Warning You may distribute the 4 through one of the other factors but not both

# Warning you may distribute the 4 through one of the

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Warning: You may distribute the “4” through one of the other factors, but not both!! We obtain: 4 x 2 + 3 x 1 = 4 x 1 ( ) x + 1 ( ) This is the kind of factorization we are typically used to, and it often helps us in simplification problems. Example Simplify x + 1 4 x 2 + 3 x 1 . Solution x + 1 4 x 2 + 3 x 1 = x + 1 1 4 x 1 ( ) x + 1 ( ) 1 , x ≠ − 1 = 1 4 x 1 , x ≠ − 1 By the way, we must complete the original problem! We had to factor f x ( ) = 4 x 3 5 x 2 7 x + 2 over C . Don’t forget the x 2 ( ) factor that we obtained earlier: 4 x 3 5 x 2 7 x + 2 = x 2 ( ) 4 x 2 + 3 x 1 ( ) = x 2 ( ) 4 x 1 ( ) x + 1 ( )
2.66 SECTION 2.6: RATIONAL FUNCTIONS PART A: ASSUMPTIONS Assume f x ( ) is rational and written in the form f x ( ) = N x ( ) D x ( ) , where N x ( ) and D x ( ) are polynomials, and D x ( ) 0 (i.e., the zero polynomial). Assume for now that N x ( ) and D x ( ) have no real zeros in common. Note: The textbook essentially makes this last assumption when it assumes that N x ( ) and D x ( ) have no common factors (over R ) aside from ± 1 , though, in Part E , we will consider what happens when we relax this assumption. Warning: Even though x 2 + x x = x + 1 x 0 ( ) , we do not consider the rational function [rule] f x ( ) = x 2 + x x to be a polynomial function [rule]. PART B: VERTICAL ASYMPTOTES (VAs) An asymptote for a graph is a line that the graph approaches. Example Let f x ( ) = 1 x 2 . Find any VAs for the graph of f . Solution Observe that 1 and x 2 have no real zeros in common. x 2 = 0 x = 2 , so the only VA has equation x = 2 .