Best method to factor like a quadratic

Hi Tim,

I start by creating a factor tree for 2304. I don't necessarily go down to the prime factors every time, though that's OK. For example, I get 2304 = (4 * 4 * 12 * 12) or (2^8 * 3^2).

Then I look at these numbers and think, "How do I mix and match these factors to get two numbers that sum to/differ by 100?" While there's still a bit of trial and error involved, having the factors in mind and starting in the middle (i.e. the square root of 2304 = (2^4)(3^1) = 48) can help.

First try: 48 and 48. Sum of 96, difference of 0. But 96 is close to 100....so swap a 3 and a 2 between the numbers for a small change. That is, 48/3 * 2 and 48/2 * 3.

Second try: 32 and 72. Sum of 104, difference of 40. 104 is still close to 100, but we went too far. Go back to 48 and 48 and swap a 3 and a 4 instead. That is, 48/4 * 3 and 48/3 *4.

Third try: 36 and 64. Sum of 100, difference of 28. Bingo.

The sign of the coefficients plays a role, of course. Since +2304 was positive in the quadratic, the factors will have the same sign. That tells us to focus on getting the sum of the factors to be 100. If it had been -2304, the factors would have opposite signs and we would pay attention to the difference as we try to match the middle coefficient.