JEE Mathematics 2026 Question with Solution

Given: $S = \{x^3 + ax^2 + bx + c : a, b, c \in \mathbb{N} \text{ and } a, b, c \le 20\}$.

Find: The number of polynomials in $S$ divisible by $x^2 + 2$.

If a polynomial of degree $3$ is divisible by a polynomial of degree $2$, then the quotient must be a linear polynomial.

Assume

$$x^3 + ax^2 + bx + c = (x^2 + 2)(x + p)$$

where $p \in \mathbb{N}$.

Now expand:

$$(x^2 + 2)(x + p) = x^3 + px^2 + 2x + 2p$$

Comparing coefficients with $x^3 + ax^2 + bx + c$, we get

$$a = p, \quad b = 2, \quad c = 2p$$

Using the condition $a, b, c \le 20$:

• $b = 2$ satisfies the bound.
• $c = 2p \le 20 \Rightarrow p \le 10$.
• Since $a = p \in \mathbb{N}$, possible values are $p = 1, 2, 3, \ldots, 10$.

Each value of $p$ gives one distinct polynomial.

$$\text{Total number of polynomials} = 10$$

Therefore, the correct option is B.

Given: The polynomial is $x^3 + ax^2 + bx + c$ with $a, b, c \in \mathbb{N}$ and $a, b, c \le 20$.

Find: How many such polynomials are divisible by $x^2 + 2$.

For divisibility by $x^2 + 2$, write the cubic as

$$(x^2 + 2)(x + p)$$

because the quotient must be linear.

Expanding,

$$\begin{aligned} (x^2 + 2)(x + p) &= x^2 \cdot x + x^2 \cdot p + 2 \cdot x + 2 \cdot p \\ &= x^3 + px^2 + 2x + 2p \end{aligned}$$

Now compare term by term with

$$x^3 + ax^2 + bx + c$$

So,

$$a = p, \quad b = 2, \quad c = 2p$$

Since $c \le 20$,

$$2p \le 20 \Rightarrow p \le 10$$

Also $p \in \mathbb{N}$, hence

$$p = 1, 2, 3, \ldots, 10$$

This gives exactly $10$ valid choices.

Therefore, the number of such polynomials is $10$, so the correct option is B.