GCF and Factoring Trinomials

Factoring Out the Greatest Common Factor (GCF)

Consider the example

        x (x - 3)  =  x² - 3x 

We can do the process in reverse:

Look at 

        x² - 3x 

and notice that both have a common factor of x.  

We can pull out the x terms (use the distributive property in reverse).

        x² - 3x  =  x (x - 3).

Example:  

Consider the expression 

        2x³ - 4x²  

1. Find the GCF
   

2. Pull it out.

Solution:

1. The GCF is 2x²
   

2. We can write 
   
           2x³ - 4x²  =  2x² (x - 2)

Exercises:

1. Find the GCF
   
   

2. Pull it out.
   
   

   1. x² - 5x                              
      
      

   2. 5xy + 25y                         
      
      

   3. 3x(2 - x) + 4(2 - x)            
      
      

   4. 6x²y³ - 8x²y²  + 10x⁴y⁴    

Factoring by Grouping

Consider the expression

        3x² + 6x - 4x - 8

We can factor this by grouping two at a time:

        (3x² + 6x) - (4x + 8)

We now pull out the GCF of each:

        3x (x + 2) - 4 (x + 2)

Pull out the GCF again:

        (3x - 4) (x + 2)

Exercises

Factor the following by grouping:

1. 3x² - 6x + x - 2       
   
   

2. x³ - 3x² + x - 3       
   

Factoring Trinomials With constant Leading Coefficient

Consider the trinomial:

        x² + 5x + 4

We want to factor this trinomial, that is do FOIL in reverse.  Since the leading coefficient is 1 we can write:

        x² + 5x + 4 = (x + a) (x + b)  

Where a and b are unknown numbers.

FOIL the right expression out to get:

        x² + 5x + 4 

        =  x² + bx + ax + ab 

        =  x² + (a + b) x + ab

Hence we are searching for two numbers such that their product is 4 and their sum is 5.  

Note that the only pairs of number whose product is 4 are

        (2,2) and (1,4).

Of these two pairs only (1,4) add up to 5.

Hence

        x² + 5x + 4  =  (x + 1) (x + 4)  

Two factor a trinomial with leading coefficient 1 we ask ourselves the following questions:

What two numbers multiply to the last coefficient and add to the middle coefficient?

Example: 

Factor

        x² - 3x - 10

The pairs that multiply to -10 are

        (1,-10)     and     (-1,10)    and     (5,-2)   and    (2,-5)

Only the last pair adds to -3 hence

        x² - 3x - 10  =  (x + 2) (x - 5)

Exercises: 

Factor if possible

1. x² + 8x + 15       
   
   

2. x² + 6x + 8         
   
   

3. x² - 4x - 5           
   
   

4. x² + 3x - 18        
   
   

5. x² - 7x +12         
   
   

6. x² - 4x - 9           
   

The AC Method

What can we do when the leading coefficient is not 1?

We use an extension of factoring by grouping called the AC method.

Step by Step method for factoring    Ax² + Bx + C

• Step 1.  Multiply together AC and list the factors of AC.
  
  

• Step 2.  Find a pair that adds to B.  If you cannot find such a pair then the trinomial does not factor.
  
  

• Step 3.  Rewrite the middle term as a sum of terms whose coefficients are the chosen pair.
  
  

• Step 4.  Factor by grouping.

   

Remember you should always first pull out the GCF

Example:  

Factor 

        2x² + 5x - 25

1. AC = (2)(-25) = -50 
   
   the pairs are

           (1,-50)     (-1,50)     (2,-25)     (-2,25)      (5,-10)   and   (-5,10).
   

2. We see that 

           -5 + 10  =  5   hence we choose the pair (-5,10)
   

3. We write  
   
           2x² - 5x + 10x - 25
   

4. (2x² - 5x) + (10x - 25) 
   
   =  x (2x - 5) + 5 (2x - 5) 
   
   =  (x + 5) (2x - 5)

Exercises:  

Factor the following

1. 15x² - x - 6                
   
   

2. 7x² - 20x + 12           
   
   

3. 30x³ + 25x² + 5x       
   
   

4. 4x⁴ + 8x² + 3            

For an interactive applet on the AC method click here

For an alternate method for the AC method click here

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