1. **State the problem:** Factor the monic quadratic trinomials and determine if they are prime over the integers. 2. **Recall the formula and rules:** For a trinomial $x^2 + bx + c$, we look for two numbers that multiply to $c$ and add to $b$. 3. **First trinomial: $x^2 + 5x - 14$** - Sum needed: 5 - Product needed: -14 - Factors of -14: $(1, -14), (-1, 14), (2, -7), (-2, 7)$ - Check sums: - $1 + (-14) = -13$ - $-1 + 14 = 13$ - $2 + (-7) = -5$ - $-2 + 7 = 5$ (matches sum) - So the pair is $-2$ and $7$. - Factorization: $$x^2 + 5x - 14 = (x - 2)(x + 7)$$ 4. **Second trinomial: $x^2 - 3x - 4$** - Sum needed: -3 - Product needed: -4 - Factors of -4: $(1, -4), (-1, 4), (2, -2), (-2, 2)$ - Check sums: - $1 + (-4) = -3$ (matches sum) - $-1 + 4 = 3$ - $2 + (-2) = 0$ - So the pair is $1$ and $-4$. - Factorization: $$x^2 - 3x - 4 = (x + 1)(x - 4)$$ 5. **Third trinomial: $x^2 - 4x + 8$** - Sum needed: -4 - Product needed: 8 - Factors of 8: $(1, 8), (2, 4), (-1, -8), (-2, -4)$ - Check sums: - $1 + 8 = 9$ - $2 + 4 = 6$ - $-1 + (-8) = -9$ - $-2 + (-4) = -6$ - None of these sums equal -4. - Therefore, the trinomial is prime over the integers. **Final answers:** $$x^2 + 5x - 14 = (x - 2)(x + 7)$$ $$x^2 - 3x - 4 = (x + 1)(x - 4)$$ $$x^2 - 4x + 8 \text{ is prime over the integers}$$