GATE CSE 2022 Q3: Value of (r+2)(r+3)(r+4)(r+5) for a root of x² + 2x + 6 = 0

Let $r$ be a root of the equation $x^2 + 2x + 6 = 0$ .

Then the value of the expression $(r + 2)(r + 3)(r + 4)(r + 5)$ is

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Answer: D (official GATE 2022 final answer key)

Concept used: Pair the factors so each product contains $r^2 + 7r$ , then use $r^2 = -2r - 6$ .

First step: $(r+2)(r+5) = r^2 + 7r + 10$ and $(r+3)(r+4) = r^2 + 7r + 12$ ; substitute $r^2 = -2r - 6$ in each.

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Source: official GATE 2022 CS question paper and final answer key, organised by IIT Kharagpur. Question paper · Answer key