# The Fundamental Theorem of Algebra

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1 The Fundmentl Theorem of Alger Jeremy J. Fries In prtil fulfillment of the requirements for the Mster of Arts in Teching with Speciliztion in the Teching of Middle Level Mthemtics in the Deprtment of Mthemtics. Gordon Woodwrd, Advisor July 009

2 Fundmentl Theorem of Alger The Fundmentl Theorem of Alger Sometimes theorem is of such importnce tht it ecomes known s the fundmentl theorem in rnch of mthemtics. This pper will eplore the mthemticl ides tht one needs to understnd to pprecite the Fundmentl Theorem of Alger. The Fundmentl Theorem of Alger FTA) encompsses one of the ig ides in mthemticl discourse nd hs led to its own rnch of mthemtics, minly the rnch involving roots of polynomils. One of the ig questions we sk ourselves when deling with polynomils is how mny zeros roots) does the function hve. The FTA tells us tht every comple polynomil of degree n hs precisely n comple roots, lthough there re mny vritions to this sttement tht siclly men the sme generl ide. It cn e very helpful in mthemtics to find roots of polynomils nd the FTA is centered on this ide. In order to understnd the FTA fully nd know ectly wht is ment, some ckground informtion needs to e reviewed nd eplined in detil so one my understnd how the FTA ws developed nd is used tody. Polynomils Since the FTA dels with polynomils, it mkes sense tht we should hve n ecellent understnding of polynomil functions nd their properties. Polynomils re used in wide rnge of prolems, from elementry word prolems to clculus, chemistry, economics, physics, nd numericl nlysis. In order to understnd the mke-up of polynomils, one first needs to understnd wht clssifies s monomil. Monomil Definition. A single term tht consists of numer, vrile, or the product of numer nd vrile to positive integer eponent is monomil. Emples: 5, -,,, nd

3 Fundmentl Theorem of Alger Understnding monomils helps one to understnd polynomils ecuse monomil is polynomil with only one term. We define polynomil s follows: Polynomil Definition. A polynomil is the ddition or sutrction of monomils written in the form: p n n n ) n n n... 0 where n is non-negtive integer nd,..., 0,, n re constnt coefficients. Emples: g ) 6 7 nd h ) Polynomils cn lso e given specific nmes to identify the numer of terms such s monomil term), inomil terms), nd trinomil terms). Polynomils with more thn terms re usully just referred to s polynomils. Emples of inomils re f ) 5 nd g ) 5. Notice tht ech individul term stisfies the definition of monomil nd then is joined to nother monomil y either ddition or sutrction. The net ide ssocited with polynomils tht one needs to understnd in order to comprehend the FTA is the degree of polynomil. The degree of polynomil is found y identifying the lrgest eponent. Therefore, if you hd inomil like f ) 5, its degree is ecuse the lrgest eponent is. In clss, we hve witnessed nd used polynomils of different degrees. A polynomil such s f ) is liner nd hs degree of. Another polynomil such s g ) 5is qudrtic function nd hs degree. Usully terms of the polynomil re written in descending order of degree, lthough they cn e written in ny order. Some polynomils hve specil nmes sed on their degrees nd they re:

4 Fundmentl Theorem of Alger Degree Nme Emple 0 Constnt y = 7 Liner y = + Qudrtic y = - - Cuic y = Qurtic y = Quintic y = 5 5 The grph elow shows how some of the emples would grph on coordinte plne. We will now tke closer look t how some of the sic mthemticl opertions ddition, sutrction, multipliction, nd division) work on polynomils. The ddition or sutrction of two polynomils is commonly used mthemticl opertion in lger clsses

5 Fundmentl Theorem of Alger tody. The rule for dding or sutrcting polynomils is tht you dd or sutrct the coefficients of ll of the terms tht shre vrile of the sme power nd the new polynomil s degree will e the sme or smller thn the lrgest degree polynomil term in the sum. An emple is 5) 8) nd the resulting polynomil is nd thus the degree of it is the sme s the lrgest degree polynomil term in the sum. Another emple is 6 5) 5 ) nd the resulting polynomil is 8 nd thus the degree of this polynomil is smller thn the lrgest degree polynomil term. There will never e cse where the new degree will e lrger thn ny of the degrees of the polynomil terms when dding or sutrcting polynomils. The product of two polynomils is nother common opertion in lger clsses nd so we know this holds true, ut wht re the rules? When multiplying two polynomils, the following eplins why degree of the new polynomil is the sum of the degrees of the two polynomils. Product of Polynomils: If two polynomils re multiplied together, the degree of the new polynomil will e the sum of the degrees of the two polynomils written s: p ) d ) f ) Degree of p) = Degree of d)) + Degree of f)) An emple would e if you hd 5) 5 ) nd multiplied y using the distriutive property to get Since the first polynomil hd degree of nd the second polynomil hd degree of, they comine to mke polynomil of degree. This concept holds true for the multipliction of ll polynomils ecuse unlike the opertions of ddition nd sutrction, with multipliction there is no wy to cncel the highest degree terms.

6 Fundmentl Theorem of Alger 5 So is there division of polynomils? Division is the inverse opertion to multipliction. The question is, when you divide two polynomils, do you get nother polynomil? The nswer is yes, in some cses, ut it does not lwys hold true. At this point we need to shift wy from polynomils for quick review of other key concepts efore eplining how to divide polynomils nd why the division of polynomils does not lwys produce nother polynomil. Other Opertions In order uild up to the FTA, it is essentil tht we understnd some of the other mthemticl terms nd opertions. To egin with, we will review the division lgorithm for integers. Most people will rememer the division lgorithm s something lerned in grde school; given n emple of 00 divided y, you would hve seen something like the picture to the right. This lgorithm works for ll numers, ut the question is why ectly does this model work nd wht hppens when the numers divided do not come out evenly? This even works when things do not divide evenly ecuse you end up with reminders or deciml nswers. The division lgorithm s presented in the elementry model ove is derived from the following theorem: Division Algorithm Given two integers nd d, with d 0, there eist unique integers q nd r such tht qd r nd 0 r < d, where d denotes the solute vlue of d. In this cse: is the dividend q is the quotient d is the divisor r is the reminder

7 Fundmentl Theorem of Alger 6 Since r is reminder, it mkes sense tht r must e etween 0 nd the solute vlue of the divisor ecuse if r ws igger thn tht, then the divisor would go into the numer one more time. Therefore, if you tke the dividend nd divide y the divisor, you get the quotient nd possile reminder left over. An emple is 0 divided y 7 which looks like this: 7 0 qd r 0 q7 r 0 )7 Thus, we would get quotient of with reminder of, nd if we wnted to find the deciml pproimtion, we would continue the sme type of steps, ut some mnipultion would need to e done, s the following illustrtes. 0 We egin with our nswer of nd rewrite this s. By rewriting, we cn 7 07 now find wht 0 is y the sme method s efore. Thus, we get 0 )7 nd get with 7 reminder of. Therefore, our new nswer is 0 7 ecuse we replced 0 with 7. Net we cn distriute nd get nd now we need to rewrite so we cn divide, nd so, we get. We cn now use the division lgorithm for 0 nd get 0 7)7 nd this gives us 7 with reminder of. 7 7 After mking this sustitution, we hve nd this of course is.7. We could continue the division lgorithm nd eventully tke this 700 to the numer of digits we prefer or until the reminder is zero. The long division lgorithm is

8 Fundmentl Theorem of Alger 7 shortened version of this. The process is very importnt when deling with polynomils s we shll see soon. The net importnt topic to discuss when trying to fully understnd the FTA is the numer system nd the definition of field. The commonly used numer systems re defined s follows: N = Nturl Numers N = {,,,, 5, 6, } sometimes clled the counting numers) W = Whole Numers W = {0,,,,, 5, } i.e. we include zero) Z = Integers Z = { -, -, -, -, 0,,,,, } i.e. we include negtives) Q = Rtionl Numers Q = {, re in Z nd 0 } i.e. frctions) I = Irrtionl Numers I = hs non-repeting nd non-terminting decimls i.e,, e ) R = Rel Numers The rtionl numers together with the irrtionl numers. Wikipedi notes tht the concept of field ws used implicitly y Niels Henrik Ael nd Evriste Glois in their work on the solvility of polynomil equtions with rtionl coefficients of degree 5 or higher. A field is n lgeric structure with notions of ddition nd multipliction, which stisfy certin ioms. The ioms re things like closure, commuttive, nd ssocitive properties nd re listed in more depth nd defined elow. Also indicted elow re the numer systems tht stisfy the ioms. Here re the common field ioms tht every field X stisfies. Where true Addition Properties X is closed with respect to ddition. N, W, Z, Q, R o If, X, then + X Addition in X is commuttive. N, W, Z, Q, R o If, X, then + = + Addition in X is ssocitive. N, W, Z, Q, R o If,, c X, then + ) + c = + + c) X hs n dditive identity, nmely 0. W, Z, Q, R o If X, then + 0 = 0 + = In X every element hs n dditive inverse. Z, Q, R o If X, then there eists X so tht + = + = 0 Multipliction Properties X is closed with respect to multipliction. N, W, Z, Q, R o If, X, then X

9 Fundmentl Theorem of Alger 8 Multipliction in X is commuttive. N, W, Z, Q, R o If, X, then =. Multipliction in X is ssocitive. N, W, Z, Q, R o If,, c X, then ) c = c) X hs multiplictive identity, nmely. N, W, Z, Q, R o If X, then = = In X every element ecept zero) hs multiplictive inverse. Q, R o If X, 0, then there eists X so tht = = Addition nd multipliction re connected In X, multipliction distriutes over ddition. N, W, Z, Q, R o If,, c X, then + c) = + c If ll properties hold true, then we sy the set of numers long with the given opertions form field. From ove we cn tell tht the rtionl nd rel numers form field. In certin instnces, like when solving equtions, some fields re just not ig enough. To solve the eqution 5, we cn sty in the field of rtionl numers ecuse we get n nswer of =/. But if we hve n eqution like, we get n nswer of =. This is not rtionl nd thus the eqution hs solution in the field of rel numers. Not even the rel field is lrge enough sometimes ecuse the eqution doesn t hve rel solution nd thus we introduce the field of comple or imginry numers in which it does hve solution. So how does this ide of field pply to our investigtion of the FTA? One thing to understnd is tht ddition, sutrction, nd multipliction hold true for the integers, even though the integers re not field. The sme three properties hold true for polynomils s well, s ws mentioned erlier when we defined polynomils. But wht out division nd why re neither integers nor polynomils closed under division? The property of multiplictive inverses doesn t hold true for integers ecuse you cn tke n integer nd divide y nother integer nd get rtionl numer tht is not n integer for emple ). Therefore integers re not closed under division. To understnd why divisiility does not lwys work for polynomils, we will tke look t few emples.

10 Fundmentl Theorem of Alger 9 One wy to divide polynomils is through the use of long division. This method llows us to tke polynomil dividend) nd divide y nother polynomil divisor). Just like the division lgorithm presented erlier of qd r ws used s the model to divide numers, we cn modify this slightly to divide polynomils. Division Algorithm for Polynomils Given two polynomils p) nd d), where d) is non-zero, then there eists polynomils q) nd r) such tht: p ) q ) d ) r ) p) is the dividend q) is the quotient d) is the divisor r) is the reminder where 0 degree of r) degree of d) For emple if we wnted to let p ) 5nd divide y d ) This mens nd we cn now do the sme thing to try nd cncel the net term.

11 Fundmentl Theorem of Alger 0 This mens nd we cn now do the sme thing to try nd cncel the net term. This mens , where the reminder is 7 7. There re certin instnces where dividing polynomils cn lso give us reminder of zero, s in the cse of p ) d ) where p ) nd d ). This division gives us n nswer of nd we hve no reminder. Wht this relly tells us is tht, so tht is fctor of. From these emples, we see tht it would e helpful to know wht polynomils divide polynomil evenly. The method of how to find these fctors my e more pprent in the lst emple ecuse it is written in form we hve tlked out in clss. If we set the eqution equl to zero nd solve for the solutions tht mke this true, we re doing something clled finding the zeros. In this cse we get 0 nd the roots or solutions or zeros re = nd =. Therefore,

12 Fundmentl Theorem of Alger finding the roots of polynomils will essentilly help us to identify fctors of them tht we would like to use s our divisor. This rings us to the concept of polynomil eing irreducile. Irreducile cn e est descried y the following definition. Definition of Irreducile Polynomil: A polynomil is irreducile over field K if it cnnot e written s the product of two polynomils of lesser degree whose coefficients come from K. This siclly mens tht polynomil is irreducile if it hs een fctored completely. For instnce, cn e fctored into ) nd thus this polynomil is not irreducile. The hrdest prt out fctoring polynomils nd determining if they re irreducile is tht this is not often ovious. The polynomil 8 looks irreducile nd is, in fct, irreducile over the rtionl numers), however, over the rel numers it cn e fctored s 8 8. The polynomil gin is irreducile over the rtionl numers nd so, looks irreducile) ut if we llow this to e viewed over the comple field, then it cn e fctored into. Some emples of irreducile polynomils over the rel field include nd. We will discuss irreducile polynomils in more depth fter discussing how roots will help us. So wht if we hve f ) 5 6, nd we wnt to find something tht divides it evenly so we do not get reminder. Is there even polynomil tht eists tht is fctor of f )? We do know tht since the degree of product of polynomils is the sum of the degrees of the fctors, nd since the degree of f) is three, ny fctoriztion into polynomils of lower degree must hve one fctor e liner. This siclly mens if you were to fctor polynomil with leding coefficient of, it must fctor into nd ecuse + =. This

13 Fundmentl Theorem of Alger essentilly tells us tht if we were to going to try nd fctor f ) 5 6, we need to find just one liner fctor of it first. To find which liner polynomil might work, we will utilize something clled the rtionl root theorem. The rtionl root theorem is quick wy to see possile vlues of the constnt to use in long division tht my divide the two polynomils evenly. Rtionl Root Theorem Given the eqution for polynomil in the form p n n n ) n n n... 0 n, where 0,,,..., re constnt integer coefficients, we cn find possile roots written s frction p in lowest terms q given tht: p is integer fctors of the constnt term 0 q is integer fctors of the leding coefficient n The root theorem is nother theorem tht helps with the division of polynomils. If we let f) e polynomil of degree or greter nd let e numer. The Division Algorithm tells us tht f ) q ) ) r ), where degree of r) is 0; hence r) = c, constnt. If is root of f), then 0 f ) q ) ) c. It follows tht c = 0 nd thus tht f ) q ) ) nd f) hs ) s fctor. Conversely, if c = 0, then must e root of f). This proves: Root Theorem If f) is polynomil with coefficients in field, nd if is in the field with f ) 0, then divides f). Given the sics of the rtionl root theorem, we will now look t n emple of how the theorem works using the emple of 5 6. If we identify vlues for p nd q we come up with

14 Fundmentl Theorem of Alger p,6 nd q. Therefore our possile vlues for re,6 We end up with the. following possile solutions:,,6, 6. We cn now pick one of these s our constnt, for emple, 6, nd use long division to see if we hve found something tht divides 5 6 with reminder zero. From the long division performed to the right, one cn tell tht 6does evenly nd the fctoriztion is 6 divide 5 6. The root theorem works ecuse it essentilly gives you the possile fctors tht could multiply out to give the constnt of f) t the end. For instnce, we knew tht we could solve 5 6 y fctoring nd doing so we would sk ourselves the fctors of 6 tht dd up to 5. The fctors tht work re nd 6, nd these were our possile vlues of in the root theorem we used. The Root Theorem cn e generlized to more comple theorems. The Horner Scheme is one tht is helpful in eliminting roots t different stges, ut tht is eyond the scope of this pper. Solving Polynomils Now tht we ve discussed how to simplify polynomils, we need to etend this to the solution of equtions involving polynomils. Most of us re fmilir with some of the wys we cn go out solving polynomil equtions. When setting the polynomil eqution equl to zero nd solving, we re doing something clled finding the roots of the polynomil. Grphiclly, this is where the grph of the polynomil crosses the -is. If we re solving for liner function such s f ), we cn solve this few different wys: Solve using sutrction nd division f ) 0 Solve y fctoring f ) 0 0 ) Here we find tht when = -, we hve the solution!

15 Fundmentl Theorem of Alger If we re solving for qudrtic function such s 5 6 ) f, we gin cn solve this in few different wys. We could solve y fctoring, y grphing, y completing the squre, or y using the qudrtic formul. In this cse, I will use completing the squre to solve this function. Solve y completing the squre ) f Set function equl to zero Crete gp to complete the squre Tke hlf of middle term, squre, dd to oth sides Fctor left hnd side Rewrite left hnd side Tke the squre root of oth sides Set up two equtions nd solve We otin the solutions of = - nd = -5, mening tht 5 6 fctors into 5. It ws previously mentioned tht nother wy to solve qudrtic equtions ws to use the qudrtic formul. So why does the qudrtic formul work nd from where does it come? Its proof is given elow.

16 Fundmentl Theorem of Alger 5 0 c c c c c c c c c We strt with generl form of the eqution nd solve y completing the squre in the steps elow. Begin y sutrcting c from oth sides nd dividing y. Complete the squre y tking hlf of the middle term, squring it, nd dding it to oth sides. Fctor the left hnd side Squre the fr right term on the right side nd rewrite the left hnd side Squre the denomintor on the fr right hnd side Multiply the first term on the right side y c c c c Rerrnge terms since the right side hs common denomintors Tke the squre root of oth sides Simplify the ottom term on the right hnd side Sutrct from oth sides nd end with the qudrtic formul! The qudrtic formul proves very helpful in solving qudrtic eqution. It tells you ectly if qudrtic eqution hs 0,, or rel solutions, nd this is indicted y oserving the sign of the rdicnd, which is the numer under the squre root; in this contet, the rdicnd is clled the discriminnt. If the discriminnt is zero, there will e only one rel solution. If the discriminnt is negtive, there re no rel solutions. Finlly, if the discriminnt is positive, then there will e two rel solutions to the qudrtic eqution. To understnd why the qudrtic equtions cn hve

17 Fundmentl Theorem of Alger 6 0,, or rel solutions, we will tke look t few emples, nmely ) f, ) g, nd 0 ) g. ) f Solve y fctoring ) ) 0 0 ) f Here we find tht when = nd =, we hve our solutions! The grph to the right verifies the two solutions. ) g Solve y fctoring ) ) 0 0 ) f Here we find tht the solution is = From the grph to the right we cn see there is only one root nd it occurs t =. ) f

18 Fundmentl Theorem of Alger 7 Solve using qudrtic formul f ) 0 c ) ) )0) 0 6 Here we find tht we cn t get solution over the rel numers ecuse we hve squre root of negtive numer. From the grph to the right we cn see tht we would epect no nswer ecuse the grph of the polynomil doesn t cross the -is. So how do we solve polynomil with degree lrger thn? One of the wys is to use the rtionl root theorem s previously mentioned to find ny rtionl root nd thus otin the fctors of the polynomil. Alterntively, we could use grphing clcultor to estimte the roots. For emple if you re given the following eqution nd grph to the right, the rtionl root theorem gives us only,,, 6 nd none of these work. So we cn pproimte the roots y using the tle function on grphing clcultor. You egin y plcing the eqution in the y = window nd then looking t the

19 Fundmentl Theorem of Alger 8 grph you cn get n estimte of the vlue where the line crosses the -is. Once you hve n ide of the vlue, you cn use the tle feture to pinpoint the ect vlue more closely. The tle llows you to chnge how closely you would like to nlyze the function. The top tle to the right is n emple of how you could zoom in on the vlues of the eqution grphed ove. We could continue to mke the vlues relly smll until we got y to e essentilly zero nd this would give us the three roots of this eqution. The tle to the right shows how we cn zoom in nd get relly close the vlues of tht mke y zero. Wht this now tells us is tht our solutions re pproimtely = -.55, -.575, nd.085. This mens tht if we fctored 6 we would get pproimtely fctoriztion of the polynomil. The previous function ws cse in which polynomil of degree three gve us ectly three rel roots. If we look t nother polynomil of degree three, we cn see from the grph we hve only one rel root nd it occurs t =. The rtionl root theorem would lso give us = s possile root of the polynomil. If we fctor out, we get:, which is pproimtely the complete From this it is not intuitively ovious, ut cn fctor into the comple fctors of, more commonly written s i i. Since this function hs two comple roots, there is only one rel root. This emple rings us to n interesting point tht should e mde out comple roots. In order to chieve negtive squre root, the fctors of polynomil with rel coefficients must pper in conjugte pirs. Therefore, comple roots of polynomils with rel coefficieints will lwys pper in conjugte pirs. This mens tht if one

20 Fundmentl Theorem of Alger 9 comple fctor of polynomil is 5, or 5i, then there must e nother fctor of the polynomil tht is 5, or equivlently, 5i. In similr fshion, the degree doesn t necessrily tell ectly how mny different roots eist. Tke for instnce the eqution nd grph t the right. If we were to fctor this, we could divide y nd get: This shows tht ll three roots re rel solutions nd they re ectly the sme. This concept of hving roots repet is clled lgeric multiplicity. It cn occur with comple roots s well, ut the degree of the polynomil would hve to e t lest four since the comple roots will come in conjugte pirs. Thus you would hve to hve two pirs of ech comple fctor. So how do we know which functions hve roots nd how mny they hve? We now hve reched point where we cn now introduce the Fundmentl Theorem of Alger. The development of the FTA in erly yers only llowed rel vlues nd so the FTA ws not very relevnt. It ws Girolmo Crdno ) in 55 tht relized he could get nswers tht were not rel numers nd he egn the ide of comple solutions. In 57, Rfel Bomelli 56 57) set out to crete rule for defining these comple numers. Alert Girrd, Flemish mthemticin, ws the first to clim tht there re lwys n solutions to polynomil of degree n in 69 in his ook L'invention en lger. In 67, Descrtes conjectured tht for every polynomil of degree n, there were n roots. He ws unle to etend his discovery much pst this ecuse he noted the imgined roots did not hve rel quntity. There were mny other mthemticins who tckled the FTA, ut it ws Guss who is credited with the first proof

21 Fundmentl Theorem of Alger 0 in his doctorl thesis of 799. He spotted errors in previous proofs ecuse mny erlier ttempts ssumed too mny things. Guss lter pulished other proofs nd mde his fourth nd finl proof in 89, fifty yers fter his first proof. This time he mde his proof complete nd ccurte. The FTA hs few different versions, ut it is most commonly presented s: A polynomil eqution of degree n with comple coefficients cn e written s product of n liner first degree) fctors. The fundmentl theorem does not tell how to fctor polynomil nor does it tell how to solve for the roots, which cn e esily solved for from the st degree fctors. To fctor polynomil we need to use some of the previously mentioned methods, such s nlyzing the grphs nd trying to pull out liner fctor or using the root theorem. For emple, if we wnt to fctor, then it would fctor into nd since we hve three liner fctors, we cnnot fctor nymore. Another emple shows how cn fctor into. Here is liner nd is irreducile over the rel field. This is why the fundmentl theorem includes the comple roots prt. Thus over the comple field, cn ctully e reduced to two liner comple fctors. Through the use of the qudrtic formul we cn find these to e: ) )) c which mens =, nd we hve fctors of which simplifies to i i.

22 Fundmentl Theorem of Alger From this these emples, the FTA mkes more sense nd should help one understnd why polynomil eqution of degree n with comple coefficients hs n comple roots. No mthemticin hs een le to come with n ect formul for how to fctor polynomils. Much of the fctoring done in high school lger is to know the ptterns tht eist nd then use the guess nd check method. We hve techniques to help fctor liner functions e.g. + ) nd qudrtic function e.g. + + ). There re even formuls found to find the roots of qudrtic, cuic, nd qurtic functions, which in turn would llow us to fctor these. The cuic nd qurtic formuls re much more complicted thn the qudrtic formul nd would tke too much time to eplin in the content of this pper. However, there does not eist n lgorithm for fctoring polynomil of degree five or higher. In fct, the mthemticin Glois proved tht there will never e formul to solve the generl polynomil of degree five or higher. His work is known s Glois Theory.

23 Fundmentl Theorem of Alger Biliogrphy Cut-the-Knot 009). Fundmentl theorem of lger sttement nd significnce. Retrieved on June 0, 009 from From Numers to Numer Systems 009). Electronic document courtesy of Jim Lewis. UNL Mthemtics Deprtment. Fundmentl Theorem of Alger 009). History of the fundmentl theorem of lger. Retrieved July 7, 009 from The Mth Pge 009). Topics in preclculus. Retrieved on July, 009 from Wikipedi 009). Online dictionry. Retrieved on June 0, 009 from

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