1. The problem is to factor the trinomial $x^2 - 12x - 45$ or determine if it is prime. 2. The general form of a quadratic trinomial is $ax^2 + bx + c$. Here, $a=1$, $b=-12$, and $c=-45$. 3. To factor, we look for two numbers that multiply to $ac = 1 \times (-45) = -45$ and add to $b = -12$. 4. The pairs of factors of $-45$ are $(1, -45)$, $(-1, 45)$, $(3, -15)$, $(-3, 15)$, $(5, -9)$, and $(-5, 9)$. 5. Among these, $3$ and $-15$ add to $-12$ because $3 + (-15) = -12$. 6. Rewrite the middle term using these numbers: $$x^2 + 3x - 15x - 45$$ 7. Group terms: $$(x^2 + 3x) + (-15x - 45)$$ 8. Factor each group: $$x(x + 3) - 15(x + 3)$$ 9. Factor out the common binomial: $$(x - 15)(x + 3)$$ 10. Therefore, the factorization is: $$x^2 - 12x - 45 = (x - 15)(x + 3)$$