Sometimes it is impossible to factor a polynomial by finding the greatest common factor.  For instance, the polynomial \(3xy - 24x^2 - 7y +56x\) has no greatest common factor.  In this case we can try searching the polynomial for factors that are common to some of the terms.  Then we can attempt a method known as grouping.
 
Take the polynomial and separate it into two groups.  We have

\(3xy - 24x^2 - 7y + 56x = (3xy - 24x^2) + (-7y + 56x)\)

Now check each group for any common factors.  In the binomial \((3xy - 24x^2)\) we see that there is a common factor of \(3x\). We factor out this common term to obtain \( 3x(y -8x)\). Checking the second binomial\((-7y + 56x)\) we see that \(-7\) is a common factor.  We factor it out to obtain \(-7(y - 8x)\). Then our polynomial becomes

\(3xy - 24x^2 - 7y + 56x = (3xy - 24x^2) + (-7y + 56x) = 3x(y - 8x) - 7(y - 8x)\)

In effect, we have created a new greatest common factor of this polynomial…\((y - 8x)\). We factor it out of both terms to obtain

\(3x(y - 8x) - 7(y - 8x) = (y - 8x)(3x - 7)\)

Finally we see that the correct factorization of the original polynomial is

\(3xy - 24x^2 - 7y + 56x = (y - 8x)(3x - 7)\)

Always check you work to see that the factorization is true.  We have just factored by grouping!

Let’s try another example: Consider the polynomial

\(120uv + 192u + 100v + 160\)

Before we attempt to factor by grouping, we see that there is a factor of \(4\) common to each term in this polynomial.  We factor it out and have

\(120uv + 192u + 100v + 160 = 4(30uv + 48u + 25v + 40)\)

Now we attempt the grouping method.  Separate the polynomial into two “groups.” 

\(4(30uv + 48u + 25v + 40) = 4[(30uv + 48u) + (25v + 40)]\)

Check for a common factor in the first group \((30uv + 48u)\). We see that \(6u\) is a common factor.  Factor it out to obtain \(6u(5v + 8)\).

Check for a common factor in the second group \((25v + 40)\). We see that \(5\) is a common factor.  Factor it out to obtain \(5(5v + 8)\).

Then our polynomial factors as:

\(120uv + 192u + 100v + 160 = 4(30uv + 48u + 25v + 40)\)

\(= 4[(30uv + 48u) + (25v + 40)]\)

\(= 4[6u(5v + 8) + 5(5v + 8)]\)

Do you catch the common factor we’ve created?  It’s \( 5v + 8\)!

Factor this out and we’re done.

\(4[6u(5v + 8) + 5(5v + 8)] = 4(5v + 8)(6u + 5)\)

So then our final factorization is

\(120uv + 192u + 100v + 160 = 4(5v + 8)(6u + 5)\)

Bellow you can download some free math worksheets and practice.