Factoring by Substitution

So far, the polynomials that we have factored, without common factors, have all been of degree 2 or 3. Some polynomials of higher degree can be factored by substituting a single variable for a variable with a higher power. After factoring, we replace the single variable by the higher-power variable. This method is called substitution.

 

Example 1

Factoring by substitution

Factor each polynomial.

a) x⁴ - 9

b) y⁸ - 14y⁴ + 49

Solution

a) We recognize x⁴ - 9 as a difference of two squares in which x⁴ = (x²)2 and 9 = 3². If we let w = x², then w² = x⁴. So we can replace x⁴ by w² and factor:

b) We recognize b) y⁸ - 14y⁴ + 49 as a perfect square trinomial in which y⁸ = (y⁴)² and 49 = 7². We let w = y⁴ and w² = y⁸:

 

Helpful hint

It is not actually necessary to perform the substitution step. If you can recognize that x⁴ - 9 = (x² - 3)(x² + 3) then skip the substitution.

 

Caution

The polynomials that we factor by substitution must contain just the right powers of the variable. We can factor y⁸ - 14y⁴ + 49 because (y⁴)² = y⁸, but we cannot factor y⁷ - 14y⁴ + 49 by substitution. In the next example we use substitution to factor polynomials that have variables as exponents.

 

Example 2

Polynomials with variable exponents

Factor completely. The variables used in the exponents represent positive integers.

a) x^2m - y²

b) z^{2n + 1} - 6z^{n + 1} + 9z

Solution

a) Notice that x^2m = (x^m)². So if we let w = x^m, then w² = x^2m:

b) First factor out the common factor z :