Have any of you heard of the Berry method of factoring? Does it always work?

Question

anonymous · Answer

I might know it but not with that name

anonymous · Answer

Is it a variation of the grouping method or the trial and error method??

anonymous · Answer

Ok, say you have $$8s ^{2}+26s-45$$
1st, write (8s    )(8s    )
2nd, multiply 8*(-45) which equals (-360)
3rd, take the factors of (-360) which add to +26
So, (+36)+(-10)=26
4th, (8s+36)(8s-10)
5th, factor=4(2s+9) 2(4s-5)

anonymous · Answer

if you are looking for the zeros of a quadratic equation, 

s= -9/2
s=  5/4

anonymous · Answer

I have seen this ... my teacher will not let us use it since each of the steps are not linked. For example, in your first step

8s times 8s is 64s^2

but you had 8s^2 when you started, you will divide out the extra 8 later, but we are not allowed to use Berry's method

anonymous · Answer

It is the fastest way that I have found to factor quadratics. However, when I had $$2x^2-7x-1=0$$, using this method didn't work.

anonymous · Answer

I was wondering if that was related to the fact that the answer isn't a "real" number

anonymous · Answer

Yes,  2x^2 - 7x - 1 cannot be factored; therefore; Berry's factoring method would not work.  2x^2 - 7x - 1 is prime

anonymous · Answer

Oooh K. Thanks! Sometimes I forget to look and see if it even needs to be factored, I guess.