5) \( n^{2}-10 n+9 \)

To factor the quadratic expression \( n^{2} - 10n + 9 \), follow these steps: 1. **Identify the coefficients:** - The quadratic expression is in the form \( an^{2} + bn + c \). - Here, \( a = 1 \), \( b = -10 \), and \( c = 9 \). 2. **Find two numbers that multiply to \( a \times c = 1 \times 9 = 9 \) and add up to \( b = -10 \):** - The numbers are **-1** and **-9** because: - \( -1 \times -9 = 9 \) - \( -1 + (-9) = -10 \) 3. **Rewrite the middle term using these numbers:** \[ n^{2} - 1n - 9n + 9 \] 4. **Factor by grouping:** \[ (n^{2} - 1n) + (-9n + 9) \] \[ n(n - 1) - 9(n - 1) \] 5. **Factor out the common binomial factor \((n - 1)\):** \[ (n - 1)(n - 9) \] **Final Factored Form:** \[ n^{2} - 10n + 9 = (n - 1)(n - 9) \] **Verification:** To ensure the factoring is correct, you can expand the factored form: \[ (n - 1)(n - 9) = n \cdot n + n \cdot (-9) - 1 \cdot n - 1 \cdot (-9) = n^{2} - 9n - n + 9 = n^{2} - 10n + 9 \] This matches the original expression, confirming that the factorization is correct.