1. The first problem is to factor the trinomial $x^2 + x - 12$. 2. To factor a trinomial of the form $ax^2 + bx + c$, we look for two numbers that multiply to $ac$ and add to $b$. 3. Here, $a=1$, $b=1$, and $c=-12$. We need two numbers that multiply to $-12$ and add to $1$. 4. The numbers $4$ and $-3$ satisfy this because $4 \times (-3) = -12$ and $4 + (-3) = 1$. 5. Therefore, the factorization is $$(x + 4)(x - 3).$$ 6. The second problem is to factor and solve the quadratic equation $x^2 - 3x - 28 = 0$. 7. Again, we look for two numbers that multiply to $-28$ and add to $-3$. 8. The numbers $7$ and $-4$ satisfy this because $7 \times (-4) = -28$ and $7 + (-4) = 3$, but since the middle term is $-3x$, we use $-7$ and $4$ instead. 9. So, the factorization is $$(x - 7)(x + 4) = 0.$$ 10. Setting each factor equal to zero gives the solutions: $$x - 7 = 0 \Rightarrow x = 7,$$ $$x + 4 = 0 \Rightarrow x = -4.$$