### By grouping

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

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

7711 x

Factor each completely.

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7721 x

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:

5066 x

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:

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