












To find out more about how to factor these special binomials, click on one of the links below:  




Note: The sum of squares is not factorable unless there is a common factor.  


The sum of squares: 

cannot be factored with integer coefficients.  
Many
beginners believe that the sum of squares is factorable as follows:

This
misconception can be easily dispelled by checking. That is, multiply out
the factored form:


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 
Factoring the difference of squares



EXAMPLE 1  EXAMPLE 2  



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 
Factoring the sum of cubes 


EXAMPLE 1  EXAMPLE 2  



The factorization of x^{3} + y^{3} 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:
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 
Factoring the difference of cubes 



EXAMPLE 1  EXAMPLE 2  



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:





Trinomials 
The product of two binomials is usually a trinomial. A trinomial is an expression containing three terms. For example:
There
are two types of products.
The second example is a trinomial with a leading coefficient not equal to one:
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: 


(x
+ 4)(x
 5)
x^{2}  5x + 4x  20 x^{2}  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: x^{2} + 5x + 6


EXAMPLE 1  EXAMPLE 2  EXAMPLE 3  EXAMPLE 4  
a^{2} +2a  15  y^{2}  7y + 12  x^{2}  4x  21  x^{2} + xy  6y^{2}  
a^{2} +2a  15 
y^{2}  7y + 12 
x^{2}  4x  21 /\ /\ x x (21)(+1) (+21)(1) (+7)(3) (7)(+3) 
x^{2}
+ xy  6y^{2} ^{ }/\ /\ x x (6y)(+y) (+6y)(y) (+3y)(2y) (3y)(+2y) 

3 and +5 yield +2  3 and 4 yield 7  7 and + 3 yield 3  +3y and 2y yield 6y^{2}  
(a  3)(a + 5)  (y  4)(y  3)  (x  7)(x + 3)  ^{(x + 3y)( x  2y)}  
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: 


(2x
+ 4)(3x
 5)
6x^{2}  5(2x) + 4(3x)  20 6x^{2}  10x + 12x  20 6x^{2} + 2x  20


Working
backwards to factor can be a challenge. Always check your answers.
Patience and practice will go a long way. Consider the following:
Factor: 6x^{2} + 7x  20
Here are two more examples to review: 

Example 1 
Example 2 

Factor:
3x^{2}  4x  4
try (x)(3x) with (2)(2) (x)(2)
and (3x)(2)
(x
 2)(3x + 2) 
Factor:
5x^{2} + 13x + 6
try (x)(5x) with (2)(3) (x)(3)
and (5x)(2) (x + 2)(5x + 3) 





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 Notice
that the GROUP
is now a common (x + y)(2 + a) þ




EXAMPLE 
Factor: ac + bc  ay  by 

ac
+ bc  ay  by
(ac
+ bc)
+ (
ay 
by) c(a
+ b) y(a
+ b) 
Create
two groups. Insert a + between the groups. Factor out the greatest common factor from each group. Factor out the common group. 


