Zeros of Polynomials

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1 Math Zeros of Polyomials I these sectios we will study polyomials algebraically. Most of our work will be cocered with fidig the solutios of polyomial equatios of ay degree that is, equatios of the form 1 f x = ax + a x ax + ax+ a = 0 (1) ( ) Defiitio A root or solutio of equatio (1) is a umber c that whe substituted for x leads to a true statemet. Thus, c is a root of equatio (1) provided f ( c ) = 0. We also refer to the umber c i this case as a zero of the fuctio f. Note: If a root is repeated times, we call it a root of multiplicity Divisio of Polyomials The process of log divisio for polyomials follows the same four-step cycle used i ordiary log divisio of umbers: divide, multiply, subtract, brig dow. Notice that i settig up the divisio, we write both the divided ad divisor i decreasig powers of x. Exercise #1 Divide + + by x +. 4x 3x x 1 The result of the divisio ca be writte as: or Note 1) Secod equatio is valid for all real umbers x, whereas first equatio carries implicit restrictios that x my ot equal -. For this reaso, we ofte prefer to write our results i the form of the secod equatio. ) The degree of the remaider is less tha the degree of the divisor. This is very similar to the situatio with ordiary divisio of positive itegers, where the remaider is always less tha the divisor.

2 The Divisio Algorithm Let f ( x) ad g( x) be polyomials with g( x) of lower degree tha f ( x ) ad assume that g( x) 0. The there are uique polyomials ( ) r( x) such that f ( x) = g( x) q( x) + r( x) q x ad where r( x ) = 0 or the degree of r( x) is less tha the degree of ( ) The polyomials f ( x) ad ( ) respectively, q( x) is the quotiet, ad r( x) is the remaider. g x. g x are called the divided ad divisor, Whe r( x ) = 0, we have f ( x) = g( x) q( x) ad we say that g( x ) ad q( x ) are factors of ( ) f x. Exercise # Usig log divisio to fid a quotiet ad a remaider. 4 Divide 5x x + x by x +. Sythetic Divisio - Sythetic divisio is a quick method of dividig polyomials. - It ca be used whe the divisor is of the form x k. - I the sythetic divisio we write dow oly the essetial parts of the log divisio table (the coefficiets). Exercise #3 Use sythetic divisio to perform the followig divisios: a) 3x x + x x + If f ( x) = 3x x + x, evaluate ( ) f. What do you observe?

3 3 b) 1 1 x x + x x f x = x x + x+ 1, evaluate If ( ) f 1.What do you observe? 3 The Remaider Theorem Whe we divide a polyomial f ( x) by x c, the remaider is f ( c ). Proof The Factor Theorem The polyomial x c is a factor of the polyomial f ( x) if ad oly if f ( k ) = 0 (that is, if ad oly if c is a zero of f ( x ) Exercise #3 a) Let ( ) f x x x = + 6. i) Evaluate f ( ). ii) Is x a factor of ( ) f x x x = + 6?

4 b) Let ( ) f x = x + x i) Fid the remaider whe f ( x ) is divided by x + 3. ii) Is x + a factor of f ( x) = 4x + 5x + 8? 4 Exercise #4 Let ( ) f x = x 4x + x 1. a) What is the remaider whe dividig the give polyomial by x? I how may ways ca you fid the remaider? Which method is the easiest oe? b) Is f x? x a factor of ( ) x a factor of ( ) c) Is 1 f x? The Cojugate Zeros Theorem If f ( x) is a polyomial fuctio with real coefficiets ad if a+ bi is a zero of f ( x ), the its cojugate a bi is also a zero of f ( x ). Exercise #5 For each polyomial, oe zero is give. Fid all the others. f x x x x a) ( ) = ; 5i

5 5 4 f x x x x x b) ( ) = ; 1+ 3i The Fudametal Theorem of Algebra Every polyomial equatio of the form 1 f ( x) = ax + a 1x ax + ax 1 + a0= 0 ( 1, a 0) has at least oe complex zero. (This root may be a real umber). The Liear Factors Theorem (Complete Factorizatio Theorem) Every polyomial of degree ca be expressed as a product of liear factors. ( ) ( )( ) ( ) f x = a x x1 x x... x x, where a is the leadig coefficiet ad x i are the zeros of the polyomial. The Zeros Theorem Every polyomial of degree 1 has exactly zeros, where a zero of multiplicity k is couted k times. Exercise #6 Write each polyomial as a product of liear factors. f x = x x a) ( ) 3 5

6 b) F( x) = x 5 6 g x x x c) ( ) = Exercise #7 Fidig polyomial equatios satisfyig give coditios. I each case, fid a polyomial equatio with real coefficiets f ( x ) = 0 satisfyig the give coditios. If there is o such equatio, say so. a) Fid a polyomial fuctio of degree 3 havig the umbers -3, 1, ad 4 as roots ad f = 30. satisfyig ( ) b) Fid a polyomial fuctio of degree 4 havig the followig zeros: 4 as a zero of multiplicity, ad 3+i.

7 The Number ad Locatio of Real Zeros 7 Descartes Rule of Sigs I some cases, the followig rule discovered by the Frech philosopher ad mathematicia Ree Descartes aroud 1637 is helpful i elimiatig cadidates from legthy lists of possible ratioal roots. x is a polyomial with real coefficiet, writte with descedig powers of x (ad omittig powers with coefficiet 0), the a variatio i sig is a chage from positive to egative or egative to positive i successive terms of the polyomial (adjacet coefficiets have opposite sigs). To describe this rule, we eed the cocept of variatio i sig. If f ( ) Example How may variatios i sig occur i the followig polyomial? ( ) f x = x x x + x + x Descartes Rule of Sigs Let f ( x ) be a polyomial with real coefficiets ad a ozero costat term. a) The umber of positive real zeros of f ( x) is either equal to the umber of variatios i sig i f ( x ) or is less tha that by a eve whole umber. b) The umber of egative real zeros of f ( x) is either equal to the umber of variatios i sig i f ( x) or is less tha that by a eve whole umber. Exercise #8 Use Descartes rule of sigs to determie the possible umber of positive real zeros ad egative real zeros for each fuctio. f x = x x + x+ a) ( ) g x x x x x b) ( ) =

8 The Upper ad Lower Bouds Theorem 8 Let f ( x ) be a polyomial with real coefficiets. 1. If we divide f ( x ) by x c (with c > 0 ) usig sythetic divisio ad if the row that cotais the quotiet ad remaider has o egative etry ( all etries are positive or zero), the c is a upper boud for the real zeros of f ( x ).. If we divide f ( x ) by x c (with c < 0 ) usig sythetic divisio ad if the row that cotais the quotiet ad remaider has etries that alterate positive (or 0) ad egative (or 0), the c is a lower boud for the real zeros of f ( x ). Fidig all the ratioal zeros of a polyomial The Factor Theorem tells us that fidig the zeros of a polyomial is really the same thig as factorig it ito liear factors. We ow study a method for fidig all the ratioal zeros of a polyomial. Example Cosider the polyomial f x = x x 3 x+ 4 Factored form ( ) ( )( )( ) = x x x Expaded form. What are the zeros of f ( x )? What relatioship exists betwee the zeros ad the costat term of the polyomial? The ext theorem geeralizes this observatio. 1 The Ratioal Zeros Theore m If the polyomial f ( x) = ax + a 1x ax + ax 1 + a0 ) has iteger coefficiets, the every ratioal zero of f ( x ) ( a0 0, a 0 is of the form p q where p is a factor of the costat coefficiet a 0 q is a factor of the leadig coefficiet a. Note: The Ratioal Zeros Theorem gives oly POSSIBLE ratioal zeros. It does ot tell us whether these ratioal umbers are actual zeros.

9 Exercise 9 Usig the Ratioal Zeros Theorem Do each of the followig for the polyomial fuctio defied by 4 f x = 6x + 7x 1x 3x+. ( ) a) List all possible ratioal zeros. 9 b) Fid all ratioal zeros ad factor f ( x ) ito liear factors. Fidig the Ratioal Zeros of a Polyomial 1. List all possible ratioal zeros usig the Ratioal Zeros Theorem.. Use sythetic divisio to evaluate the polyomial at each of the cadidates for ratioal zeros that you foud i Step 1. Whe the remaider is 0, ote the quotiet you have obtaied. 3. Repeat Steps 1 ad for the quotiet. Stop whe you reach a quotiet that is a quadratic or factors easily, ad use the quadratic formula or factor to fid the remaiig zeros. Exercise #10 For each polyomial fuctio a) ( ) i) List the maximum umber of real zeros; ii) List the umber of positive real zeros ad egative real zeros; iii) list all possible ratioal zeros; iv) fid all ratioal zeros; v) factor f ( x ) over the real umbers. f x x x x x 4 =

10 10 h x = x + x + x + x+ b) ( ) Exercise #11 a) Fid all the complex zeros of ( ) Write f i factored form. f x x x x x 4 = b) Fid all the solutios of 5 4 x 3x 5x 15x 07x =.

11 Itermediate Value Theorem Let f deote a polyomial fuctio. If a b f b are of opposite sig, there is at least oe real zero of f betwee a ad b. < ad if f ( a) ad ( ) 11 4 Exercise #1 Show that f ( x) = 8x x + 5x 1has a zero i the iterval [ ] 0,1. Exercise # = has a solutio i the iterval [ 1,0] 4 The equatio x 8x x 0 solutio to two decimal places.. Approximate this Exercise #14 Each polyomial fuctio has exactly oe positive zero. Fid a approximatio of the zero. f x x x x a) ( ) b) ( ) = g x = x x x Exercise #15 Graph each polyomial fuctio. f x = x + x + x+ a) ( ) b) ( ) 1 4 g x = x 3x 4 4 h x x x x x c) ( ) = Exercise #16 Fid k such that ( ) f x x kx kx x +. 4 = has the factor Exercise #17 What is the remaider whe ( ) = + + is divided by x + 1? f x 3x x x x Exercise #18 Prove that x+ cis a factor of x + c if 1 is a odd iteger. Exercise #19 Let ( ) 4 f x = x + 1. a) Factor f ito the product of two irreducible quadratics. b) Fid the zeros of f.

12 Aswers to selected problems: 1 10a) 1, -1/, ± 10b) -1/ -1/3, ± i 11a) -3, ±, i 3± i 11b) -3, 4, 1/, ± 3i 13) a) b).53 16) -17/1 17) 1

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