The difference of perfect squares can be generalized as a factoring technique. By extension, any difference between terms that are raised to an even power like can be factored using the difference of perfect squares technique. This is because even powers can always be written as perfect squares: .

What about the sum or difference of terms with matching odd powers? How can those be factored?

#### Watch This

http://www.youtube.com/watch?v=55wm2c1xkp0 James Sousa: Factoring: Trinomials using Trial and Error and Grouping

#### Guidance

Factoring a trinomial of the form is much more difficult when . In Examples A and B, you will see how the following expression can be factored using educated guessing and checking and the quadratic formula. Additionally, you will see an algorithm (a step by step procedure) for factoring these types of polynomials without guessing. The proof of the algorithm is beyond the scope of this book, but is a reliable technique for getting a handle on tricky factoring questions of the form:

When you compare the computational difficulty of the three methods, you will see that the algorithm described in Example A is the most efficient.

A second type of advanced factoring technique is factoring by grouping. Suppose you start with an expression already in factored form:

Usually when you multiply the factored form of a polynomial, two terms can be combined because they are like terms. In this case, there are no like terms that can be combined. In Example C, you will see how to factor by grouping.

The last method of advanced factoring is the sum or difference of terms with matching odd powers. The pattern is:

This method is shown in the guided practice and the pattern is fully explored in the exercises.

**Example A**

Factor the following expression:

**Solution: ** While this trinomial can be factored by using the quadratic formula or by guessing and checking, it can also be factored using a factoring algorithm. Here, you will learn how this algorithm works.

First, multiply the first coefficient with the last coefficient:

Second, factor as you normally would with :

Third, divide the second half of each binomial by the coefficient that was multiplied originally:

Fourth, simplify each fraction completely:

Lastly, move the denominator of each fraction to become the coefficient of :

Note that this is a procedural algorithm that has not been proved in this text. It does work and can be a great time saving tool.

**Example B**

Factor the following expression using two methods different from the method used in Example A:

**Solution:** The educated guess and check method can be time consuming, but since there are a finite number of possibilities, it is still possible to check them all. The 6 can be factored into the following four pairs:

1, 6

2, 3

-1, -6

-2, -3

The -28 can be factored into the following twelve pairs:

1, -28 or -28, 1

-1, 28 or 28, -1

2, -14 or -14, 2

-2, 14 or 14, -2

4, -7 or -7, 4

-4, -7 or -7, -4

The correctly factored expression will need a pair from the top list and a pair from the bottom list. This is 48 possible combinations to try.

If you try the first pair from each list and multiply out you will see that the first and the last coefficients are correct but the coefficient does not.

A systematic approach to every one of the 48 possible combinations is the best way to avoid missing the correct pair. In this case it is:

This method can be extremely long and rely heavily on good guessing which is why the algorithm in Example A is provided and preferable.

An alternative method is using the quadratic formula as a clue even though this is not an equation set equal to zero.

This means that when set equal to zero, this expression is equivalent to

Multiplying by 2 and multiplying by 3 only changes the left hand side of the equation because the right hand side will remain 0. This has the effect of shifting the coefficient from the denominator of the fraction to be in front of the just like in Example A.

**Example C**

Factor the following expression using grouping:

**Solution: ** Notice that the first two terms are divisible by both 4 and and the last two terms are divisible by . First, factor out these common factors and then notice that there emerges a second layer of common factors. The binomial is now common to both terms and can be factored out just as before.

**Concept Problem Revisited**

The sum or difference of terms with matching odd powers can be factored in a precise pattern because when multiplied out, all intermediate terms cancel each other out.

When is distributed:

When is distributed:

Notice all the inside terms cancel:

#### Vocabulary

** To factor** means to rewrite a polynomial expression given as a sum of terms into a product of factors.

#### Guided Practice

1. Show how factors by checking the result given in the guidance section.

2. Show how factors by checking the result given in the guidance section.

3. Factor the following expression without using the quadratic formula or trial and error:

**Answers:**

1. Factoring,

2. Factoring,

3. Use the algorithm described in Example A.

#### Practice

Factor each expression completely.

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8.

Expand the following expressions. What do you notice?

9.

10.

11. Describe in words the pattern of the signs for factoring the difference of two terms with matching odd powers.

12. Describe in words the pattern of the signs for factoring the sum of two terms with matching odd powers.

Factor each expression completely.

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### Answers for Explore More Problems

To view the Explore More answers, open this PDF file and look for section 2.2.