Fastest way to factor out. Is there any?

Since the sum of the two numbers is 1, put down the + and - and take the square root of 1332 on a calculator. You know the two numbers have to be close to the square root ( 36.49....)

Trial & error is the fastest way and that is what you've done. Another way is complete factoring just in case trial and error won't work.
x² + x = 1,332
x² + 1/2x = 1,332 + (1/2)²
x² + 1/2x = 5,328/4 + 1/4
(x + 1/2)² = 5,329/4
x + 1/2 = 73/2

1st factor:
= x + 1/2 - 73/2
= x - 72/2
= x - 36

2nd factor:
= x + 1/2 + 73/2
= x + 74/2
= x + 37

Anyway, what you think best you may do. After all, nobody's should be bound by anybody with any rules.

x^2 + x - 1332
= x^2 + 37x - 36x - 1332
= (x^2 + 37x) - (36x + 1332)
= x(x + 37) - 36(x + 37)
= (x + 37)(x - 36)

By the way, you need to practice more so you can do it faster.

I don't really know how to explain this, but I'll try with an example.

5x^2 + 23x + 12 = (5x + 3)(x + 4)

1 .. 4 --> First term: 1(5) = 5x^2
5 .. 3 --> 2nd term: 5(4) + 1(3) = 23x
....... --> Last term: 4(3) = 12

Following that format, you just work backwards. First, start with the last term: find the factors of 1332. That may seem daunting, but notice that the middle term has a coefficient of +1. And since the 2nd is the result of adding 2 numbers together, the solution should be consecutive numbers: 36 and 37. In order for the coefficient to be +1 for the 2nd term and -1332 for the last, the solution has to be +37 and -36.

Ex: 30 has factors 1,2,3,5,6,10,15,30
x^2 + x - 30 = (x + 6)(x - 5)

Hopefully, that wasn't too confusing! XD

I think that's about the quickest way

either trial and error and practice or

quadratic equation:
to solve ax + by + c = 0
x = [-b ±√(b²-4ac)] / 2a


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