Discussion
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
Answer: D
The value is \(-126\).
Since \(r\) is a root, \(r^{2} + 2r + 6 = 0\), so \(r^{2} + 2r = -6\). Pair the factors so that each product can be reduced with this relation.
\[ \begin{aligned} (r+2)(r+3) &= r^{2} + 5r + 6 = 3r \\ (r+4)(r+5) &= r^{2} + 9r + 20 = 7r + 14 \\ \text{product} &= 3r(7r + 14) \\ &= 21(r^{2} + 2r) = 21 \times (-6) = -126 \end{aligned} \]
The roots themselves are complex (\(r = -1 \pm i\sqrt{5}\)), so substituting them directly is slow; the reduction avoids that. 126 has the wrong sign, and 51 and \(-51\) come from slips such as reducing \(r^{2}\) to \(-6\) instead of \(-2r - 6\).
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