Answer is inside the Question!

$\begin{aligned}(r+2)(r+3)(r+4)(r+5)&=(r^2+5r+6)(r^2+9r+20)\\&=(r^2+2r+6+3r)(r^2+2r+6+7r+14)\\&=3r(7r+14)\\&=21r^2+42r\\&=21(r^2+2r+6)-126\\&=-126\end{aligned}$

Nice problem Rishabh !

We have $r$ as a root of the given quadratic equation , which implies , $r^{2} +2r +6 = 0$ .

From above equation we get , $(r+2) = \frac{-6}{r}$ and $r+3 = \frac{-r^{2}}{2}$ . Hence we get , $(r+2)(r+3) = 3r$ .

Now after simplifying $(r+4)(r+5)$ we get , $r^{2} + 9r +20$ : replacing $r^{2}$ with $-2r - 6$ , we get , $7r + 14$ .

so , $(7r + 14)(3r) = 21r^{2} +42r$ , taking $21r$ common , we get $(21r)(r+2)$ . Value of $r+2$ is nothing but $\frac{-6}{r}$ .

hence we get , $21r \times \frac{-6}{r}$ and finally we get the product as $21 \times -6 = \boxed{-126}$

This is rather a very long solution , but to have more types of solutions i posted it .

Shortest solution is by division algorithm . Please check Josh Banister's solution for more details :)

An alternate is expand (x^2+5x+6)(x^2+9x+20)=(3x)(7x+14)=21(x^2+2x)=-126.

Divide $(x+2)(x+3)(x+4)(x+5) \equiv x^4 + 14x^3 + 71x^2 + 154x + 120$ by $x^2 + 2x + 6$ . This gives us a remainder of $-126$ . In other words by the division algorithm: $(x+2)(x+3)(x+4)(x+5) = (x^2 + 12x + 41) (x^2 + 2x + 6) - 126$ When $r$ satisfies $r^2 + 2r + 6 = 0$ , it means $(r+2)(r+3)(r+4)(r+5) = -126$ .

$x^2+2x+6=0$ , $r$ is the root of the equation, then $r^2+2r+6 = 0$

$r^2+2r+1=-5$ ,

$(r+1)^2=-5 ......(1)$

$(r+2)(r+5)(r+3)(r+4) = (r^2+7r+10)(r^2+7r+10+2)$ ,

Let $r^2+7r+10 = r^2+2r+6+5r+4 = 5r+4 = t$ ,

$(r+2)(r+5)(r+3)(r+4) = t(t+2)= t^2+2t = (t+1)^2 -1 =$

$(5r+5)^2-1 = 25(r+1)^2 -1 =25(-5) -1$ from $(1)$

$= -125-1$ ,

$=-126$

If r is the root then r^2+2r+6=0... Then multiply r+2×r+3 summit all values which in eq...and similarly multiply remaining .....and finally u get ur ans -126