x^3 + 9x^2 + 23x + 15 = 0

Let P(x) = x^3 + 9x^2 + 23x + 15.

By the Rational Root Theorem, possible rational roots are divisors of 15: ±1, ±3, ±5, ±15.

Testing x = -1: (-1)^3 + 9(-1)^2 + 23(-1) + 15 = -1 + 9 - 23 + 15 = 0. So, (x+1) is a factor.

Performing polynomial division or synthetic division of P(x) by (x+1) yields x^2 + 8x + 15.

Factoring the quadratic: x^2 + 8x + 15 = (x+3)(x+5).

Therefore, the roots are x = -1, x = -3, and x = -5.