Greatest Common Factor
The first method for factoring polynomials will be factoring out the greatest common factor. When factoring in general this will also be the first thing that we should try as it will often simplify the problem.
To use this method all that we do is look at all the terms and determine if there is a factor that is in common to all the terms. If there is, we will factor it out of the polynomial. Also note that in this case we are really only using the distributive law in reverse. Remember that the distributive law states that
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In factoring out the greatest common factor we do this in reverse. We notice that each term has an a in it and so we “factor” it out using the distributive law in reverse as follows,
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Let’s take a look at some examples.
Example 1 Factor out the greatest common factor from each of the following polynomials.
8x4 – 4x3 + 10x2 = 2 x2 ∙(4x2 – 2x +5)
3x6 – 9x2 + 3x = 3 x ∙ (x5– 3x + 1)
First we will notice that we can factor a 2 out of every term. Also note that we can factor an x2 out of every term. Here then is the factoring for this problem.
Factoring By Grouping
This is a method that isn’t used all that often, but when it can be used it can be somewhat useful. This method is best illustrated with an example.
Example 2
3x2 – 2x + 12x -8
In this case we group the first two terms and the final two terms as shown here,
(3x2 – 2x ) + (12x -8 )
Now, notice that we can factor an x out of the first grouping and a 4 out of the second grouping. Doing this gives,
3x2 – 2x + 12x -8=x(3x-2)+4(3x-2)
We can now see that we can factor out a common factor of 
so let’s do that to the final factored form.
3x2 – 2x + 12x -8=(3x-2)(x+4)
And we’re done. That’s all that there is to factoring by grouping. Note again that this will not always work and sometimes the only way to know if it will work or not is to try it and see what you get.
Special Forms
There are some nice special forms of some polynomials that can make factoring easier for us on occasion. This formulas was listed in “Special Factors” part. Let’s work some examples with these.
Let’s work some examples with these.
Example 3
3x2 + 6x + 9
This time we need two numbers that multiply to get 9 and add to get 6. In this case 3 and 3 will be the correct pair of numbers. Don’t forget that the two numbers can be the same number on occasion as they are here.
Here is the factored form for this polynomial.
3x2 + 6x + 9= (x+3)(x+3)=(x+3)2
Example 4
25x2 – 9
In this case all that we need to notice is that we’ve got a difference of perfect squares,
25x2 – 9 = (5x)2 – 32
So, this must be the third special form above. Here is the correct factoring for this polynomial.
25x2 – 9 = (5x+3)(5x-3)
Example 5
8x3 + 1
This problem is the sum of two perfect cubes,
8x3 + 1 = (2x)3 + (1)3
and so we know that it is the fourth special form from above. Here is the factoring for this polynomial.
8x3 + 1 = (2x + 1)(4x2 – 2x + 1)
Completing the Square
Completing the square may be used to solve any quadratic equation. For example:
Example 6
x2 + 6x + 5 = 0
1 Step
Focus on first two terms and formula (1) from chapter “Special factors” and think what should be the third term to make the perfect square. In our case it is 9, so we are adding and subtracting 9 :
x2 + 6x + 9 +5 – 9 = 0
2 Step
Now we can apply formula (3) from chapter “Special factors”:
(x+3)2 – 4 = ((x+3)-2)((x+3)+2) = (x+1)(x+5) = 0
And now you can easy to solve the equation x = -1 and x= -5
