Warning: You may distribute the “4” through one of the other factors, but not both!!We obtain:4x2+3x−1=4x−1()x+1()This is the kind of factorization we are typically used to, and it often helps us insimplification problems.ExampleSimplify x+14x2+3x−1.Solutionx+14x2+3x−1=x+114x−1()x+1()1,x≠ −1=14x−1,x≠ −1By the way, we must complete the original problem! We had to factorfx( )=4x3−5x2−7x+2over C.Don’t forget the x−2()factor that we obtained earlier:4x3−5x2−7x+2=x−2()4x2+3x−1()=x−2()4x−1()x+1()
2.66SECTION 2.6: RATIONAL FUNCTIONSPART A: ASSUMPTIONSAssume fx( )is rational and written in the form fx( )=N x( )D x( ),where N x( )and D x( )are polynomials, and D x( )≠0(i.e., the zero polynomial).Assume for nowthat 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 willconsider what happens when we relax this assumption.Warning: Even though x2+xx=x+1∀x≠0(), we do notconsider the rational function[rule] fx( )=x2+xxto be a polynomial function [rule].PART B: VERTICAL ASYMPTOTES (VAs)An asymptote for a graph is a line that the graph approaches.ExampleLet fx( )=1x−2. Find any VAs for the graph of f.SolutionObserve that 1 and x−2have no real zeros in common.x−2=0⇔x=2, so the only VA has equation x= 2.