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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. $-126$

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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