JEE Main 2026 — Jan 28, Shift 1 — Mathematics solution

JEE Main2026Jan 28, Shift 1Mathematics

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

A
20
B
6
C
120
D
10

Solution

If $x^{2} + 2$ divides $x^{3} + a x^{2} + b x + c$ , write

$x^{3} + a x^{2} + b x + c = (x^{2} + 2) (x + \frac{c}{2})$ (the quotient is monic of degree 1 with constant term $c /2$ to match the constant terms).

Expanding the right side: $x^{3} + \frac{c}{2} x^{2} + 2 x + c$ .

Match coefficients: $a = \frac{c}{2}$ and $b = 2$ .

Constraints: $b = 2$ (fixed). $c$ even, $1 \leq c \leq 20$ , and $a = c /2$ must be a natural number $\leq 20$ . So $c \in {2, 4, 6, \dots, 20}$ , giving 10 choices.

Total polynomials: $10$ . Option (4).

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