It works for some polynomials but not for others. Mostly , it works for this polynomial because the teacher, or author, or test-maker, chose a polynomial that could be factored this way.
Example 1
Factor: #3x^3+6x^3-5x-10#
I group the first two terms and take out any common factor of those two:
#(3x^3+6x^2)-5x-10=3x^2(x+2) -5x-10#
Now I'll take out any common factors in the other two terms. If i get a monomial times #(x+2)# then factoring by grouping will work. If I get something else, it won't work.
Ther common factor of #(-5x-10)# is #-5#. Taking that factor out leaves #-5(x+2)# so we know factoring by grouping will work.
#3x^3+6x^2-5x-10 = (3x^3+6x^2)+(-5x-10)#
#=3x^2(x+2)-5(x+2)#.
Now we have two terms with a common factor #C# where #C=(x-2)#. So we have #3x^2C-5C=(3x-5)C#
That is: we have #(3x^2-5)(x+2)#
We'll stop there if we're only willing to use integer (or rational) coefficients.
Example 2
Factor: #4x^3-10x^2+3x+15#
#4x^3-10x^2+3x+15=(4x^3-10x^2)+6x+15#
#=2x^2(2x-5)+6x+15#
Now if we take a common factor out of #6x+15# and get a monomial times #(2x-5)#, then we can finish factoring by grouping. If we get something else, then factoring by grouping won't work.
In this case we get #6x+15=3(2x+5)#. Almost!, But close doesn't work in factoring by grouping. So we can't finish this by grouping.
Example 3 You do the test-maker's job.
I want a problem that CAN be factored by grouping.
I start with #12x^3-28x^2# So, if it CAN be factored by grouping, the the rest of is has to look like what?
It has to be a monomial times #(3x-7)#.
So finishing with #6x-14# would work, or #15x-35#, or I could get tricky and use #-9x+21#. In fact ANY number times #(3x-7)# added to what I already have will give me a polynomial that can be factored by grouping.
#12x^3-28x^2+k3x-k7# for any #k# can be factored as:
#12x^3-28x^2+3kx-7k=4x^2(3x-7)+k(3x-7)=(4x^2+k)(3x-7)#
Final note: #k=-1# or #k=-9# would make good choices. Because then the fisrt factor is a difference of 2 squares and can be factored.
