, 14, 16, 18, 20\$\{2, 4, 6, 8, 10, 12, 14, 16, 18, 20\}$. For each such c$c$, a$a$ is uniquely determined as a = c/2$a = c/2$ and a leq 10$a \leq 10$ (which easily satisfies a leq 20$a \leq 20$). ### Step 3: Final Count The number of valid (a, b, c)$(a, b, c)$ tuples corresponds to the number of valid c$c$ values. There are exactly 10$10$ values for c$c$. Thus, the number of polynomials in S$S$ is 10$10$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 10 Mathematics: Polynomials Class 11 Mathematics: Complex Numbers and Quadratic Equations