For which values of x is x + 1, 3, 5; 2, x + 2, 5; 2, 3, x + 4 = 0? …

Step 1. Add the columns. Replace $C_1$ by $C_1 + C_2 + C_3$. Each entry of the new column $1$ is $x + 9$.

Step 2. Take out the factor. The determinant is $(x + 9)

$.

Step 3. Clear column 1. $R_2 - R_1$ gives $(0, x - 1, 0)$ and $R_3 - R_1$ gives $(0, 0, x - 1)$. The small determinant is $(x - 1)^2$.

Step 4. Solve. $(x + 9)(x - 1)^2 = 0$ gives $x = -9$ or $x = 1$.