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.

Factor each completely.

This free worksheet contains 10 assignments each with 24 questions with answers.**Example of one question:**

**Watch bellow how to solve this example:**

Factor each completely.

This free worksheet contains 10 assignments each with 24 questions with answers.**Example of one question:**

**Watch bellow how to solve this example:**

Factor each completely.

This free worksheet contains 10 assignments each with 24 questions with answers.**Example of one question:**

**Watch bellow how to solve this example:**