Multiple choice

Let $r$ be a root of the equation $x^2+2x+6=0$. The value of $(r+2)(r+3)(r+4)(r+5)$ is equal to.

  1. $51$
  2. $-51$
  3. $-126$
  4. $126$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

r^2 + 2r + 6 = 0. We want (r+2)(r+3)(r+4)(r+5). Group as [(r+2)(r+5)] * [(r+3)(r+4)] = (r^2 + 7r + 10)(r^2 + 7r + 12). Since r^2 = -2r - 6, substitute: (-2r - 6 + 7r + 10)(-2r - 6 + 7r + 12) = (5r + 4)(5r + 6) = 25r^2 + 30r + 20r + 24 = 25r^2 + 50r + 24. Substitute r^2 again: 25(-2r - 6) + 50r + 24 = -50r - 150 + 50r + 24 = -126.

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