Factoring Polynomials Completely
So far, a typical polynomial has been a product of two factors, with possibly a common
factor removed first. However, it is possible that the factors can still be factored
again. A polynomial in a single variable may have as many factors as its degree.We
have factored a polynomial completely when all of the factors are prime polynomials.
Example 1
Factoring higherdegree polynomials completely
Factor x^{4} + x^{2}  2 completely.
Solution
Two numbers with a product of 2 and a sum of 1 are 2 and 1:
x^{4} + x^{2}  2 
= (x^{2} + 2)(x^{2}  1) 


= (x^{2} + 2)(x  1)(x + 1) 
Difference of two squares 
Since x^{2} + 2, x  1, and x + 1 are prime, the polynomial is factored completely.
In the next example we factor a sixthdegree polynomial.
Example 2
Factoring completely
Factor 3x^{6}  3 completely.
Solution
To factor 3x^{6}  3, we must first factor out the common factor 3 and then recognize
that x^{6} is a perfect square: x^{6} = (x^{3})^{2}:
3x^{6}  3 
= 3(x^{6}  1) 
Factor out the common factor. 

= 3((x^{3})^{2 } 1) 
Write x^{6} as a perfect square. 

= 3(x^{3}  1)(x^{3} + 1) 
Difference of two squares 

= 3(x  1)(x^{2} + x + 1)(x + 1)(x^{2}  x
+ 1) 
Difference of two cubes and
sum of two cubes 
Since x^{2} + x + 1 and x^{2}  x + 1 are prime, the polynomial is factored completely.
In Example 2 we recognized x^{6}  1 as a difference of two squares. However,
x^{6}  1 is also a difference of two cubes, and we can factor it using the rule for the
difference of two cubes:
x^{6}  1 = (x^{2})^{3}  1 = (x^{2}  1)(x^{4}
+ x^{2} + 1)
Now we can factor x^{2}  1, but it is difficult to see how to factor x^{4} x^{2}
`` 1. (It is
not prime.) Although x^{6} can be thought of as a perfect square or a perfect cube, in
this case thinking of it as a perfect square is better.
In the next example we use substitution to simplify the polynomial before
factoring. This fourthdegree polynomial has four factors.
Example 3
Using substitution to simplify
Factor (w^{2}  1)^{2}  11(w^{2}  1) + 24 completely.
Solution
Let a = w^{2}  1 to simplify the polynomial:
(w^{2}  1)^{2}  11(w^{2}  1)
+ 24 
= a^{2}  11a + 24 
Replace w^{2}  1 by a. 

= (a  8)(a  3) 


= (w^{2}  1  8)(w^{2}  1  3) 
Replace a by w^{2}  1. 

= (w^{2}  9)(w^{2}  4) 


= (w + 3)(w  3)(w + 2)(w  2) 

