Solving for the Expression Value

We are given the quadratic equation $x^2 + 2x + 6 = 0$ and $r$ is a root of this equation. This means $r$ satisfies the equation:

$r^2 + 2r + 6 = 0$

From this, we can derive:

$r^2 + 2r = -6$

Evaluating the Expression

The expression we need to evaluate is $(r + 2)(r + 3)(r + 4)(r + 5)$.

Let's rearrange and group the terms strategically:

$[(r + 2)(r + 5)] \times [(r + 3)(r + 4)]$

Now, expand each pair:

• $(r + 2)(r + 5) = r^2 + 5r + 2r + 10 = r^2 + 7r + 10$
• $(r + 3)(r + 4) = r^2 + 4r + 3r + 12 = r^2 + 7r + 12$

Substitute these back into the expression:

$(r^2 + 7r + 10)(r^2 + 7r + 12)$\

Let $P = r^2 + 7r$. The expression becomes:

$(P + 10)(P + 12)$\

Expand this:

$(P + 10)(P + 12) = P^2 + 12P + 10P + 120 = P^2 + 22P + 120$

Substitute $P = r^2 + 7r$ back:

$(r^2 + 7r)^2 + 22(r^2 + 7r) + 120$

This looks complicated. Let's use the relation $r^2 + 2r = -6$ differently.

Alternative Simplification

Consider the grouped expression again:

$(r^2 + 7r + 10)(r^2 + 7r + 12)$\

We can write $r^2 + 7r$ as $(r^2 + 2r) + 5r$.

Substitute $r^2 + 2r = -6$: $r^2 + 7r = -6 + 5r$.

Now substitute this into the grouped expression:

$(-6 + 5r + 10)(-6 + 5r + 12)$\

Simplify the terms inside the parentheses:

$(5r + 4)(5r + 6)$\

Expand this product:

$(5r + 4)(5r + 6) = 25r^2 + 30r + 20r + 24 = 25r^2 + 50r + 24$

Factor out 25 from the first two terms:

$25(r^2 + 2r) + 24$

Now substitute the known value $r^2 + 2r = -6$:

$25(-6) + 24$

Calculate the final value:

$-150 + 24 = -126$

Therefore, the value of the expression $(r + 2)(r + 3)(r + 4)(r + 5)$ is -126.