Copyright 2007 by Pearson Education, Inc. A) We look first for a common factor: 6x 2y 4 21x 3y 5 3x 2y 6 3x 2y 4 2 7xy y2 . b) There are three terms...

Copyright 2007 by Pearson Education, Inc. A) We look first for a common factor: 6x 2y 4 21x 3y 5 3x 2y 6 3x 2y 4 2 7xy y2 . b) There are three terms in 2 7xy y 2. We determine whether the trinomial is a square. Since only y 2 is a square, we do not have a trinomial square. Can the trinomial be factored by trial and error? A key to the answer is that x is only in the term 7xy . The polynomial might be in a form like 1 y 2 y , but there would be no x in the middle term. Thus, 2 7xy y 2 cannot be factored. C) Have we factored completely?