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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

  1. $51$
  2. $-51$
  3. $126$
  4. $-126$
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Answer: D

Since $r$ is a root of the given equation $\text{}x^2+2x+6=0, \\ $ So, $r$ will satisfy this equation.  Hence, we have $r^2+2r+6=0$

$\Rightarrow \boxed{r^{2} + 2r = -6}$

The value of expression $ (r+2)(r+3)(r+4)(r+5)\\ = (r^2+5r+6)(r^2+9r+20)\\ = (r^2+2r+6+3r)(r^2+2r+6+14+7r)\\
=(3r)(7r+14)\\=
21(r^2+2r)\\ =
21 \times -6 \\ =
\color{blue}{-126}.$
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