1. **State the problem:** Factor the monic quadratic trinomials $x^2 + 5x - 14$ and $x^2 - 6x + 8$ and determine if they are prime over the integers. 2. **Recall the factoring rules:** For a quadratic trinomial $x^2 + bx + c$: - The factors will be of the form $(x + m)(x + n)$ where $m$ and $n$ are integers. - The product $m \times n = c$. - The sum $m + n = b$. - If $c$ is negative, $m$ and $n$ must have opposite signs. - If $c$ is positive and $b$ is negative, both $m$ and $n$ must be negative. 3. **Factor $x^2 + 5x - 14$:** - Here, $b = 5$, $c = -14$. - Since $c$ is negative, $m$ and $n$ have opposite signs. - Find factor pairs of 14: $(1,14), (2,7)$. - Check sums with opposite signs: - $1 + (-14) = -13$ (no) - $-1 + 14 = 13$ (no) - $2 + (-7) = -5$ (no) - $-2 + 7 = 5$ (yes) - So, $m = -2$, $n = 7$. - Factorization: $$x^2 + 5x - 14 = (x - 2)(x + 7)$$ 4. **Check by expanding:** $$ (x - 2)(x + 7) = x^2 + 7x - 2x - 14 = x^2 + 5x - 14 $$ 5. **Factor $x^2 - 6x + 8$:** - Here, $b = -6$, $c = 8$. - Since $c$ is positive and $b$ is negative, both $m$ and $n$ are negative. - Find factor pairs of 8: $(1,8), (2,4)$. - Check sums with both negative: - $-1 + (-8) = -9$ (no) - $-2 + (-4) = -6$ (yes) - So, $m = -2$, $n = -4$. - Factorization: $$x^2 - 6x + 8 = (x - 2)(x - 4)$$ 6. **Check by expanding:** $$ (x - 2)(x - 4) = x^2 - 4x - 2x + 8 = x^2 - 6x + 8 $$ 7. **Summary:** - For $x^2 + 5x - 14$, factors add to 5 and multiply to -14. - For $x^2 - 6x + 8$, factors add to -6 and multiply to 8. - Both trinomials factor over the integers and are not prime.