Algera Chapter : Polnomial and Rational Functions Chapter : Polnomial and Rational Functions - Polnomial Functions and Their Graphs Polnomial Functions: - a function that consists of a polnomial epression in a form of P() = a n (n ) c (n ) d (n ) constant term where n W Leading Coefficient (a): - the coefficient of the largest degree term of a polnomial function. Root: - commonl known as solution, or zero. - the value for when P() =, which is the -intercept of the polnomial function (when = ). Graphs of Simple Monomials and their Transformations Eample : Sketch the following functions and their transformations. Lael at least two points a. = and f() =. = and f() = ( ) c. = and f() = ( ) (, ) (, ) (, ) (, ) f() = = = f() = ( ) (, ) (, ) (, ) (, ) f() = ( ) = (, ) (, ) (, ) (, ) d. = and f() = ( ) e. = and f() = f. = and f() = f() = f() = ( ) = = = (, ) (, ) (, ) (, ) (, ) (, ) (, ) (, ) Note how the end ehaviours of Even and Odd Degree Polnomial Functions change with the Sign of the Leading Coefficient (see the net page). f() = (, ) (, ) (, ) (, ) Coprighted Gariel Tang B.Ed., B.Sc. Page.
Chapter : Polnomial and Rational Functions Algera Graphs of Polnomials: - a polnomial graph is smooth and continuous. reak corner reak corner cusp Not a Polnomial Function Not a Polnomial Function Corners and Discontinuous Continuous ut Not Smooth End Behaviours and Leading Terms Polnomial Function Smooth and Continuous Odd Degree Polnomial Function and Positive Leading Coefficient, a > Odd Degree Polnomial Function and Negative Leading Coefficient, a < Even Degree Polnomial Function and Positive Leading Coefficient, a > Even Degree Polnomial Function and Negative Leading Coefficient, a < Odd Degree Polnomial Functions When a >, Left is Downward ( as ) and Right is Upward ( as ). When a <, Left is Upward ( as ) and Right is Downward ( as ). Even Degree Polnomial Functions When a >, Left is Upward ( as ) and Right is Upward ( as ). When a <, Left is Downward ( as ) and Right is Downward ( as ). Multiplicit: - when a factored polnomial epression has eponents on the factor that is greater than. c Even Multiplicit or Multiplicities of,, means -intercept is tangent to the -ais Odd Multiplicit or Multiplicities of, 5, means inflection at -int c P() = a( ),,,,, ( c) P() = a( ), 5, ( c) Page. Coprighted Gariel Tang B.Ed., B.Sc.
Algera Chapter : Polnomial and Rational Functions Graphing Polnomial Functions:. FACTOR the Polnomial. Otain the -intercepts letting each Factor EQUAL to Zero and Solve.. Lael the -intercepts. Note an Multiplicities.. Analze the Polnomial Function and determine its End Behaviours. Sketch the Graph using the End Behaviours as well as the -intercepts.. To verif, find the -intecept letting =. (Alternativel, the constant term from the epanded polnomial function is the -intercept.) If -int =, let equal to a simple numer and solve for. Eample : Sketch the graph of the polnomial function. Lael all intercepts and eplain the end ehaviours. a. P() = ¼ ( ) ( ). P() = ( )( ) ( ) P() = ¼ ( ) ( ) =, ( ) =, ( ) = -ints = (multi-three); (multi-two) and inflection tangent to -ais a < (a = ¼) Overall Degree = = - Even End Behaviour: Left is Downward ( as ) and Right is Downward ( as ). P() = ( )( ) ( ) ( ) =, ( ) =, ( ) = -ints = ; (multi-two) and (multi-two) tangent to -ais tangent to -ais a > (a = ) Overall Degree = = 5 - Odd End Behaviour: Left is Downward ( as ) and Right is Upward ( as ). P() (, ) P() (, ) Verif: Let = (since -int = ) P() = ¼() (() ) (() ) P() = ¼()() () P() = ¼()()() P() = We can see that the graph we sketched has a negative -value when < <. Verif: Let = for -int P() = (() )(() ) (() ) P() = ()() () P() = ()(9)(9) P() = We can see that the graph we sketched has a negative -value when < <. Coprighted Gariel Tang B.Ed., B.Sc. Page.
Chapter : Polnomial and Rational Functions Algera Eample : Factor the polnomials function, P() = 9. Without sketching the graph, descrie the functions its intercepts and eplain its end ehaviours. P() = 9 P() = ( 9 ) ( ) Factor Grouping P() = ( 9) ( 9) GCF of each racket P() = ( 9)( ) Factor out common racket P() = ( 9)( )( ) Factor Diff of Squares a > (a = ) Overall Degree = Odd End Behaviour: Left is Downward ( as ) and Right is Upward ( as ). ( 9) =, ( ) =, ( ) = Let = for -int -ints = 9,, and (No Multiplicities) P() = () 9() () P() = Eample : Factor the polnomials functions elow. Sketch their graphs laelling all intercepts and eplain their end ehaviours. a. P() =. P() = P() = P() = ( ) Take out common factor P() = ( 5)( ) Factor Trinomial =, ( 5) =, ( ) = -ints =, 5, and a < (a = ) Overall Degree = - Odd End Behaviour: Left is Upward ( as ) and Right is Downward ( as ). P() (, ) 5 P() = P() = ( ) ( ) Factor Grouping P() = ( ) ( ) GCF of each racket P() = ( )( ) Factor out common racket P() = ( )( )( 9) Factor Diff of Cues Note that the last trinomial is not factorale nor it eilds a set of real roots using the quadratic formula (the discriminant, ac < ). Hence, there will not no -ints associate with that factor (more in.). P() = ( ) ( 9) ( ) = -ints = (multi-two) - tangent to -ais a > (a = ) Overall Degree = - Even End Behaviour: Left is Upward ( as ) and Right is Upward ( as ). (, ) P() Verif: Let = (since -int = ) P() = () () () P() = () () P() = We can see that the graph we sketched has a positive -value when < <. Page. Note the umps here indicating that there are a set of comple roots. (If the remaining trinomial has real roots, the would have come down to the -ais and generate two -ints.) Verif: Let = for -int P() = () () () P() = Note the constant term of a polnomial function is its -int. The graph we sketched has a positive -value at an. Coprighted Gariel Tang B.Ed., B.Sc.
Algera Chapter : Polnomial and Rational Functions Eample 5: The graph elow represents the polnomial function, P() with the smallest possile degree. Write out the function in its factored form. a. (, ) From the graph: -ints = (multi-two) and (multi-three) tangent to -ais inflection Hence the factors are: = () = ( ) P() = a ( ) Overall Degree = = 5 - Odd End Behaviour: Left is Downward ( as ) and Right is Upward ( as ). (Therefore, we epect a to e positive.) Using point (, ) to solve for a = a() (() ) = a()() = a()() = 5a = a a = 5 P() = ( ). (, ) From the graph: -ints = (multi-two) - tangent to -ais, and Hence the factors are: = ( ) = ( ) = ( ) P() = a( ) ( )( ) Overall Degree = = - Even End Behaviour: Left is Downward ( as ) and Right is Downward ( as ). (Therefore, we epect a to e negative.) Using the -int at (, ) to solve for a = a(() ) (() ) (() ) = a() ()() = a(9)()() = a = a a = P() = ( ) ( )( ) - Assignment: Intro to Graphing Polnomials Worksheet and pg. 9 #5 to (all),, 5,, 9,,,, 5,, 9; Honours: # and Coprighted Gariel Tang B.Ed., B.Sc. Page 5.
Chapter : Polnomial and Rational Functions Algera - Dividing Polnomials Consider 5. 5 R We can sa that Original Numer Original Numer 5 = Divisor Quotient Divisor Divisor or, we can sa 5 = () () Remainder Quotient Remainder Polnomial Function Divisor Function In general, for P() D(), we can write P( ) R = Q( ) or P() = D() Q() R D( ) D( ) Restriction: D() Long Division to Divide Polnomials: Quotient Function Remainder Eample : Divide 9 5 9 5 9 5 = ( ) 9 ( ) 5 5 ( ) ( ) R= 5 You cannot divide a monomial a polnomial! Dividing Polnomial is onl possile Long Division! 9 5 = ( ) or 5 9 5 = ( )( ) 5 Page. Coprighted Gariel Tang B.Ed., B.Sc.
Algera Chapter : Polnomial and Rational Functions Coprighted Gariel Tang B.Ed., B.Sc. Page. Eample : Divide 5 Eample : Divide Eample : Divide 5 = ( 9) or 5 = ( )( 9) 9 9 5 9 5 = R = ( ) 9 or = ( )( ) 9 = Missing Term from Decreasing Degree! 9 = R = ( ) or = ( )( ) ( ) = Missing Term from Decreasing Degree! = R
Chapter : Polnomial and Rational Functions Algera Snthetic Division: - a simplified method to divide polnomial a inomial linear divisor (a ). Snthetic Division onl works well with divisor that is in a form of a, where Leading Coefficient of Divisor is (that is a = ). Eample 5: Divide Root of the Divisor 5 5 9 Coefficients and constant of Polnomial Add Numers in each column Remainder = Quotient = 9 (one degree less than the original polnomial) Eample : Divide 9 Remainder = 9 Quotient = Eample : Divide 5 5 9 Remainder = Quotient = 9 - Assignment: pg. #, 9,,,, 5,,, 9, 5, 59,, 5; Honours: # and Page. Coprighted Gariel Tang B.Ed., B.Sc.
Algera Chapter : Polnomial and Rational Functions - Real Zeros of Polnomials Root: - commonl known as solution, or zero. - the value for when P() =, which is the -intercept of the polnomial function (when = ). Consider, the quotient is, the remainder is. Hence we can sa that is a factor of. Similarl, if P() ( ) and the Remainder =, ( ) is a factor of P(), and is a root of P(). If R = when P ( ), then ( ) is a factor of P() and P() =. P() = D() Q() P() = Original Polnomial D() = Divisor (Factor) Q() = Quotient If R when P ( ) Eample : For the polnomial function, P() =, a. Determine if the divisors ( ) and ( ) are factors using snthetic or long divisions. Epress each dividends in a form of P() = D() Q() R().. Evaluate P() and P(). What is the relationship etween them and the remainder determined in part a.? a. For P() ( ),. For P() ( ), 5, then ( ) is NOT a factor of P(). P() = D() Q() R() P() = ( ) P() = ( )( 9) Eample : For P() = 5, eplain the relationship etween a. the order pair (, 5), the remainder of P() ( ) is 5, and P() = 5.. the -intercept is, the remainder of P() ( ) is, and P() =.. P() = () () () P() = () () P() = P() = () () () P() = () () P() = When P() ( ), the Remainder is the same as the value of P(). a. For (, 5), it means that when =, = 5. Hence, it has the same meaning as P() = 5, ecause P() =. The remainder of P() ( ) is 5 matches the - value of the point (, 5) as well as P(). (, ). For -intercept =, the order pair is (, ), it means that when =, =. Hence, it has the same meaning as P() =, ecause P() =. The remainder of P() ( ) is matches the -value of the point (, ) as well as P(). (, 5) Coprighted Gariel Tang B.Ed., B.Sc. Page 9.
Chapter : Polnomial and Rational Functions Algera Eample : Determine whether = and = i are roots for the function P() =. To test if = is a root of P(), we evaluate P(). P() = () () () P() = () (9) P() = Since P() =, = is a root of P(). To test if = i is a root of P(), we evaluate P(i). P(i) = (i) (i) (i) P(i) = (i ) (9i ) i Recall, P(i) = (i) (9) i i = P(i) = i 9 i = i Since P(i), = i is not a root of P(). The Remainder Theorem: P( ) To find the remainder of : Sustitute from the Divisor, ( ), into the Polnomial, P(). P( ) In general, when, P() = Remainder. To find the remainder of P( ) : Sustitute a a from the Divisor, (a ), into the Polnomial, P(). In general, when P( ) a, P a = Remainder. Eample : Find the remainder of the followings using the remainder theorem. a. 5 Dividing ( ) means sustituting = in the numerator for the remainder. R = () () 5() R = () () R = R = = =. 5 Dividing ( ) means sustituting = in the numerator for the remainder. R = () () 5() R = () (9) 5 R = 5 5 R = = = c. Dividing ( ) means sustituting = in the numerator for the remainder. R = ( ) ( ) ( ) R = ( ) ( 9 ) R = 9 R = = = Eample 5: When P() = k is divided, the remainder is. Find the value of k. P() ( ) with remainder = means that P() =. = () k() () = () k() = 9 k 9 = k = k k = 5 Page. Coprighted Gariel Tang B.Ed., B.Sc.
Algera Chapter : Polnomial and Rational Functions The Factor Theorem: P( ). If gives a Remainder of, then ( ) is the Factor of P(). OR If P() =, then ( ) is the Factor of P().. If P( ) gives a Remainder of, then (a ) is the Factor of P(). a OR If P =, then (a ) is the Factor of P(). a Eample : Determine whether ( ) is a factor for the function, P() = 9 9. To test if ( ) is a factor of P(), we evaluate P(). P() = () () 9() 9 P() = () (9) 9 P() = Since P() =, = is a root of P() and ( ) is a factor of P(). Eample : Determine whether ( ) is a factor for the function, P() =. To test if ( ) is a factor of P(), we evaluate P(⅓). P(⅓) = (⅓) (⅓) (⅓) P(⅓) = ( ) ( 9 ) Eample : If P() = k k and is a root of P(), find the value of k. Rational Roots: - an roots of a polnomial function that elong in a set of rational numers (an numers that can e epressed as a fractions of integers). Rational Roots Theorem: For a polnomial P(), a List of POTENTIAL Rational Roots can e generated Dividing ALL the Factors of its Constant Term ALL the Factors of its Leading Coefficient. Potential Rational Zeros of P() = = = = = ⅓ P(⅓) = Since P(⅓), = ⅓ is not a root of P() and ( ) is not a factor of P(). If = is a root of P(), it means that when P() ( ), the remainder = and P() =. = () k() k() = () k(9) k = k = k = k k = ALL Factors of the Constant Term ALL Factors of the Leading Coefficient Coprighted Gariel Tang B.Ed., B.Sc. Page.
Chapter : Polnomial and Rational Functions Algera Eample 9: List the potential rational roots for the following polnomials a. P() =. P() = Potential Rational Zeros = Potential Rational Zeros = Factors of Constant Term Factors of Leading Coefficient ±,,, 5,, ± Potential Rational Zeros = ±,,, 5,, To Factor Third Degree Polnomials: Potential Rational Zeros =. Generate a List of Potential Rational Zeros from the polnomial function, P().. Pick some Potential Integral Zeroes (Integers tpe zeros) and use the Factor Theorem to Test them for an actual root of the polnomial function.. Using Snthetic Division, find the Quotient dividing the polnomial with the root found in the previous step.. Find the Remaining Factors or Roots from the quotient (usuall a quadratic function) either using regular factoring methods or the quadratic formula. Eample : Completel factor the following polnomial functions and graph. ±,,,, 9, ±,,, = ±,,,, 9,, ½,,,, 9,, ¼,,,, 9,, ⅛,,,, 9, Possile Zeros = ±,,,, 9,, ½,, 9, ¼,, 9, ⅛,, 9 a. P() = 5 Potential Rational Zeros = ±,,, Test Potential Rational Zeros for Actual Roots. P() = () 5() () = ( ) is a factor Use Snthetic Division to find the Quotient. 5 P() = ( )( ) Factor the Quadratic ( nd degree) Quotient and Graph. -intercept = P() = ( )( )( ) -ints =, and. P() = 5 Potential Rational Zeros = ±,, 5, 5 (, ) Test Potential Rational Zeros for Actual Roots. P() = 5() () () = ( ) is a factor Use Snthetic Division to find the Quotient. Page. 5 5 5 P() = ( )(5 ) Factor the Quadratic ( nd degree) Quotient and Graph. P() = ( )(5 ) P() = ( ) (5 ) -intercept = -ints = 5 and (Multi-of-twos) (, ) Coprighted Gariel Tang B.Ed., B.Sc.
Algera Chapter : Polnomial and Rational Functions To Factor Higher Degree Polnomials (more than rd degree):. Generate a List of Potential Rational Zeros from the polnomial function, P().. Pick some Potential Integral Zeroes (Integers tpe zeros) and use the Factor Theorem to Test them for an actual root of the polnomial function.. Using Snthetic Division, find the Quotient dividing the polnomial with the root found in the previous step.. Find the Remaining Factors or Roots from the quotient (usuall a function with degree one less than the original polnomial) repeating steps through until a quadratic quotient appears. Then factor it using regular factoring methods or the quadratic formula. Eample : Completel factor the polnomial function, P() = 5 9 and graph. Potential Rational Zeros = ±,,,,, Test Potential Rational Zeros for Actual Roots. P() = () 5() () 9() = is not a root P() = () 5() () 9() = ( ) is a factor Use Snthetic Division to find the Quotient. 5 9 5 5 P() = ( )( 5 ) (, ) Since the Quotient is a rd degree polnomial, we have to repeat the entire process again until we get to a quotient that is a nd degree. Potential Rational Zeros of the Quotient = ±,,,,, Test Potential Rational Zeros for Actual Roots. P() = () () 5() = is not a root P() = () () 5() = ( ) is a factor Use Snthetic Division to find the Quotient. 5 P() = ( )( )( ) Factor the Quadratic ( nd degree) Quotient and Graph. P() = ( )( )( )( ) P() = ( ) ( ) -intercept = -ints = and (Multi-of-threes) - Assignment: pg. #, 9, 5, 9, 9,, 5, 55, 9, 5, 9, 95 (ou ma use our calculator for this); pg 5 # 55, 5, 9, 5 Honours: pg. #5, 5 and Coprighted Gariel Tang B.Ed., B.Sc. Page.
Chapter : Polnomial and Rational Functions Algera - Comple Zeros Comple Zeros: - an roots that are not real numers. - it occurs when a quadratic quotient has a negative discriminant ( ac) in the quadratic formula. - it can also occur when none of the potential rational zeros of an even degree polnomial turn out to e the actual -intercept of the polnomial. - graphs of polnomials with comple zeros can e recognized their local / asolute etremes (umps) that never lower or rise to meet the -ais. Eamples: P() = P() = 5 P() = The Zero Theorem P() = Note: All odd degree polnomials have at least one real zero (the line must e ale to go through the -ais). Even degree polnomials can have all or some comple zeros. In general, there are (n ) umps for an n th degree polnomial (an inflection counts as two umps). There are n numer of solutions (comple, real or oth) for an n th degree polnomial function accounting that that a zero with multiplicit of k is counted k times. Eample : Find the roots of the 5 th degree polnomial function, P() = ( )( 5 ). State the multiplicit of each root. We check if the Quadratic Factor is factorale. P() = ( )[ 5 ] P() = ( )[( )( )] P() = ( )( ) ( ) P() = ( ) ( ) -ints = (Multi-of-threes) and ½ (Multi-of-twos) Page. Coprighted Gariel Tang B.Ed., B.Sc.
Algera Chapter : Polnomial and Rational Functions Eample : Write a polnomial function with 5 as a root of multiplicit of, and as a root of multiplicit of, along with as a root. -intercepts = 5 (Multiplicit-of-Two), (Multiplicit-of-Three), and. P() = a ( 5) ( ) ( ) P() = a ( 5) ( ) Some Irrational Real Roots and All Comple Roots eist as Conjugate Pairs. If the irrational in a form of a c is a root, then a c is also a root. If i is a root, then i is also a root. If a i is a root, then a i is also a root. Eample : Given a th degree polnomial with some of the zeros equal to, i, i, and its other zeros. Zeros: = (real rational zero); = i (comple zero) = i = i (comple zero) = i = 5 (real irrational zero) = 5 Multipling Factors with Conjugate Irrational Numer Pairs:. Set up the factors from the roots. Be careful with the signs. If the roots are = (a c ) and 5, find = (a c ), then the factors are ( (a c ) ) and ( (a c ) ).. The Products of these Two Irrational Factors will ield a Quadratic Epression d e where d = a and e = (a c ). Basicall, the Coefficient of the Linear Term is the Negative Sum of these irrational roots; the Constant Term is the Product of these roots. Multipling Factors with Conjugate Comple Numer Pairs:. Set up the factors from the roots. Be careful with the signs. If the roots are = (a i) and = (a i), then the factors are ( (a i) ) and ( (a i) ).. The Products of these Two Comple Factors will ield a Quadratic Epression d e where d = a and e = (a ). Basicall, the Coefficient of the Linear Term is the Negative Sum of these comple roots; the Constant Term is the Product of these roots. Eample : A quadratic polnomial function has a root 5i. Determine the other root and the equation of this polnomial function. Zeros: = 5i (comple zero) = 5i P() = ( ( 5i) ) ( ( 5i) ) The Long Wa: Multipl them out using FOIL. P() = ( ( 5i) ) ( ( 5i) ) P() = ( 5i) ( 5i) P() = 5i 9 5i 5i 5i 5i P() = 9 5() P() = Note that we must add, a, as leading coefficient ecause we do not know if the graph has een stretched or reflected verticall. The Short Wa: P() = d e Coefficient of Linear Term, d = Neg Sum of roots d = [( 5i) ( 5i)] d = Constant Term, e = Product of roots e = ( 5i)( 5i) e = 9 5i 5i 5i e = 9 5() e = Coprighted Gariel Tang B.Ed., B.Sc. Page 5.
Chapter : Polnomial and Rational Functions Algera Eample 5: Find all roots of P() = 5 has a root of i. Zeros: = i (comple zero) = i P() = ( i)( (i)) Q() P() = ( ) Q() To find Q(), we can do Long Division. 5 5 ( ) 5 ( 5 ) ( ) R= The Short Wa: Eample : Find a lowest degree polnomial function that has ( i)( i) = d e Coefficient of Linear Term, d = Neg Sum of roots d = [(i) (i)] d = Constant Term, e = Product of roots e = (i)(i) = i = () e = P() = ( )( 5 ) P() = ( )( )( ) and i as roots. Zeros: = (irrational zero) = = i (comple zero) = i For ( ( ) )( ( ) ) Coefficient of Linear Term, d = Neg Sum of roots d = [( ) ( )] d = Constant Term, e = Product of roots e = ( )( ) e = ( ) e = e = ( ( ))( ( )) = ( ) For ( ( i) )( ( i) ) Coefficient of Linear Term, d = Neg Sum of roots d = [( i) ( i)] d = Constant Term, e = Product of roots e = ( i)( i) e = 9 i i i e = 9 () e = ( ( i))( ( i)) = ( ) P() = ( ( ))( ( ))( ( i))( ( i)) P() = ( )( ) Eample : Find all the zeros of the polnomial function, P() =. Potential Rational Zeros = ±,,, Test Potential Rational Zeros for Actual Roots. P() = () () () P() = () () () = ( ) is a factor Use Snthetic Division to find the Quotient. P() = ( )( ) The Quadratic ( nd degree) Quotient cannot e factored. We need to use the quadratic formula. = ± ac = a ± = = ± i i Zeros =,, and ± ( )( ) ( ) i Page. - Assignment: pg. #,, 5, 9,,,,, Honours: pg. #9, 9, 55 and Coprighted Gariel Tang B.Ed., B.Sc.