Group the terms:(x³ + 2x²) – (5x + 6)Factor out common factors from each group:x²(x + 2) – 1(5x + 6) → This doesn’t match yet, so let’s test possible roots.Possible factors: ±1, ±2, ±3, ±6.Test x = 2:2³ + 2(2²) – 5(2) – 6 = 8 + 8 – 10 – 6 = 0 → So (x – 2) is a factor.Use synthetic division:Coefficients: 1 | 2 | -5 | -6Bring down 1 → multiply by 2 → 2 → add to 2 = 4 → multiply by 2 → 8 → add to -5 = 3 → multiply by 2 → 6 → add to -6 = 0Result: x² + 4x + 3Factor x² + 4x + 3:(x + 1)(x + 3)Final factors:(x – 2)(x + 1)(x + 3)