Make the given changes in the indicated examples of this section and then factor.

In Example 2, change −ax² to +a⁴x².

Data from Example 2

In factoring ax⁵ − ax², we first note that each term has a common factor of ax 2. This is factored out to get ax²(x³ − 1). However, the expression is not completely factored because 1 = 13, which means that x³ − 1 is the difference of cubes [see Example 1(b)]. We complete the factoring by the use of Eq. (6.9). Therefore,

ax5 ax = ax(x - 1) ax(x - 1)(x2 + x + 1)