Chapter 6.1, Problem 58E

In Exercises 55–62, factor the given expressions by grouping as Illustrated in Example 10.

2y – y² – 6y⁴ + 12y³

EXAMPLE 10 Factoring by grouping

Factor: 2x ‒ 2y + ax ‒ ay.

We see that there is no common factor to all four terms, but that each of the first two terms contains a factor of 2, and each of the third and fourth terms contains a factor of a. Grouping terms this way and then factoring each group, we have

$\begin{array}{ccc}2x-2y+ax-ay& =& (2x-2y)+(ax-ay)\\ & =& 2(x-y)+a(x-y)\\ & =& (x-y)(2+a)\end{array}$    now note the common factor of (x – y)