Algebra – Polynomials – JEE Main 28 Jan 2026 Shift 1

Solution:

Let $P(x) = x^3 + ax^2 + bx + c$.
Since $P(x)$ is divisible by $x^2 + 2$, we can write:
$$x^3 + ax^2 + bx + c = (x^2 + 2)(x + k)$$
Comparing the coefficients:
$$x^3 + ax^2 + bx + c = x^3 + kx^2 + 2x + 2k$$

Equating coefficients of like powers of $x$:
1. Coefficient of $x^2$: $a = k$
2. Coefficient of $x$: $b = 2$
3. Constant term: $c = 2k$

From (1) and (3), we get $c = 2a$.
So the conditions on $a, b, c$ are:
1. $b = 2$ (fixed)
2. $c = 2a$
3. $a, b, c \in \mathbb{N}$ and $a, b, c \le 20$

Since $c = 2a$ and $c \le 20$, we have $2a \le 20 \Rightarrow a \le 10$.
Also $a \in \mathbb{N}$, so possible values for $a$ are $\{1, 2, 3, …, 10\}$.
For each valid $a$, $c$ is determined uniquely ($c = 2a$) and is within the range $[2, 20]$.

The number of such polynomials is equal to the number of possible values for $a$, which is 10.

Ans. (4)