Question

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

• A. 51
• B. $-51$
• C. 126
• D. $-126$

Hint:

When evaluating expressions involving roots, use the original quadratic to eliminate $r^2$ and simplify the product efficiently.

Answer 1

The Correct Option is D

Given equation:
\[ r^2 + 2r + 6 = 0 \]
We rewrite the expression $(r+2)(r+3)(r+4)(r+5)$ by grouping terms:
\[ (r+2)(r+5) \cdot (r+3)(r+4) \]
Compute the first pair:
\[ (r+2)(r+5) = r^2 + 7r + 10 \]
Compute the second pair:
\[ (r+3)(r+4) = r^2 + 7r + 12 \]
Let
\[ A = r^2 + 7r \]
Then the product becomes:
\[ (A + 10)(A + 12) \]
Expand:
\[ A^2 + 22A + 120 \]
Now compute $A = r^2 + 7r$ using the given equation.
From the quadratic:
\[ r^2 = -2r - 6 \]
Substitute into $A$:
\[ A = (-2r - 6) + 7r = 5r - 6 \]
Now compute $A^2$:
\[ A^2 = (5r - 6)^2 = 25r^2 - 60r + 36 \]
Substitute $r^2 = -2r - 6$:
\[ A^2 = 25(-2r - 6) - 60r + 36 \]
\[ A^2 = -50r - 150 - 60r + 36 \]
\[ A^2 = -110r - 114 \]
Now compute the full expression:
\[ A^2 + 22A + 120 = (-110r - 114) + 22(5r - 6) + 120 \]
Compute the $22A$ term:
\[ 22(5r - 6) = 110r - 132 \]
Combine all:
\[ (-110r - 114) + (110r - 132) + 120 \]
The $r$ terms cancel out:
\[ -114 - 132 + 120 = -126 \]
Thus, the value of the expression is −126.