Extension to factoring, when the trinomials do not factor into a square (it also works with squares).
Sum-product-method
Say you have an expression like #x^2+15x+36#
Then you try to write #36# as the product of two numbers, and #15# as the sum (or difference) of the same two numbers. In this case (with both being positive) it's not so hard. You take the sum.
You can write #36=1*36=2*18=3*12=4*9=6*6#
Sums of these are #37,20,15,13,12# respectively
Differences are #35,16,9,5,0# respectively
#15=+3+12# will do. So the factoring becomes:
#(x+3)(x+12)#
Check your answer! #=x^2+12x+3x+36#
It's a bit harder when one or two of the numbers are negative, let's take #x^2-15x+36#
Same as the first, only now both factors are negative
#(x-3)(x-12)=x^2-12x-3x+36=# the original
Extra
If the last number (#36#) is negative, you will have to work with the difference of the factors. Check the next one yourself:
#x^2+5x-36=(x+9)(x-4)=?#
And now try: #x^2-5x-36=?#