FACTORING TECHNIQUES: Binomials

types

A  binomial is an expression containing two terms. Aside from factoring out the greatest common factor, there are three types of special binomials that can be factored using special techniques. These special binomials are:

1. The Difference of Squares:

• the first term is a perfect square: AxA
• the second term is a perfect square: BxB
• the two terms are being subtracted

2. The Sum of Cubes:

• the first term is a perfect cube: AxAxA
• the second term is a perfect cube: BxBxB
• the two terms are being added.

3. The Difference of Cubes:

• the first term is a perfect cube: AxAxA
• the second term is a perfect cube: BxBxB
• the two terms are being subtracted

The sum of squares:

The sum of squares can be factored with real numbers only if there is a common factor. Here is an example:

FACTORING TECHNIQUES: Difference of Squares

Difference of squares

To factor the difference of two squares it is useful to know the integers that are perfect squares. Here are the first 20:

FACTORING TECHNIQUES: Sum of Cubes

Sum of cubes

The factorization of x³ + y³ has a first factor of x + y, where x and y are the roots or the numbers that must be cubed to obtain each term. The second factor has three terms with the following pattern: 

Looking at the first factor, x + y

• the first term of the second factors is the square of the first term of the first factor.

• the second term of the second factor is the negative product of the two terms of the first factor

• the third term of the second factor is the square of the second term of the first factor. 

To factor the sum of two cubes it is useful to know the integers that are perfect cubes. Here are the first 10:

FACTORING TECHNIQUES: Difference of Cubes

difference of cubes

The factorization of the difference of cubes is similar to the factorization of the sum of cubes. The only difference between the two are the signs. Here are some observations to keep in mind:

1. The first factor in each will always have the same sign as the original problem. The second term of the second factor will always have the opposite sign. The last term of the second factor will always be positive.

2. The middle term of the second factor is the product of the terms of the first factor.

FACTORING TECHNIQUES: Trinomials

Trinomials

The product of two binomials is usually a trinomial. A trinomial is an expression containing three terms. For example:

1.  (x + 4)(x - 5) = x² - 5x + 4x - 20 = x² - x - 20
   

2.  (3x + 5)(2x - 7) = 6x² -21x + 10x - 35 = 6x² - 11x - 35

There are two types of products. 
The first example is a trinomial with a leading coefficient of 1:

x² - x - 20 = 1x² - x - 20

The second example is a trinomial with a leading coefficient not equal to one:

6x² - 11x - 35 = 6x² - 11x - 35

The technique used for factoring trinomials will depend on whether or not the leading coefficient is a one. 

Factoring Trinomials with a leading coefficient of one

It is important to understand where these trinomials come from. The patterns that produce them will be the patterns that we use to reverse the process. Notice the following pattern when multiplying two binomials:  

• The first two terms are identical and multiply to make x²
• The second two terms multiply to make the last term of the trinomial. They are factors of the last term.
• The middle term is the sum of the factors of the last term.

x² - 5x + 4x - 20

x² - x - 20

These observations can be used to go backwards. To factor a trinomial with leading coefficient of one, we need to look at the two binomials that generated this product. Consider, for example, the following: 

Factor: x² + 5x + 6         

step 1: 
Open up two sets of parentheses. The first term in each must multiply together to get x². This must be x and x.

step 2: 
Determine the factors of the last term. It is helpful to make a list to refer to. Include the appropriate sign. If a term is positive, the factors will have the same sign: (+)(+) or (-)(-). If the term is negative, the factors will have opposite signs. (+)(-). In our example, we need like signs. 
x² + 5x + 6
           /  \
          (+3)(+2) or (-3)(-2)
          (+6)(+1) 0r (-3)(-2)

step 3: 
Choose the factors whose sum will produce the coefficient of the middle term. If the last term is positive, the middle number will be the result of an addition. If the last term is negative, the middle number will be the result of a subtraction.

In this case +3 and +2 will multiply to make +6 and combine to make +5. Here are a few more example to review. Pay close attention to the signs.

       a² +2a - 15
      /\           /\
     a a       (+3)(-5)
                (-3)(+5)
               (-1)(+15)
               (+1)(-15)

     y² - 7y + 12
     /\           /\
    y y      (-1)(-12)
              (-1)(-12)
              (+1)(+12)
              (-3)(-4)
              (-6)(-2)
              (+6)(+2)

Factoring Trinomials when the  leading coefficient is not one

We will begin by looking a the pattern of multiplication of two binomials. Notice the following:

• The first two terms multiply to make 6x²
• The second two terms multiply to make the last term of the trinomial. They are factors of the last term.
• The middle term, is much more complicated. It is the sum of the outer and inner products of the binomial factors. This will be the key to finding the correct factors. 

6x² - 5(2x) + 4(3x) - 20

6x² - 10x + 12x - 20

6x² + 2x - 20

Factor:  6x² + 7x - 20

step 1: 
Open up two sets of parentheses. The first term in each must multiply together to get 6x². List all the options for the first terms. Do not place them in the parentheses.

                                                               

step 2: 
Skip to the last term.  List all the factor options for the last term. Do not place them in the parentheses. Ignore the signs of your terms for now. This narrows our choices down for finding the middle term. 

step 3: 
Begin a trial and error process. Choose one pair of factors from the first list and one pair of factors from the second list. 

For example we might choose (2x)(3x) and (20)(1). 

Multiply one factor from list one with one factor of list two, then multiply the "partners". Check to see if the products will combine to make the middle term. 

For example:  (2x)(1)= 2x and (3x)(20) = 60x but 2x and 60x will not make 7x by addition or subtraction. 

Note that (2x)(20)= 40x and (3x)(1) = 3x yields another option, yet one that again does not work. Continue pairing off factors until you find two products that will produce the middle term. 

Helpful HintS: 

Try the smaller numbers first. Usually the numbers that differ by a  smaller quantity are more likely to be the factors. 

For example, here we might want to try (2x)(3x) in combination with (5)(4).  

Also, try to make a product that is close to the middle number. Here, for example, we might choose (2x)(5x) or (3x)(5x)

(2x)(5) = 10
(3x)(4) = 12
__________________
(2x)(4) = 8
 (3x)(5) = 15

These are the two possibilities: 

OR
    

step 4: 
Once you have found the correct combination, place the appropriate sign with the second term according to the desired product. We need the middle term to be positive, so the larger product [15x] must be positive. Since we need 15x to be positive, make the 5 positive. We need the 8x to be negative so the 4 must be negative. 

step 5: 
Check by multiplying the two binomials.

(2x + 5)(3x - 4) = 6x² - 8x + 15x - 20 = 6x² + 7x  - 20 þ

Here are two more examples to review:

Example 1

Example 2

try (x)(3x) with (2)(2)

(x)(2) and (3x)(2)
2x and 6x will yield 4x

(x - 2)(3x + 2)

try (x)(5x) with (2)(3)

(x)(3) and (5x)(2)
3x and 10x will yield 13x

(x + 2)(5x + 3)

FACTORING TECHNIQUES: Grouping

Four Terms

If a polynomial has four terms, the expression may be factorable by grouping. This technique uses the technique of factoring out a common factor. For example, suppose we wish to factor the following expression: 2x + 2y + ax + ay

There is no common factor to all terms. However, a common factor can be found between the first two terms. A different common factor can be found between the second two terms. We GROUP the terms and factor out the common factor.

Group 1     Group 2
(2x + 2y) + (ax + ay)
2(x + y) + a(x + y)
We now have two terms:
term 1     term 2
2(x + y) + a(x + y)

Notice that the GROUP is now a common
factor between the two terms:

2(x + y) + a(x + y)


(x + y)(2 + a)

(x + y)(2 + a)    þ

step 1:
Create two groups by grouping two terms and two terms. Note: the groups should always be separated by addition. If the third term is negative, the negative sign goes in the second group. Insert a "+" between the groups.

step 2:
Factor out the common factor from each group.

step 3:
Factor out the common group as a common factor.

step 4:
Check your answer by multiplication.

EXAMPLE

Factor:  ac + bc - ay - by

 ac + bc - ay - by
ac + bc - ay - by
(ac + bc) + (- ay - by)

(ac + bc) + (- ay - by)
c(a + b) -y(a + b)

c(a + b) -y(a + b)
(a + b)(c - y)
(a + b)(c - y)

Factor out the greatest common factor from each group.

Factor out the common group.