KVPY #4

First note that as $r$ is a root of $x^{2} + 2x + 6$ we have that $r^{2} + 2r + 6 = 0 \Longrightarrow r^{2} + 2r = -6$ .

Now expanding the given expression, we have that

$(r + 2)(r + 3)(r + 4)(r + 5) = (r^{2} + 5r + 6)(r^{2} + 9r + 20) =$

$((r^{2} + 2r + 6) + 3r)((r^{2} + 2r + 6) + 7r + 14) = 3r(7r + 14) = 21(r^{2} + 2r) = 21 \times (-6) = \boxed{-126}$ .

If $r$ is a root of $x^2 + 2x + 6$ , then

$r^2 + 2r + 6 = 0$ $\color{#D61F06}\implies$ $\boxed{r^2 = -2r - 6}$ $\color{#20A900}(1)$

$(r+2)(r+3)(r+4)(r+5) = (r^2+3r+2r+6)(r^2+5r+4r+20)$ $=(r^2+5r+6)(r^2+9r+20)$

Substituting $\color{#20A900}(1)$ , we have

$(r+2)(r+3)(r+4)(r+5) =$ $(-2r - 6 + 5r + 6)(-2r - 6+9r+20) = (3r)(7r+14) = 21r^2 + 42r$

Substituting $\color{#20A900}(1)$ , we have

$(r+2)(r+3)(r+4)(r+5) =$ $21(-2r-6) + 42r = -42r - 126 + 42r=$ $\boxed{\color{#69047E}\large-126}$

First find the value of x by quadratic formula. Since r is the root of given equation. So x = r . Remember this is going to be a complex number.

Now put this value in given equation

(r+2)(r+3)(r+4)(r+5)

And answer will be -126 .