Polynomial Functions. Polynomial functions in one variable can be written in expanded form as ( )


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1 Polynomil Functions Polynomil functions in one vrible cn be written in expnded form s n n 1 n 2 2 f x = x + x + x + + x + x+ n n 1 n Exmples of polynomils in expnded form re nd = f x x x Polynomils cn be written in fctored form s r1 r2 f x = 1.7x + 3.1x 5x + 3x 2 f x = x c x c x c, r + r + r = n n 1 2 k 1 2 k where the c's re the rel or complex roots of the polynomil. Exmples of polynomils in fctored form re = nd f ( x) = x( x + 3) ( x 4)( x+ 5) f x x x x 1. Polynomils only hve vribles with powers. Tht mens no rdicls or division involving the vrible re possible. 2. The domin of polynomil is ll. 3. In expnded form, the degree of the polynomil is the exponent. 4. In fctored form, the degree of the polynomil is the of the exponents. If ny of the fctors re nonliner, then use rules of exponents to determine the proper degree. 5. The leding term is the term in expnded form with the exponent. In fctored form, the leding terms of ech fctor need to be together to find the leding term. 6. The leding coefficient is the of the leding term. 7. In expnded form, the constnt is esy to spot. In fctored form, the constnt is found by the constnts in ech fctor. Be sure to wtch out for powers other thn 1. r k
2 Prctice: Write the leding term, degree, leding coefficient, nd constnt for ech polynomil. Polynomil 6 3 f x = 5x 4x + 2x f x = 3x + 5x = 3( 2) ( + 5)( 2 3) f x x x x 2 = 4 ( + 1)( 3)( + 2) 3 f x x x x x Leding Term Degree Leding Coeff. 8. According to the Fundmentl Theorem of Algebr, there is t lest rel or complex zero of polynomil. 9. According to the Corollry to the Fundmentl Theorem of Algebr, the number of rel or complex zeros is equl to the of the polynomil, lthough the zeros my not be unique. 10. The mximum number of rel roots is equl to the of the polynomil. The number of rel roots my decrese by. Constnt 11. The mximum number of extrem / turns (reltive mximums nd reltive minimums) is equl to thn the degree of the polynomil. The number of turns my decrese by. x + y + y 12. The right side of the grph (s ) will go if the leding coefficient is nd ( leding coefficient is. ) if the 13. The left side of the grph (s x ) will do the s the right side if the degree is nd the of the right side if the degree is. 14. The y intercept of polynomil is the.
3 15. Polynomils re, they cn be drwn without lifting your pencil. 16. Polynomils re, they hve no shrp turns. Grphs for #19 = ( 3)( + 2) 2 f x x x x= 17. If is root of polynomil, then is fctor of the polynomil. 18. Roots re lso clled nd. Rel roots (but not complex roots) re lso clled. x f ( x) 19. Assume tht is fctor of nd tht x= is rel number. = ( 3) ( + 3) f x x x x 3 2. The grph will cross the xxis nd chnge sides t x= if the exponent on the fctor is. b. The grph will touch the xxis but sty on the sme side t x= if the exponent on the fctor is. c. The lrger the exponent on fctor, the the grph will be ner tht root. d. The only plce tht polynomil (ny continuous function) cn chnge signs is t n. Prctice: Write polynomil function, in fctored form, tht is negtive on the fr right side, crosses the xxis t x=3, nd touches the xxis t x=1. Prctice: Write polynomil function, in fctored form, tht is positive on the fr right side, touches the xxis t x=4 nd x=4, nd crosses the xxis t x=0.
4 20. Sign Chrts. Look t the function in form nd find ll the zeros. Put these on number line in the proper order. b. When mking sign chrt, lwys strt on the nd work your wy to the. This is becuse the sign on the side cn be determined by looking t the leding coefficient. c. Look t the on the fctor corresponding to ech root nd decide whether the sign on the left will be the sme s the sign on the right or the opposite of the sign on the right. 2 3 Prctice: Mke sign chrt for f ( x) = 2( x 3)( x+ 4) ( x+ 1) 21. Descrtes Rule of Signs (requires rel coefficients). The mximum number of positive rel roots is equl to the number of sign chnges in. b. The mximum number of negtive rel roots is equl to the number of sign chnges in. c. Either of these vlues my independently decrese by, s long s they remin nonnegtive. f ( x) d. When finding the signs for, only the signs on the powered terms will chnge, the signs on the powered terms will remin the sme.
5 22. Fctoriztion (requires rel coefficients). Every polynomil cn be fctored using fctors only, but the fctors my involve complex numbers. + bi b. Complex solutions lwys come in. If is solution, then is lso solution. c. If the coefficients re rtionl, then irrtionl solutions involving squre roots lwys come in. If + b is solution, then is lso solution. d. Every polynomil cn be fctored using nd fctors. 23. Rtionl Root Theorem:. If polynomil hs integer coefficients, then ny rtionl zero will be of the form of fctor of the over fctor of the. b. The Rtionl Root Theorem is n existence theorem, it does not tht there will be ny rtionl roots, it only sys tht if there re ny, then they will be of the form indicted. f x = 3x 5x + 2x 4 Prctice: List ll possible rtionl fctors of 4 3 Prctice: There is specil cse when there is no constnt. In tht cse, x is fctor of ll the terms nd x=0 is one rel root. Fctor out the x first nd then use the constnt tht remins to list the remining possible rtionl roots of f x = 2x 5x + 4x + 3x
6 24. Polynomil Division.. Polynomil division is used to divide polynomils. It is performed similr to. b. Much of the writing in polynomil division is redundnt nd cn be eliminted. The technique used to do this is clled. 25. Synthetic Division. Synthetic division needs fctor with leding coefficient of. b. Write down the coefficients in the dividend nd leve blnk line below them. If there re ny missing powers, you will need to write s plceholder. c. To the left of the vlues, write the root tht corresponds to the fctor. If is the fctor, then write to the left. x d. The process begins by bringing the coefficient down. e. The number in the bottom row is by the vlue to the left nd this product is written in the next column. f. The vlues in the next column re nd the sum is written t the bottom of tht column. These lst two steps re repeted until the division is done. g. When dividing syntheticlly by, the lst number in the bottom row is the nd the numbers before tht re the. The exponents on the quotient re less thn the exponents on the dividend. x Prctice. Use synthetic division. ( 3x 4 5x 3 + 4x 2 2x+ 2) ( x 2)
7 26. Reminder Theorem: x. If polynomil is divided by, then the is f. b. Synthetic division cn be used to polynomil. To find, divide syntheticlly by. 27. Fctor Theorem: f x. A vlue is root of polynomil if nd only if the reminder is. x b. If the reminder is zero, then is one fctor of the polynomil. The other fctor is the from the synthetic division. 28. Upper nd Lower Bound Theorems: (requires rel coefficients nd positive leding coefficient  fctor 1 out if leding coefficient is negtive). Upper bound: If synthetic division is performed with vlue nd ll the coefficients in the quotient re, then there re no rel roots greter thn tht vlue, tht vlue is n upper bound. b. Lower bound: If synthetic division is performed with vlue nd the coefficients in the quotient in sign, then there re no rel roots less thn tht vlue, tht vlue is lower bound. c. cn be considered s either positive or negtive s needed to mke the theorem work. d. Another wy to stte the upper bound theorem (so tht you don t hve to worry bout the leding coefficient being negtive) is tht the vlue is n upper bound if ll the vlues in the quotient re the.
8 e. The first thing to do when checking for upper or lower bounds is to look t the vlue being evluted. The vlue cn only be n upper bound if it is nd lower bound if it is. 29. Determine which of the following re upper bounds, lower bounds, or neither.. x = 2 is bound becuse the vlue is nd the signs. There re no roots thn. If x = 2 f x 2x 3x x = + ( 2) b. x = 5 is n bound becuse the vlue is nd the signs re ll the. There re no roots thn. If x = 5 f x 2x 3x 18x 62 f =, then. 3 2 = ( 5) f =, then. c. x = 2 is bound becuse the vlue is nd the signs. In this cse, we need the to be. There re no roots thn f x 2x 3x 14x 5 f 2 = 3 2 = d. x = 3 is not n upper bound or lower bound. It is vlue, but the signs re not ll the. There my be roots thn f x = x 3x 4x x = f ( 3) =. 3. If, then x = 2. If, then
9 e. x = 3 is not n upper bound or lower bound becuse it is but the signs ren t ll the. Alternting signs only pply when the vlue is. x = 3 is root becuse the is zero. There my be roots thn. f. x = 5 is not n upper bound or lower bound becuse the vlue is but the signs don t. Hving the numbers ll x = 3 the sme sign pplies only when the vlue is. There my be roots thn. x = 5 f 5 = 3 2 = f x x x x If, then. 30. Intermedite Vlue Theorem:. A polynomil (ny continuous) function will ssume vlue between f() nd f(b) on the intervl [,b]. b. Of prticulr use to us is when f() nd f(b) re different signs. In this cse, there will be n of the function somewhere between x= nd x=b Drw function between the points without lifting your pencil. Wht hs to hppen on the intervl?
10 31. Use the informtion given to nswer the questions bout the polynomil function f x = 6x 25x + 38x 46x + 44x + 8x f x = 6x + 25x + 38x + 46x + 44x 8x 16 f x x x x x 2 2 = ( 2 + 1)( 3 2)( 2) ( + 2). How mny rel or complex zeros re there? b. Wht is the mximum number of extrem (turns)? c. As x +, wht does y pproch? d. As x, wht does y pproch? e. Any rtionl zeros will be fctor of over fctor of. f. Wht is the mximum number of positive rel roots? g. Wht is the mximum number of negtive rel roots? h. List ll rel nd complex zeros of the function. i. Where will the grph of the function cross the xxis? j. Where will the grph of the function touch the xxis? k. Wht is the yintercept of the grph of the function? l. Wht is the domin of the function? m. Mke sign chrt for the function. n. Sketch the grph of the function.
11 32. Use the informtion given to nswer the questions bout the polynomil function. = + + f x = x 19x + 99x 81x f x x x x 2 f x 7 x 5 19x 3 99x 81x = ( 1)( 9). How mny rel or complex zeros re there? b. Wht is the mximum number of extrem (turns)? c. As x +, wht does y pproch? d. As x, wht does y pproch? e. Any rtionl zeros will be fctor of over fctor of. f. Wht is the mximum number of positive rel roots? g. Wht is the mximum number of negtive rel roots? h. List ll rel nd complex zeros of the function. i. Where will the grph of the function cross the xxis? j. Where will the grph of the function touch the xxis? k. Wht is the yintercept of the grph of the function? l. Wht is the domin of the function? m. Mke sign chrt for the function. n. Sketch the grph of the function.
12 33. Putting it ll together to find the roots of polynomil. Consider the polynomil eqution x + 7x 6x 12x+ 8= 0. According to the Fundmentl Theorem of Algebr, there re rel or complex roots. b. According to Descrtes' Rule of Signs, the mximum number of positive rel roots is nd the mximum number of negtive rel roots is. c. According to the Rtionl Root theorem, ny rtionl roots will be of the form of fctor of over fctor of. When you list ll of the possible rtionl roots in order, you get:, , 4, 2,, 1,,,,,1,, 2,, 4, Perform synthetic division using vlue from the list of possible rtionl roots. Begin with something simple nd sort of in the middle. works well x =1 x =1 d. Since we got reminder of, tht mens tht is nd x 1 is. We cnnot use the bound theorem becuse the signs re not ll the sme. ( x )( x 3 x 2 x ) = 0 We cn fctor the eqution s. Now we try nother fctor, but we use the reduced polynomil 3 2 insted of the originl one. Let's try. 3x + 10x + 4x 8 x = x = 2 8 not to try ny of the vlues 3 e. is not solution, but it is n bound, so we know. This is why we usully don't
13 try extreme vlues like 8. They're unlikely to work nd we re not ble to eliminte ny vlues if they don't. Now we try something else. There's still possibility of 1 positive from Descrtes' Rule of Signs nd we've rn out of nice vlues, so let's try negtive like x = 1 f. It didn't work nd it's not bound, so let's move on nd try something else, sy x = 2 x = 2 g. Woohoo! It worked. Tht mens tht is nd is. x = 0 We cn now fctor the eqution s ( x )( x )( x 2 x ) h. The other fctor is qudrtic fctor. When you get down to qudrtic fctor, you cn try,, the, or using the to find the remining roots. 2 3x 4x 4 + ( 3x 2)( x+ 2) In this cse, fctors s. ( x )( x ) 2 ( x ) = 0 x = x = x = i. The entire eqution fctors s, so the roots re,, nd. j. If you re performing synthetic division nd you get zero s reminder, it is possible tht the vlue you used my work gin. In this cse, x = 2, ws solution twice. You my wnt to keep trying root until you know it doesn't work or you cn use the Rule of Signs to eliminte some nswers.
14 34. Write polynomil function tht hs the given sign chrt.. b. c. 35. Write function tht hs the grph shown.. b. c.
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