I ran across a neat little trick today for factoring quadratics.

Let’s start simple: factor x^2 + 11x + 24. To this, most people would start by finding factors of 24 because we need two numbers that multiply to +24 while adding to +11. Well 1,24 would sum too high and 4,6 would sum just a little too low. Nearby, 3,8 works out. x^2 + 11x + 24 = (x + 3)(x + 8).

How about factoring 6x^2 – 11x + 10? This is a bit more annoying because now we have to factor both 6 and 10, check pairwise combinations, and try both permutations of which factor of 6 multiplies which factor of 10. There are a lot more combinations to check.

However let’s transform this problem into a similar problem: factor 6(6x^2 – 11x – 10) = (6x)^2 + 11(6x) – 60 = y^2 + 11y – 60, where y = 6x. This factors to (y – 4)(y + 15) or (6x – 4)(6x + 15). However, there is an extra factor of 6 in there, which we remove to get (3x – 2)(2x + 5).

In summary, we turn an annoying problem we’d solve by trial and error to one that can be solved by less trial and error. Okay, we’re replacing factoring two small numbers with factoring one large number, but we have a heuristic to guide the latter in that the sum/difference of the two factors behaves in a simple manner.

tl;dr taking the time to turn a hard problem into an easier problem may save you time in the long run.

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