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 higher-degree polynomials completely

Factor x⁴ + x² - 2 completely.

Solution

Two numbers with a product of -2 and a sum of 1 are 2 and -1:

Since x² + 2, x - 1, and x + 1 are prime, the polynomial is factored completely.

In the next example we factor a sixth-degree polynomial.

Example 2

Factoring completely

Factor 3x⁶ - 3 completely.

Solution

To factor 3x⁶ - 3, we must first factor out the common factor 3 and then recognize that x⁶ is a perfect square: x⁶ = (x³)²:

Since x² + x + 1 and x² - x + 1 are prime, the polynomial is factored completely.

In Example 2 we recognized x⁶ - 1 as a difference of two squares. However, x⁶ - 1 is also a difference of two cubes, and we can factor it using the rule for the difference of two cubes:

x⁶ - 1 = (x²)³ - 1 = (x² - 1)(x⁴ + x² + 1)

Now we can factor x² - 1, but it is difficult to see how to factor x⁴  x² `` 1. (It is not prime.) Although x⁶ 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 fourth-degree polynomial has four factors.

Example 3

Using substitution to simplify

Factor (w² - 1)² - 11(w² - 1) + 24 completely.

Solution

Let a = w² - 1 to simplify the polynomial: