Jonah,

You probably looked for integers a, b, c and d so that 9x² + 6x + 4 = (ax + b)(cx + d). I did so also without success. The next technique to try is to complete the square

9x² + 6x + 4 = 9(x² + 2/3x + 4/9) = 9(x² + 2/3x + 1/9 - 1/9 + 4/9) = 9[(x + 1/3)² + 1/3]

The challenge now is that (x + 1/3)² + 1/3 doesn't factor, or at least doesn't factor over the real numbers. If you will settle for factors over the complex numbers then, since i² = -1 you can write

(x + 1/3)² + 1/3 = (x + 1/3)² - (1/3)i²

This I can factor as a difference of squares

(x + 1/3)² - (1/3)i² = [x + 1/3 - (1/√3)i][x + 1/3 + (1/√3)i]

So

9x² + 6x + 4 = 9(x + 1/3 - i/√3)(x + 1/3 + i/√3)

Harley