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Question from Jonah, a student:

how can i solve this by factoring: 9x^2 + 6x + 4

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

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