Manual Algebra Basica B Factorizacion

8/13/2019 Manual Algebra Basica B Factorizacion

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CHAPTER IIIFACTORING

18. Definition of , factoring. Factohng is the process offinding the twb or rnore algebraic expressions whose -product is equI to a given expression.fn *lultiplication we have two factors given and are required tofind .their product. fn division we have'the product and one factorgiven and. are required to find the oher factor. In factori-ng, hoy.ever, the problem is a, Iittle more d.ifficult, for we have only tleproduct,given, and our experience in multiplication and division iscallef. upon t'o enable us to determine the factors.19. Rational expressions.' A rational algebraic expressionis one "rmhich- can be written without the use.of indicatedroots of. the letters involved.Thus 2, 5 r, 3 y -\f2, and. of ur" rational expressions. In thischapter factors which involve rad.icals wiII not be sought.

'a, rational expression canan indicated division inin a denominator, it is' ,'Thus 3, .7 a,3t and,4 r - 3 arnintegral expressions.brfactors whieh i*.qrolve fratioas wilt not be sought.

.tIn this chapprime when

,l:r o

expresslonsffi,;,$l. Prime factors. An integral expression isis'the product of no two rational integralbxaqt, itsolf,, ax.d 1.PE -i


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' ft must be remembered that to facor an integrl expres-sio means to resolve it into its prime factors.The rnethods of this chapter enable one to factor integral rational: xprssions in one letter which are not priffie, as well as some ofthe simpler expressions in two letters. No attempt is rnade even todefine what is r-neant by prime factors of expressions which are notrational and integral.There is no simple operation the performance of whichmakes us sure that we have found the pfme factcrrs of agiven expression. Only insight and experience enable usto find prime factors with certainty.A partial check that may be applied to all the exer-cises in factoring consists in actually multiplying togetherthe factors that have been found. ff the result is theoriginal expression, co-rrect factors have been found, thoughthey may not be prime factors.22. Polynomials with a common monomial f,actor. ThetYPe form is ab + ec - ad.Factoring,' b+ec -,d: (b* c - d).

ORAL EXERCISES

f

,*

Factor:1. 3a*6.2. 5n {I5.3.'a,2 * a.4.2c-6.c2.5. 9n2-3r.6, cd * o'd'.7. afrz - azn.8. 4an - 8 c'.9. 14 h - 2I lbzk.

10.11.L2.13.{ L4.15.16.L718.

5an-2e,tr7.2 o * 4 - 2 cd.4a-t0a2-2e,8.6 ac - 3 bc { 3 c.io*+1bnB-5*y.14aa - T as + T q,, - T . ,3cz 6 c4+9cs-L5el,

;'l

.;'::*"


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"may'socomlnon factsred.,5y gr' _ ."tbinomial factor.. (rt+sg+bx+w"

Facoringo, ' (r,n'+ eU * bn + by: (an + Y> * (br + by)= a( + y)'+ b'(* + Y,' ;. n:rnricrsbst ': I 'Separate into PolYnomial factors :l. 2("* b)#r'(a * b). 5. a(c - d) -b(t - d)'2.3( + 5) { a( + 5). 6.'2(at - Y)- n(n - y)'3. 5 n(u - t)+ y(o - t). 7. h(* * 3 ") - 2k(nu + 3n\,1. Z a(r,-Zy) +b(n -2y). 8. 2r,(5n-&y)-9s(dr -*i

*a

*, 11'. Za(c-3d)tr(3d-t)' '1 12. 5r(3rn-Znf-Zr(zn-3*)' :

.n: i,-,

IS- cd,-g.qf+2d-6f:l1?. a,h,r + abr - ahs - u,lcs,',,18' fs * 2rs - 3ft * 6'rt.r '19.t* ,hrt' + 8 hn * 6 k'm - LZ kn'20 jtV 2 e - a,c { cn.gl . n8,-3nyz 3n'y*93/t.N, a,n + y * bn + by * cn + cY.i*2r*rbs-,2s* ffit-2t-;,, . ,. .4


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SCOND COIA;S,E

Separat ino binomial ,factors :1. u,2+Zar*r"..2. rz L,Z *y * yr,,Sr m2 Zmn*n'. '7. TP - 10 rs * 25 s2.,8.9+6ala2. - ,' 'q9. 16 - 9an * a'n,.

i,25. A binsmialr.Iorru rs; the difference of. d'- b2.Facoring, a,2 bs - (+ ) {a;'b).More generally,a,2+2ab+b2-&+Zcd-dz:, &2 + 2 eb + bz - (& - 2 c,i+,d$

1J.1{Ir1.{r


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',Fa,eor.

: .. : . . .r..r o- .... .) ;, ,FCTOR,INORAL EXERCISES

10 25-36b2e.11. 36 azbz - 49 czdt.12. 6 ; 25 b4.13. nnyn -- $.LlL. *'yr - 64 za .15. 8L a2-- 100 nru.,'16. &4 - L6.IlrNr. Fihd thsee factors.L7-. ,a.,a - b4. , '

s, EXERCISES

*,'

18' d4_19. lr90. #,;*1.' ' 1,22' 628. 824. fiz25. oLz26. o,

.1,.1.'81.a .

- fr.4.,.*y"#- bzn' s4b

-..

-g*-b-d'

b:a

+

c&r&"i _.

:

1

n..'in ab.'

*.1*1(-C

b-h*m

6-,25-1-,Zffi .-2 lt'l,,-

1$. W + ,? rn?, + ,z :* (i' * 2 *y + .16. fr2 * 2 *,y *,y?* s,2 - 2 ab - lf ,i$';',.'l?, n2 + G n *g - yz + Z y* - 2.18.r.gs';+','frT a + I?L -'bz + 20 bc - 100c119. 2 - 8r2+ S n4 + 2y'- 8 yz - 822.


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"r,ffi:.SECOND COURSE I1{

. .Q1.,,..:: t'{l+:'

'l/

ALGEBRA {qp. 49#- 49yn I4yT1.. 28. +rs -(f -t').'zta. LzLfr|-1 - 78y-BLy'. zg.'nLz'- n2 - m, - n. l.25. a,2 -bz -(g'- b). . 30. nL + n - rLz + n?.:' 27. f -?.s(rn-t). 3.p, f -T* 3s-9s2.

. t"*I \ 26. The quadratic trinomial. The type form is i- ? - l*1 \,r') l xz+bx*c. { ""dli..ii ' .+ -T- vJl -T- 1,. .Since (* * h) (" + k)-: n2 + (h + k)r * Irk,I

it fotlows that # + bn * c Ean be factored into the bino-mid,l factors (* + h) (ry + k) if two numbers h and k canbe found which have the sum and. the' product c. Themethod. of determining these factors is illustrated, in theEXAMPLEFactor nz - 2 n -I5.

"Solution. Here -15:1. -- 15 or -1.15 or *5. -3 or +3.5.f'these pairs of factors of - 15 only the pair - 5 and + 3 give thesufo - 2.' 'ilIencea

'a .1 '':

'For fe,ctonng expressions of the ty + b* + c we havetheEule. Eind, two numbers whose algebraic ltroduct is * c and,' ' '

whsse algebraic surn is + b.Write fo, the factors two binomials both of which haue x"fo, tle,ir rtrst terms and these numbers fw the se'cond, terrns.EXERCISESfactors':a2+8q,*12.d'+7 d + 18.2+I},a+25.

-r+- (

,a..:i

a.i'I s'

Separate into binomial1. n2+5n*6. 4.2. 8+ T n *72. E.' $. n'-ff"T'fr + 6. 6.

, .'.'':, "., .:,' ' .


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. . wq:*:* .'.': :FACTORINGX ,="

Itr'. j,' -i, 10. f-r- 9q.\ 11. rnz-3m-18.

2. n2-2*:24.' 13.9-10r+n'.14.. r2 * frs * 3s'.15. m,2-Tn'LnlI0n2.

li'.lG.1 5n*u*'. )L7. I+2n-24n2--

18. 2+9-10.19. (" + b)' -. 2 (" + b)- b. t20, (n - y)n + a@ - y) +.3.2 . a,2n + \Z"qw + gf. i,u22. cto - 7 co - 18. '23. mz't - ILyn - 12.24. a,4* - 5,2* - 6.25. bav - Zbzs - 35. .{ \a \. _:,\i' -t-"

For ma,ny trinornials of this type two binomial factorsof the form (h* + k) (** + n) may be found. The method,I of factoring such trinomials is illustrated- in the\--- E*aMPLEFactor 2#+Tnr 5. ?r*7' Solution. 2r2*7r-15-('?r*?)(?r*?). ]r+?

, To frnd the proper factors we must suppLy 2 I,' * 1" { ,such numbers $r the interrogation points in = ,, 1 1" ;1?s G)t*l

+

?(1) and (2) as will give 2q2 + 7 r:I5 (3)- r.., . r2 uz for the product of the first two terms of the binomials,; ;l l:, lli ll"*:l ;JJ::"":'lJlffJ-s of the binsm ials'l[ow 2n2-2fr'frt (4)-15' : = 1. + 15; 1. -15; + 3'- 5; - 3'+ 5. (5The'factors,.LaJ-^'lLv7'factors sf. 2 and - li5 from (4) and (5) may. be substitutedrdor the interrogation pointg in (1) and (2) to forin the following*; il

' ,-:' ,]-i.s .. ,, .,.,,pa,irs of binomid,ls, each having a prod.uct containing the first and2*,; 2x*15 2r*L 2r-L5 2n*3 2r-5 2r-B 2x*6-n+is -;-i t*-Ll r*1 r'-5 r*3 " n-'3./r^'w'.

'a.\

.:'t:':

_ : :


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SECOND COURS.Eproducts, which gives the midrile erm. of the trinomial.. Therefore 2 rz * T u- lb _ (Z n -g)(, + b).

Eacbvl,:1_: 2n2+ 5 n * Z:, 4. 4 +Ta, +9.'sq 7. Anz+ g n {3;,z:Znz+Tn+6., 5. Brz+18nlIZ. g.6f +Tr+2.3:..2q,2+9ccl;l-O. 6. Br2+rTnlL0. g. zbzlE. +2.10. Tnz - 8 n + 5.11. 6c2+Tc\+2." 12.3tr2-11 n+6.. 14 . .4n2'- 1g n I L0.'15. t0 ffiz - 2g n + 10.*., 16., 12 nz -,- 11 ny + Z gp.t 17.2a2+3u-2..f'18;'3f +r-2. \ ,

' .a:1AfteralittIepracticeitwilIusual}yilrooounneceSsarytowrite down i,lt of the pairs of binomiut, thut.do ""; ;;;J;r- therequired prod.uct. * sIf none of the pairs gives the required, product, thegiven trinomial s pri,me. .If an expression of the form a,xz * bx*.c .is' not prime,'it can be factored. b)' *pplying ther.t RuIe. Iind two binomials, such thrt ', ' *I. TIte product of the f,rst terms is axz ;If. Th,e product of the last terms is' I c ;

*

. --;X

FEXERCISES(' ,,.e. 11. 9a2+Sa-2.22.72."B +10r-LZ.t-23. 10 rs:*-- 79 rs - 15 s2.

30. 3 arzn - 10 6n , _$' ,.,;


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TA TSBII{G

on+a2l2+bn.aa * a2b2 + b4 : e,4 + 2 n?q, t bo r &2b2

2. :Factor 49 l,,n + 34 hzk' + 2 k:.: *1"- ': ""' ' 'Solution. If .36 hzlcz *is added, thb expression becomes aluare. 4,4ing and subtracting 36 hzkz, we.haves-4gh4+34h2tc2+"25ha=,49la+70l2lc2+25ka-.3Bh?h2.#,"

tA

*EXA1TIPLES

Hrwr. Laa^*1:4q4+4a2+1-1a2' '4'. 15. e4.+4d,0. i, 16. 64 aanz + #. -.,

,&8.44t,.'&i.. :t.


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SE6ND COURSE IN ALGHBRA

: 9. A binqmiat' the sm or,thg difference'of two cubes.The type form is d + ff.*a $a divided by o + b gives the quotient aP - ab bz,oj and a,3 - 3 divided by a,, - igio** the quotie nt az * ab + bz.'.;i ,Therefore a,s +'bB : (o + b) (o, - ab * bz),3 - bs : (a,'* b) (o, * ab + br),* and. Formulas (1) and (2) above may be applied as the

t ) EXAMPLES i.1. Fact or as + zT . :_-- r__Solution. aB + 27 = aB * lu: (o * B)(o, _ a.B + Bz)

(1)(2)

_*t.": .;

Factor:L.. n: + f.-2. hr+do.' 8. (Ls + 2r.'t-4. cs + 5'.5. ds +" 8.6. dB + 2.7.7. rs +I25.,8. gf * y?. ^

EXERCISES9. &8 - 28.10. as - 27.11. mB - 64.

L2. TL' - (2 n)'.18. 8 nB'- y'.14. r8. - 2T s8.15. 64 - &'.16.'" I25 - s;8..25.af;f+n-y.2$,. ms-nt-i *Tn.

Y28.29.30.

tg.ojr


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*' ;.

-ory

a?7 I FACTORII{-G" 30. The Remainder Theorem. If any ratioaal integral. .. I -1 ' :-1 - il t---expression in n be divided by n - %; the rernainder is the

same as the original expression with n substituted for #,-This ,fact is illustrated in the','EXAMPLE5n * 6 by ffi - ?.

.:'i,t";,tii,,.i'll'i;Divid,e fiZ -Solution.e t

n2 - 5 n * 6 - RemainderHere the remaind.er r2 - 5 n *6 is thexame as # - 5'n | 6;the given expressi'on, when n is substituted. for n.; ExERcrsEs' -i 1. Divide n2+k+cby n-r and show that the remain-der is nz+bn|c.2. Divid.e nz + bn * c by n - and find the remairider.. ,a.' 3*. Divid.e n8+ 'n?+br * cby n n and. find. the remaind.er.,.rsA. Io'(rt * n' - 5 n * 3)* (n - 2) find. the remainder (") by.division] (a by. the Remainder Theorem. '5. In (# - n * 5) * (n -"3) find. the remainder (") by diviso sion , (O) by, the Remaind.er Theorem. I the rem.ainders in By.we of the Rernaind.er Theorerq fint

*

t

"i:

12-" 5r*6lrp"-n*12- ntr lt+(n-5


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86 ,'o SN6P COURSE IN ATGEBR,A31. Factor Theorem, By substiuting 2 f.or r in - 5 *+ 6wb obain 4 10 + 6, or 0. Hence n- 2 is an"exact divisor(or factor) of - 5 r * 6, Again, if 3 is substituted for rin n2-5r*6, the expression equals zero. Hence tr-3 isa factor of I - 5 r * 6. These examples illustrate theTheorem. If *y rationul integral erpression in x becornesz,ero when a ntnnher n is substituted, fo, x) tlen x - n is- afactor of the erpression..The Factor Theorem niay be used to factor some of the

lreceding exercises and, in adclitiorr, many others which arevery difficul to factor by previous methods.\trorn. By means of the Factor Theorem we are able to solvecubic and higher equations when the roots are integers. The solu-tion of the general cubic equation is one of the faurous problerns ofmathematics and one which is -accompanied by many interestingapplications. This problem was first solved by the ltalian, Tartaglia,about 1530; but was published by Cardan, to whom 'Iartaglia ex-plained his solution on the pteage that he would not divulge it. Formany ye4rs the credit for the discovery was given to Cardan, and tothis day i is usually called Cardan's Solution.

When searchitg for the values of r which will maktnexpression qe.roe only integral divisors of the last term of theu*ptt*sion (arranged accrditg to the descencling powersof. n) need be tried, for the last term of the factor musttbe n integral divisor of the last term of the expression.

#

-,d rlf

'\t**

l

t: /...-

EXAMPLEFactor n8'+2n-3.Solution. If. n - rz is a facor of rs t 2 *- 3, then z must be.:a,integral divisor of 3. Now the factbrs of - 3 are 1, - 1, 3, and - 3.If L is put for r,then rs * 2 r - 3 equals zero, hence n -1 is a factorof rs*2r-3. Diqfudingrs *2r-3 by n-1,we obtainthe quotiea{

rz* r*3. Sinee 8+ r*3 is prime, the$actors of r8+2r$are x-L n. r2*tr+3. ';,.*


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ra\*cr t\

{E* TjACTORTNGoRAL #XERCTSES. i .. ':,'' :1; Is -1 a factor of ns + 3n - 4?,2. Is r 2 a factor of 2n8 *n'*20?

4. Is n -1 a factor of rs + 3a2 - 4?1? 5. Is r +1 a factor of ?"8 - f - 4'r +1??. Is s *1 a factor of 3.ss - d,"t + 8 ?'8. Is k - 3 a factor of lt' - 5 tcz -g?

. EXERCISES .Factor:2.na*2r*3. *9. fra.-78-6n.;..,3. a,B + a,2 -.36. 10. ns -T frz + 4n {L2.4. fr'+n 10.o" 11. z*s-znz n-6.5. d8+d,'-72. L2. fr3-n2-4.t 6. fry-2#.-5r+6.- i' 13. 3n8 2fr2+Tfr,'_'$- .#$. fr'-ufro + 4n -4. I L4. &4- 6 a,8 +LI.* 6;.'il F82. The stim or difference of two like eoffi,"Th typg.formis " d+.

l'u;s

fies,in which s," =.b" is divisible by'n* or a*,'*he ea#. ;or Theorem.?bs, irr sn -Wr'n being either an odd or*antgerr,ubstitute b for a, Then b" becomes 'refore a, * is always a factor of an - bn.

orof d'-bn. '

,oeven ln-- bru t"'- Q.


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snccn coun,sn rN aLGEBRA,.ri".sff ,, .'t'',' . r i'b or -b for d,.Then n + b" becomes b" + ", rvhictr is not' zero. Therefore,ffi ** is nerrer divisible by o* or a-b wher n is even.

'

r is od.d a* is a divisor of a,n + br'..',tl* . I. an - Dn is al,wags divisible bJ a - b.vli .ff. o'- V, when n is trun; is divi$ble both by o'+'--e+f{r a'- b'. III; a" * ff is neaer divisible by a, - b.i ,fV. + r, when n s odd,, is divisible by a+ b.i. ORAL EXERCISES

For each of the following, statd a binomial f,aetor :

:bab

',b{t

'.; .. .. .,- a,,1. n' - y'.'g. n8 - 58.3. 2T - as.4. r5 - 2s.5. 32 n6.n it:6.-n. 21.3',G22. 10s -,.1.

Factor nu * yu.

11. ms '21, 16. I' - rr.L2. rs + 8. L7. L - rnr.6. *' - y'-7. n7 - 27.8. 27n7 . y'. 13. n|+gf. lE. l+f.9. 4 - tn,k ; .14 . r5 + 2u. 19. 1 + ,q.10. 8 - Ad: {'s, a6 32. 20. bt + 123. rs 10'.jF 1 divisibte by 11 ?24. Is 10e 1 divisible bv 9 ?{I',iEXAMPLE u

.., Solution. By division,


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:: . ry: r; }t

4.alii 'FACITORING/,F

{.' }'actor :fr6 + a5.2. n6 +L.3. 5 + 2t.4. rc6 32.5, (or)t+(br)t.6. fr6 - 25.HrNr. Find. the sec-ord factor by divisionand observe the signsof the terms and. theorder in which the ex-. ponents occur.7. 5 - 26.

EXERCISES.

e,5 - 32 n6.#(2*)'-243t'.e,7 - n7.aL*f.n7 - 28.nto * yt'.LCI + 32 nL.n7 + a,7.L+f.L28 r\ * t.

18. L - ro'.HrNr. Write only theflrst five terms and thelast term of the poly-nomial factor.19. L + rn'.20. nu - yu.$rNr. Factor firsas the difference of twosquares.2L. frr - yt.2,2. a,Lo - bL|.2b. aLz - bP.

8.9.

10.11.L2.

16.L7.

Yttr.-ii.:

dt {"{d,-&wl at' ,&i

13.t4.15.

\\

33. General directions for factoring. The followirg sug=.rr I a 1t a . n r .gestions will prove helpful in factoring:i Eir,st' look fo, (L cornmon, n?,onomial fact.or, and, if thereis one (other than 7), seprte th,e enpress'on, into itp_ grqq=Wy-t. M -ts-efipr, .qnd' t* qpM@;ffi$ #W*ie] fq.c7gL:-*Tfl- 'nen from the form of th,ffiolgmomi fa.frpr d,etermine**n uthich of the fohiwr"g iypni it should,' ffiffi ed, and'u,r, the method,s of factorng ifpticable to tlmi it.

1. ax+oA+bx*W.2. a2 +Zab + ffi.B. a2 - 2. ''5. (xrz*bx*c..6. +"he'+bn.7. d+ff.4. * + bx + c. 8. an*.= * ti* t''i f. Proc,eed agan aslin ff wh each polynomal "faeurpbwrined,- Qfrrrtil the orginal efrpression has been seprt&'i'4tt& ,' " ,':

,':@pfue faetors.t If the preceding steps ft; try -th.e Xactor ,Wfi,wpyv",i,-,. , 'li;';j,:.F


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t ,,'' f$.,t,.45 nt'y - 2o ny'. ' 4L-\-'.'-lJ t*. T'l,stc + 4 h,kz - 30 hlt'. 42.-.r'tt. Ba2,-10a,b*3b2. , .48.

snaf,xpREVIEW EXERCISES,{

-fiz-18*+n)'+10 25+r)r-3(, 5. \mB .-- 3 m,2 -I\m. -$ 33. -8n2-,fr6.2n2+3an{az. , 34. *'27 a'2..7.rfr4 -7 n'y' +Iyn..,' 35. d?+2d8.tr-*' 8.. aao - a. c' 36. - abzcz * abu9.2*'y )z*yJ. 37. +'-6@+ )z10. fr5-2ffi4''-9fr2. 38. *,n)u - I6(m11. a,a + Zbc - ad - Zbd. 39. -Znt;t12. 18'rs -'24* * 8rs2. 40. ao - 75 aloz - TO

* g'blF. tt'''

COUR,SE IN ALGEBRAfr8 3nz - 4lYr -(a-Qr

ono-a8-64^t-4b*(*' I(o-2r.8104*-"1art 1#b4'(e

oAr' -fr2:6#fr9t#-o**#-e-&,{a,

##*.aa*E

3n2,)'+

rOn)'-3n2-ZT a,t+2abzc,)t *

*)u --2n',-1ntnbtl.trt5,#,,:'*,)1#l0 arir(ii-133n2,*?fuL2abo+cd+- )ef-,*it.,r ,

+:ia.;f*t-8.

bsc-'- y.)w->, fr.T

l

..,...'(*'tb -;:fbz'

'ft1-,i

:

U.T: -1i*'.&F:

6)',*

rl 18

1, #+TC2+16. . 44.'" 1?' o-: Yoi: ,,' ct :u^' *'aiurte('I8,s, " T46.^IS- fnrn"ff&n'. , ': 47.= 0., '*r + 4&- 5. 48. # -*].Bn t-6."?.1 . a,fiv-4u*3n2-12. 49.\A,*afbs- Lezbz{Aab,. 50. .; *-,-#f**&,-*12a*36-- 8 n2 +IT n -2b'. a,28 - cz a.,2 + 1. 51. *72a+36.

=2cd+*?*."+)w1-r-


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e$jc,FACTORING

'1 69.I ?0.. '?1.

17Q^73.K.75.

3 a2-anrn +n'yu *n'*,10n - 3. r

4 a.(b-c)*S (b{t)'-4. Inz|6.

. 98. B* - ,8'. '*'gg. di,'' + bt".

U.W,' 94, fr2?L',#: n,**;ri*t6. (r-"1+ .otffiiF i)===, 1;ig?. a,B* + brff , :,i .'. .r;--.'-:-r..


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' ;, .'".*'iB ,sncKt couRSE rN aLeEBRA ; +U, 5 a,y - lllry - 5 a' + 3 ab "' 0.zrd,a-3ud,:bz-3bd.fr,cn-2dn:c2-4"d2.

Solve forSolve forSolve forSolve for-Solve forSolve forSolve forSolve forSolve for

aSolve forSolve for,Solve forSolve forSolve for

Saqh of the facors may beone factop is sufficient to Lake

nL, cLrrl, - n'r, + 1 - 2 - 0.y, bU ,* 3 dy - b2 + 9 d2 : 0.r r(2 o - 7 c) : 4 a2 - 28 a,c + 49 c2.ffit ffi (3 o - c)- 9 ad, : 2 ec- 6 o, - B cd.n, fr- 3 bn - a,2 : 3b2 - Aab.s,2s-Ts*LBu-2a221.t, 2te - 2 e +15 - 8e2+ 3 t : 0.AtA+2Id,n+4d:Tdy+1.n, d(1 - 3 n)* n+ 3 cbc - c + 3an.?, a,z * ae - 2 ec + 2 cd, - 2 ca I ud,.A - 3 - a.y - 3 y - ?(" - 3).

| -f-

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Solv4 for n,5 n-h c*-(5- 2 t)-5 +5-2 c Z ac.Solve fot n, &zfr - fr + n - a,s -1 - 0.Solve for fu, cB - czfr - 2 cd,n- 8 ds - 4 d,zn.Solve.for ffit ffi (a - 2) (o' + 4) - 4 - 16.Solve for s, 16 s - 8 a,s + 4 azs - 2 ass * ans - BZ : ,6.*

a . sr'-a

34. Solution of equations by factoring. The"_"methods u9l.Jggdr'g.enable.us...t.o'.so1ve'maJryequaibnsinon9-'*q*gggn- ^ fl the solution of equations by factoring useis made of the '|: : 'Wple. rf theo prod,uct of two 6r rnor.e factqrs iso,'af the factors must be zero.,?ro, bul the vanislFg of'.the product zeto. T*

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19/22

{"-$;-,...-+ .F ,t 'o# '' *t.-.r I;'1

"'g'4 ,?. .5,''Solve the equatiant ns - ffiz : 6 n. ni -*'in * ,flE'-*-/"f?

F'ACTOR,INGEXAMPLE

fj ..rf ;_'ir t ,t'f"./" -Fdx -;* s)*,

Solving the equations resulting from setting each factor separatelyequal to zero; we have fr :0,, - 2, and. + &Substituting these vIues in the originalS, - 2, and * 3 are roots of the equation.The method of solving the aboye example is stated in theRuIe. If. pe qe^s-gq z-Igw y e th,e. -e-gu-et:|qk- p-g- &gI JhSScend,membef is zero: TIen factor th,e rtr membnfr,X-pgJqc_tgrw@;*$-\, 'r.a -aru equatoqns.**brr*.t as usual. '&

'tIfn he above example the division of each member of -- rz :6 rby * gives nz - a : 6. This last equation has only the $ots.- 2 and3, while the original equation had, in addition, the root 0.Again,if theequation rcZ - 5r * 6 - r - 2 issolved by therulertheroots are2and 4. If,however,eachmemberof fiz- 5r*6- x-2is divided by r-- 2, the resulting eqtration n = 3 =d.,libas the%oot 4 :,'Qaly." I-q eachTf these,gases th.e root would not lulr" been lost.if ,*:the fuctdr used as a divisor had been set equal to zeroand the equa- " ' ,', .

ion thus obtained had been solved. The solution of an eqoatioo 'iC often simplified by the use of this method nprrne Dy tne use or tnls metnoo.

/'\^.equation, we find that

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E:(EREISffiSolve ty {toljng:1. n2 ''4.:'0-n2. fr2-.5*:0,3. n2 an+A,.'- . : ;,\'. 4, ,#? - 4a'2 = 0.

.5. 4ns': 25 n.*6' Y' :64Y'7. # +5 =6n,8. 2f * n: $.


20/22

sEcoND [o**t,*,,.,9. rnt ;'',/ll, - 2 q,2 == 0.,0. n2 Zan-82-0.11. y'-y+2-2y*.L2. frt-9a2-fr+3g,:'Q.'13. fr8 - a# -.I2 azn; 0,14. q\-annf 2 anz -ia8:0.'15, fi8 + n2 - a,,zn - an: 0..16, ay+79y', 5U':0.i?. yt-Ty-6':0.18. Un-I}y'+36:Q. '19. #-lnsi-- 4n.Po. *l'n - 7 n2: :- 6 n. r,o$1. }rs + 24 + 48r _ 0.

I}{ ALGE,BRA &,25.26.27.28.29.30.31.32.33,34.35.36.37.38.39.40.

8n2+TLns-9na-0. -,..,(2 *- 3)' - (5 n * 6)': 0. ':t"(n - r)'- (n - s)2 - 0. .(r- 3)'- 2(*-S):g. vn8+50- 25n-2n2:0.'2y'--2y' -8 y+8:.0. , ,*.,ns - arz - 4 a,n {A 8 :b:-.n8+8+ 6#+12,s:Q.y'+3f+3y.+L-0.acnz * bcn * adq +bd ^- 0.frz-Tr- 8= fr+t.'4n2*n-1-2n-7,3n2-11 n*6-3n-2.a}n|-2n-3- an-3.frs-Zaz*75fi:12-5n.6n2+7n 3- 3n-1.' 35. The highest common factor.. The highest common factor(H. C. F.) of two or m.ore monomials or polynomials is thee*p"d*sion of highest degree, rvith the greatest numerical"1 coefficient, which is a,n exact divisor*of each,

I Thus the H.C.F. of 28 azb\ and 42 a2b2 is 14 a2b2. The H.C.F. ofi *8 - 4 r and. ns - 5 12* 6 r is r (r - 2), ot'tr2 - 2 r.EXAMPLE'x.Find the H. C. F. of 9 na - 36 nz and. 3 n7 - 72 n6 + L2 e6.Solution. Factoring, we have

g,tr4 - B6 rz - g2*2 (, * 2) (* - 2),' 317-t2r6*1215:3r5(*-2Y. ,',.Therefore the H.C.f'. is 3 or(* - 2), which equals 3 us - 6 12,

22. ut#, z-622 -8-0.23. 3 lr+* 72,r + 33 x62 - Q.24. Esz&r-Brr-0.

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21/22

IIl'- -..,4b,.*

j, - : tt. -i.r:r ..., o &",'-9,.;''.,r.:;',*, .;ii''crt ,iF- '.tii \tH{dI .. .tlr f\

FACTORING+.i The method used in the preceding solutions fb findingf ra.*' the H. C. F. of two or more monomials or polynomials isstatecL in the*i Rule.' Separate i,ach enpression intp its prime factors. Tlten' find, th,e prod,uct of such factors a,s occur iTL ea,ch efrpreson,i usng , eqih prme factoi the least rrunber of tirhes 'tt,occurs n

a\ any one rpression. 'If tw.o or more polynomials have no common factoroher than 1, then 1 is their H. C; F., and the polynomials.-t - are said to be prime;to each other. ,EXERCISES I

Find the II.C.F..of the followitg:1. 12, 18, 24. " 5.. 30 c'd, 45 cd,z, T5 p2d,2.2. L5, 25, 40. 6. 28 o'bu, 42 a6, 70 a2b3.3. 24, 60, 72. 7. 66 can, 782 c'n',lffi'ot.4. 12 a2, 30 a5, 36 aa.. 8. 2 + 2 ab * b',.ffi '., r,g, 3a,2 - 3 b',9(o - b)u,}as - }bs. #'10. anz * 2 eny * a, a2fr2 .- o'y'r'Z a'frg 2 ey'.11. 2a,2m,2-2 &'rL'r 4amz-LZ amn+8 &tuZ, I0 um*L0an.

13. 3 na - 6 nty, 6 n6 - 2+nuA', 72 r5 - 96 n'y'.L4. 24 a6 - 6 onb', 48 a6 + ?4 etb, 48 a5 - 48 aab - 36 a|bz.15. 5 17 - 160 12, L5 n6 - 60 n8, 25 n7 '- 200 na.16, LB aa - 2 o'b',12 a6 -"8 a;b - 4a,4b2r 30 aab? + 10 asbs.L7. 4na - 4*'y'r 5n7 - 5*tynr SrLL - \'*'yu.18. a,5 - 3a4b +2o'b'r,q,6 - 2aab- a,sb2 +2o'b',' 7 -5a5lf +4&sba.'Notn. The most famous, and in some respects the most perfect,treatise on elementary mathematics ever written is Euclid's " EIe-ments." About one third of the material of th thirteen books rry,ts

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22/22

46 SECO}) COURSE(II{ ALGEBRAtopics which to-day would be considered arithmetical in character.In appearanoe and language, howeyer, they are all geometrical, for' Euclid represents quantities not by numerals, as we do in arithmetic,or by letters, as we do in algebra, but by lines. Book VII containsthe earliest statement of a general method for finding the G.t.D. oftwo numbers. This rnethod, though never necessary in elementarymathematical work, is so perfect and beautiful from a scientific pointof view that until recently it remained in elementary treatises ,onalgebra and arithmetic by iorce of tradition. ft is a great tribute t".-.-Euclid's genius that he was able to devise so perfect a nr ethod for -\,..the - process that aII the efforts of two thousand years have beenunable to improve it essentially. ft is of fundamental importanceih advanced portions of algebra,

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