1. **State the problem:** Simplify or factor the quadratic expression $m^2 - 10m + 11$. 2. **Recall the factoring formula:** For a quadratic $ax^2 + bx + c$, we look for two numbers that multiply to $ac$ and add to $b$. 3. Here, $a=1$, $b=-10$, and $c=11$. We need two numbers that multiply to $1 \times 11 = 11$ and add to $-10$. 4. The factors of 11 are 1 and 11. To get a sum of $-10$, both must be negative: $-1$ and $-11$. 5. Check sum: $-1 + (-11) = -12 \neq -10$, so these do not work. 6. Since no integer factors satisfy the conditions, the quadratic does not factor nicely over integers. 7. Use the quadratic formula to find roots: $$m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-(-10) \pm \sqrt{(-10)^2 - 4 \times 1 \times 11}}{2 \times 1} = \frac{10 \pm \sqrt{100 - 44}}{2} = \frac{10 \pm \sqrt{56}}{2}$$ 8. Simplify $\sqrt{56} = \sqrt{4 \times 14} = 2\sqrt{14}$: $$m = \frac{10 \pm 2\sqrt{14}}{2} = 5 \pm \sqrt{14}$$ 9. Therefore, the factorization over real numbers is: $$m^2 - 10m + 11 = (m - (5 + \sqrt{14}))(m - (5 - \sqrt{14}))$$ **Final answer:** The quadratic does not factor over integers but factors over reals as above.