$12 = \color{blue}{(4)} \ \color{indianred}{(3)},$ so $\ \color{blue}{4}\ $ #[and][y]# $\ \color{indianred}{3}$ are factors of $12.$ \\ $12 = \color{blue}{(12)} \ \color{indianred}{(1)},$ so $\ \color{blue}{12}\ $ #[and][y]# $\ \color{indianred}{1}$ are also factors of $12.$ \t \\ $2x = \color{blue}{(2)} \ \color{indianred}{(x)},$ so $\ \color{blue}{2}\ $ #[and][y]# $\ \color{indianred}{x}$ are factors of $2x.$ \t \\ $2x^2 = \color{blue}{(2)} \ \color{indianred}{(x^2)},$ so $\ \color{blue}{2}\ $ #[and][y]# $\ \color{indianred}{x^2}$ are factors of $2x^2.$ \\ $2x^2 = \color{blue}{(-2x)} \ \color{indianred}{(-x)},$ so $\ \color{blue}{-2x}\ $ #[and][y]# $\ \color{indianred}{-x}$ are also factors of $2x^2.$