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Solve (x + 1)(x + 2)(x + 3)(x + 4) = 120 for real x.

1. x = 1 or x = -6
2. x = 1 only
3. x = -6 only
4. x = 2 or x = -5

Correct answer: x = 1 or x = -6

Solution

(x+1)(x+4) = x²+5x+4 and (x+2)(x+3) = x²+5x+6. Let y = x²+5x: (y+4)(y+6) = 120 ⇒ y² + 10y + 24 = 120 ⇒ y² + 10y - 96 = 0 ⇒ y = 6 or y = -16. y = -16 gives x²+5x+16=0 (no real root). y = 6: x²+5x-6=0 ⇒ (x+6)(x-1)=0 ⇒ x = 1 or x = -6.

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