MAC1105-College Algebra Chapter -Polynomial Division & Rational Functions. Polynomial Division;The Remainder and Factor Theorems I. Long Division of Polynomials A. For f ( ) 6 19 16, a zero of f ( ) occurs at. Because is a zero of the polynomial function f, therefore, ( ) is a factor of f ( ). This means that there eists a second-degree polynomial q( ) such that f ( ) ( ) q( ). To find q( ), one can use long division. Using long division, divide f ( ) 6 19 16 by. B. If the remainder is zero, then is a factor of the polynomial epression f ( ). C. The Division Algorithm If f ( ) and d( ) are polynomials such that d( ) 0, and the degree of d( ) is less than or equal to the degree of f ( ), there eists unique polynomials q( ) and r( ) such that f( ) d( ) q( ) r( ) where r( ) 0 or the degree of r( ) is less than the degree of d( ). If the remainder r( ) is zero, d( ) divides evenly into f ( ). In summary, the Division Algorithm can be written as * f( ) q( ) d( ) r( ) d( ) f ( ) is improper because degree of f ( ) degree of d d( ) ( ). * r ( ) is proper because the degree of r( ) degree of d d( ) ( ) Try: Divide: 1. 10 by. 1 by 1 1
II. Synthetic Division (if and only if first term of the divisor is "") Eample: divide 10 by - 1 0-10 - - 9-1 - -1 1 1 Remainder: 1 Quotient: 1 Try: 1. f ( ) 6 19 16 by.. 10 by. 1 by 1 III. Remainder and Factor Theorem A. Remainder Theorem If a polynomial f( ) is divided by k, the remainder is r f( k).
To evaluate a polynomial function f ( ) when k, divide f ( ) by k. The remainder will be f ( k ). Eample: Evaluate the function f ( ) 8 5 7 at. Using synthetic division, - 8 5-7 -6 - - 1-9 Because the remainder is -9, one can conclude that f ( ) 9 Therefore, ( -, -9 ) is a point on the graph of f. Check this. B. Factor Theorem A polynomial f( ) has a factor ( k) if and only iff f( k) 0 Eample: Determine if ( ) and ( ) are factors of the polynomial: f ( ) 7 7 18 C. Results of Synthetic Division: The remainder, r, obtained in the synthetic division of f ( ) by k, provides: 1. The remainder r gives the value of f at k. That is, r f( k). If r 0, ( k ) is a factor of f( ).. If r 0, ( k,0 ) is an -intercept of the graph of f Application eample: A rectangular house has a volume of 11 cubic feet. The height of the house is 1. Find the number of square feet of floor space in the house.
.5 Rational Functions I. Domain of a Rational Function Rational function is a function f that is a quotient of two polynomials, that is f ( ) p( ), s.t. p( ) and q( ) are polynomials and q( ) 0 q( ) Domain of f consists of all inputs for which q( ) 0 1 eample: f ( ), Domain: { 5}, or ( 5, 5) ( 5, ), graph and note where the asymptote is: 5 II. Vertical Asymptote: where the restriction for domain ( value that makes the denominator zero.) is: constant try: f ( ) 11 graph, vertical asymptote(s)? 8 where is f ( )? as toward restriction try: f ( ) graph, vertical asymptote(s)? 5 where is f ( )? as toward restriction III. Horizontal Asymptote: where the function value heads toward as, y constant 8 10 1 eample: f ( ), horizontal asymptote: f ( ) 8 11 11 if the degree of numerator = degree of denominator, take the ratio of the coefficient eample: f ( ), horizontal asymptote: f ( ) 0 or y 0 if the degree of the numerator <degree of the denominator, horizontal asymptote would be y 0
eample: f ( ) =, oblique asymptote: f 1 1 ( ) if the degree of the numerator is 1 greater than the degree of the denominator, oblique asymptote may occur (if numerator divided by denominatorresult with remainder; then as,.the term for remainder will approach zero or drops out, there is the oblique asymptote.) divisor Alternatively, it is possible that the f ( ) will result in a straight line if there is no remainder ( )( 1) (eample: f ( ) results in a straight line f ( ) 1, w/a discontinuity at ) ( ) *Note: graphs can cross horizontal or oblique asymptotes, but they cannot cross vertical asymptotes. eample: f ( ) ( 1)( ) Domain: { 1, }, or (, 1) ( 1, ) (, ) Only vertical asymptote: 1 Only horizontal asymptote: y 0 Since is not in the domain of the function ( where ), therefore, f ( ) 0, no -intercept y intercept: (0, 1) Try: f ( ) 5 f ( ) 1 1 f ( ) 6 f ( ) f ( ) 9 f ( ) 1 f ( ) ( 1) f 1 ( ) 5