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Lqpp`0 @̋p@p0?`~YlCa`@F1f` LA@Ũ|GyO xg`~x3`LqOAsp|~?`|ԂԂ|?f88`q L988xՂ0L@p@x_y>~~88`@F1f` LA@ŨxqO >Xc{@?8p??Ղ3`LyGA3`̃vx:<f`` L>Ղ0L@`|GyOp8p`@F1gaNA@쁁@@08`|Ղ3`L@A3`8쁁6``@rxf`` L8<`ԊgaLOc׊x`Պ@`@~?~C|GA`̃@ԂԂF1o `ONS<W2> RAPHS OF OLYNOMIAL UNCTIONSO1dF EGREE REATER THAN WO<u3> ATIONAL UNCTIONS<4> ARTIAL RACTIONS<5> ONIC ECTIONS#XPONENTIAL AND OGARITHMICi UNCTIONS<51> XPONENTIAL UNCTIONS<F2> OGARITHMS<W3> OGARITHMIC UNCTIONIN1@WO IMENSIONS<Q2> RAPHS<b3> EFINITION OF UNCTIONS<s4> RAPHS OF UNCTIONS<5> INEAR UNCTIONS<6> OMPOSITE AND NVERSE UNCTIONS<7> ARIATIONOLYNOMIAL UNCTIONS, ATIONAL,UNCTIONS, AND ONIC ECTIONS<F1> UADRATIC UNCTI9> -RACTIONAL XPRESSIONS (ART )$QUATIONS AND NEQUALITIES<@1> INEAR QUATIONS<R2> PPLICATIONS<d3> UADRATIC QUATIONS<v4> ISCELLANEOUS QUATIONS<5> NEQUALITIES<6> ORE ON NEQUALITIESfUNCTIONS<31> OORDINATE YSTEMS !1> '!-!EAL UMBERS<02> '0-0OORDINATE INES<?3> '? -?NTEGRAL XPONENTS<N4> 'N-NADICALS<]5> ']-]ATIONAL XPONENTS<l6> 'l-lOLYNOMIALS AND LGEBRAIC-{XPRESSIONS<7> '-ACTORING<8> '-RACTIONAL XPRESSIONS (ART )<CEo oToO THE NEXT QUESTIONO:xR OBJECTIVE FRAME.#$ TO RECEIVE HELP WITH A CONCEPT. # $ TO RETURN TO THE MENUʔ.O+RT HE NUMBER WOF THE ANSWER OR MӮENU ITEM. UNDAMENTAL ONCEPTS OF LGEBRA<R '@ 0L 0Ld0Ld0Ld0Ld0Ld0Ld0Ld0Ld0Ld0Ld0Ld0Ld0Ld0Ld0Ld. *  LcLc@ @` Pp`@( H`TD@8\`~>\5hxi=4^ z||B@ Lz*L)LACE DISK IN DRIVE AND PRESS .  )  c0 /  a0  c0 .  a0  c0 .  a0  c0 .  a0  c0 /  a0  c0 .  a0  w0  X0 A Z. Q  0LY* 0 5 X* LO* . &V! X . L*ERROR # *  Q. aL* OCCURRED(  w0  / . }/ .L(!HANK YOU FOR YOUR INTEREST IN THE .  (  w0 - ,  M, - V/ . / .L$) LECTRONIC L-)TUDY L6)UIDE. M,  )  c0 /  a0  ')  c0 %-  a0  0)  w0 - ,  - '1 =0   1 - B.L& / G n N. .(U(L'SEC USER ' 0L%!L'SEC 1.7 ' 0L%!L'SEC 1.8 ' 0L%!L'SEC 3.6 ' 0L%!L(SEC 4.5  ( 0L%!L)(SEC 5.2 !( 0L%!LD( SEC FEATURES 7( 0 - -L%! k, . -LE& M, - ,, . t/ . - ,L(%OU HAVE REACHED THE END OF THIS DEMO. /  f. . 0 L&BLOAD LOGO,A$4000 &  . , L0& 1- -LE& E.LE&L&D k0 5 & L!' f. 3. 6. 9 $L`&L&'M k0 5 $' L`&L>'E k0 5 <' La( k0  n 5/ ( + k0  + D/ - + (0 L' - - L`& k0  - - ''''((4 L% , - # , - - , - # -` 0 L%BLOAD SEC MMENU,A33600 %  0 L&BLOAD SEC MMENUMAP,A &   k, a k, - k, . .- - - % - :- - B. K - , - ^- $L& , , + =- - $ G, + , - [- =- - $L& =- V, LE& L$ 9 $ , ,L! =- - $ + , - 0 L$ , - - ,L! - - L! - , - , - % - V, $LH% - - $L! - ,  + - - 0 L% , - - - - 0 (L! % - , $ + - - $0  + - - $0 L!LN% - G n - - +,$` L( L5 LD 8 XL XLR XQHPH RPhPhQ8 XL.0'L+ LfLLLLL ꥪLYԨé`Л@F1f` Lc@ŨfA@1Lc9GP OUT. E HAVE4#X21-? 8X + 9Y2~+ 36Y + 4 = 0+UR IMMEDIATE TASK IS TO COMPLETE THES9QUARE. HEREFORE, WE ARRANGE THEEGQUATION IN THE FORM(_4X2\- _8)_X+;_)P_+(9Y2s\+|_3_6Y+_)_=_-_4tEXT, FACTOR OUT THE 4 AND THE 9.3364FD(X-_D1jD)2xA+D9D(Y+D2D)2A=D3D6(NUX-`U1kU)2yRmOUmmmmmm9d`+[(UY+U2U)2RmUmmmmmm4`=[1[.JȀ{{f-z-z-z-g~-p~-~-~-~-~g(1,-2) ET'S SEE IF WE CAN HELLP WITH THIS PROBLEMHANGE 47 X2E-P 8[ X+m 9x Y2+ 3 6Y+ 4 = 0 TO STANDARD FORM.4#(X2& -1#2<#X+N#mX#)+i#9s#(Y2 +#4#Y+#m#)=#-#443(X2&0-132<3X+N31X3)+i39s3(Y20+343Y+343)=3-34+343+Wh3ah)2oemEhmmmmmm4Zs+}n(hY-h1h)2emhmmmmmm9s=n1n.ROM THIS WE SEE THAT THE CENTER(H,K) I.S (3,1).KETCH THE GRAPH OF45X2*2-85 8X + 9Y2w2+5 36Y + 4 = 0\YPE:<v1> TO CHECK YOUR GRAPH<> TO GET HEe8uY+u4u9=u0uWE FIRST CHANGE ITTO THE FORM 98(X2M-W6aX)+y4(Y2-2Y)=ĩ-Ω49EXT WE COMPLETE THE SQUARES TO GET9G1(X-_13i1)2w.+141(Y-111)2.=1316.DIVIDING BY 36 WE OBTAIN THE DESIREDFUORM:(FhX-kBmAX.)UR DISCUSSION OF ELLIPSES AND HYPER-BOLAS CAN BE EXTENDED TO THE CASE WHERET)HE CENTER IS ANY POINT ()H,K) AND THEA9XES ARE PARALLEL TO THE COORDINATEAIXES.YO SKETCH THE GRAPH OF98uX2Fr-Pu5Zu4X+ru4|uY2r-u,m275=212. IF Y=I0)I, WE HAVE X=IkI3I; THEREFORE, THEXY-INTERCEPTS OF THE HYPERBOLA ARE (3,0)AiND (-3,0). OTICE THAT THERE ARE NOYy-INTERCEPTS.OW FIND THE ASYMPTOTESAND SKETCH THEGRAPH.(MI[NbTi:p SEY=]XX]Y_cgikkkzithogjeecc_^[Yzumg`Y ET'S SEE IF WE CAN GET YOU STARTED.HE EQUATION 25X2w-9Y2-225=0 ISE1QUIVALENT TO [2 Xi,2p)mg,m9l7-|1Y,2)m-~-y-t-o-j-e-`khmhqhshwhyh}hhhhhhhhhknqtwz}}ywsqmmkkkkkkkkkkkk|kykvkskpkmkjkh- 9 Y2- 2 25= 0 IS EQUIVALENT TOX~2m|m9%-Y2mm2%5= 1 .9HE X-INTERCEPTS ARE (3,0) AND (-3,0)AIND THE ASYMPTOTES ARE Y=IkI(5/3)X.AW-~-F-------------------------mAXmAEaXT^.kWppp\ҩ-AAĀBm-B'KETCH THE GRAPH OF28A5X2M>-[A 9Y2w>-A2A25=A0AT[HEN TYPE<u1> TO CHECK YOUR SKETCH<> TO GET HEeLP WITH THIS PROBLEMHE EQUATION2i 5X2~HE RECTANGLE, DRAWN TO PASS THROUGH AAND -A# *ON THE X-AXIS AND THROUGH B AND-)B) )ON THE Y-AXIS, HAS DIAGONALS WHICHA9RE THE ASYMPTOTES OF THE HYPERBOLA 9IHE ASYMPTOTES HAVE EQUATIONS OF THEF]ORM Y=-^k7]BFXERBOLALUOOKS LIKE:1~=}>{?yAuCrCrEoGlS`===>?ACFIQ##"}!z vrok ^""!  OwRrrqpomkii`rrqpomkaǎʎ͏АԑؓܕĎRAMMING10HIS IS THE END OF THEbJEATURESN@\@\NNsOU MAY USE THE:<> KEY TO REVIEW THE SECTION< > KEY TO RETURN TO THE MENUWSTEMS OF p=INEAR QUATIONS IN #HORE THAN WO ARIABLES<U4> ATRIX OLUTIONS OF YSTEMSO#`F QUATIONS<m5> HE LGEBRA OF ATRICES<z6> ETERMINANTS<7> ROPERTIES OF ETERMINANTS<8> RAMER'S ULE<9> YSTEMS OF NEQUALITIES<10> INEAR ROGS<h4> OMMON OGARITHMS<y5> XPONENTIAL AND OGARITHMIC/QUATIONS<6> INEAR NTERPOLATION<7> OMPUTATIONS WITH OGARITHMS, YSTEMS OF QUATIONS AND ]NEQUALITIES<#1> YSTEMS OF QUATIONS<02> YSTEMS OF QUATIONS IN 2 ARIABLES<=3> YOBLEMS.1 ) Y DEFINITION, 3{ 4 = 8 1  L OG38 1= 4 2") "Y D%"EFINITION, Ls"OG5'"1mmm1(25"=#-!3# 5#-3="mmm1(253>) A) LOG4?C(F>1/64)=s>-~<3> BECAUSE 4-93;=>1>/64BNTIAL FORM.(^3) IND (A) LOG4wc(~^1/64) (B) LOG5c1^25(x4) IMPLIFY LOGAw}{3umiiXt2qmtmY}ZUxSING THE LAWSO#F LOGARITHMS.(5) OLVE FOR X: LOG2(X+3)=Ǘ5ї.HECK A'NSWERS A]FTER WORKING ALL PرRdUf_me_Wfh)od=d LOGAiUd -d LdOGAiWd(III)LBOGAW(^UCl)s = C LOGAUOW CHECK YOUR UNDERSTANDING WITH THESEPROBLEMS:('1) HANGE 34[$=f'8q'1 TO LOGARITHMIC FORM.(=2) HANGE LOG5iB(p=1/125)=-=3 TO E#LXPONESS  TO REVIEW THE LAWS OFLOGARITHMS.IMPLIFY LOGA[Xd2kYr AND LOGAZ3,Ƞ THEN TRYAGAIN.HESE ARE THE THREE LAWS OF LOG-A#RITHMS THAT WE USE:(=I) L@=OGAUB(\=UW) = LOGABU= += L=OGABW=(dII)LAdOGAVi(]IN. OUR ANSWER: b DON'T KNOW WHAT TO DO.#HAT YOU WANT TO DO IS SIMPLIFYL3OGA8X%32,0Y33/Z3H0 O3USING THOSE FORMULAS.POR EXAMPLE, BY FORMULA (II),LcOGA*hX3c2:`YAc/Z3V` ]c=bc LncOGAhXc2`Yc -c LcOGAhZc3`RETHE FORMULA INCORRECTLY.@RESS (@ /@T4@O REVIEWTz@HE LAWS OF LOGARITHMS._Y FORMULA LrOGA4wX=r2DoYKr/Z3`o gr=krLvrOGAwXr2oYr -r LrOGAwZr3oIMPLIFY LOGAUX^2eYl ANDLOGAZ3, THEN TRY AGAN - LOGASZO3L=BsLPsOGAexXls2sp zs+ LOGAxYs - LOGAxZs3p=B 2.WL^OGAsXz + LOGAY-3ʗ.ѕLؗOGAZ`BKKKKOUR ANSWER:2[(LOGA~X)(LOGAY) - 3.LOGAZ0OU ARE USING E IS CORRECT, BUT IT CANB=E SIMPLIFIED FURTHER.\OR EXAMPLE, LOGAwaX~\2Y \= 2.ZL\OGAaX\.{IMPLIFY EACH TERM AND THEN CHOOSE THECORRECT ANSWER.*XC*E*LL-*EN>*T.LNOGASX&I2-FY4Im&ImmZ-Y34V=BNLPNOGAeSXlN2sKYz><\2>22\.9ZL@\OGAUaX\\ + LOGAaY\-\3\.ZL\OGAaZ\<v3>22v(LOGAU{X\v)(LOGA{Yv) -v 3v.tLvOGA{Zv<4> 2 DON'T KNOW WHAT TO DO.OUR ANSWER:LaOGAvX}2 + LOGAY - LOGAZ30HAT YOU HAVA#(*20) =J LUOGAj(q4.~5) = LOGA4Ɓ +с L܁OGA5.OW TRY THE PROBLEM AGAIN.IMPLIFY LAOIGQAYXc2kYs/{Z3 USING THE LAWSO'F LOGARITHMS.<A1>L2AOGAGFXNA2U> \A+ LOGAFYA - LOGAFZA3J LUOGAj(q4.~5) =LOGA4ā +ρ LځOGA5.OW TRY THE PROBLEM AGAIN.OUR ANSWER:2b.240ORRY, BUT YOU ARE USING THE WRONGF@ORMULA. F@OU ARE MULTIPLYING INSTEADOPF ADDING.mSING ONE OF OUR LAWS,WmE HAVE LOG=idLwdOGAi4d + LOGAi5d=i~1w~.39 + 1.61=i3w.00ssOUR ANSWER:.b220ORRY,(0 BUT YOU ARE USING THE WRONGF@ORMULA. F@OU ARE SUBTRACTING INSTEADOPF ADDING.mSING ONE OF OUR LAWS, WE HAVE LOGA#(*20) =){ = C LOGAU#SING THE LAWS OF LOGARITHMS, EVALUATEL6OGA;(620) 56IF LO6OGAd;4k6 =w6 16.39 A6ND L6OGA;56=616.61.3].00.w222.240OO0D0!L*JOGA?O(FJ20) = LOGAO(J4.H5J) SELECT THE CORRECTANSWER. EAWS OF OGARITHMS*HERE ARE THREE FORMULAS THAT WE NEED:(&DI) LJDOGA_I(fDUW) = LOGAIUD +D LDOGAIWD(kII)LJkOGA_p(fkUqgWqqmpg){k = LOGApUk -k LkOGApWk(III)LJOGA_(fUCtSE, WHAT WE ADD TO ONE SIDE OFTHE EQUATION MUST BE ADDED TO THE OTHERSIDE.+_4_.]1_+__ ]_OW FACTOR THE TERMS ON THE LEFT ANDDIVIDE BY THE NUMBER ON THE RIGHT.F#HIS IS THE END OFM=ONIC ECTIONSIAHAH44AHA(_44_(8)_2)_(^_99^_(3_64_EaLaaassHAT NUMBER GOES IN EACH BLANK TO COM-PLETE THE SQUARE?1F_DaMa4_aassF COUR 9X)zH{H{88H{H( c3) AB - 3A - 2B + 6=Zp A(B - 3) - 2(B - 3)=Z} (B - 3)(A - 2)deett˂e( 4) 125 - C = (5 - C)(25 + 5C + C )3ayyy>HIS IS THE END OFV=ACTORINGAGBGB--GBGjOU MAY USE THE:<10X + 2129 (=2) 1 - 81X2c:(W3) AB - 3A - 2B + 6(q4) 125 - C3cnON'T GO ON UNTIL YOU HAVE FACTOREDALL OF THE PROBLEMS.ERE ARE THE ANSWERS:( #1) X27 >#+ 10X + 21 = (X + 7)(X + 3)(((( C2) 1 - 81X2a@ hC= (1 + 9X)(1 -2SXY + 3UV - 2XV - 3UY=M` (2XY - 2XV) + (mmmmmmmmmm)-`3UY + 3UVm`mmmmmmmmmzEXT, USE THE DISTRIBUTIVE PROPERTY:=M 2X(Y - V) - mm(mmmmm)3U Y - VmmmmmmmmOW, COMPLETE THE FACTORIZATION.OW FACTOR EACH OF THE FOLLOWING:(#1) X + .OO??OO dF YOU GOT A DIFFERENT ANSWER, TYPE TqO GET SOME HINTS.ORK ALONG WITH US USING PAPER AND PEN-CIL. RY TO FIGURE OUT WHAT GOES IN THEB#LANK, THEN PRESS THE SPACE BAR TOC0HECK YOUR WORK.CIRST, REARRANGE THE TERMS:C + AD = -5, THE MIDDLE COEFFICIENT.HAT IS THE FACTORED FORM OF20XY + 3UV - 2XV - 3UY?(CINT: TRY GROUPING.)<*d112d> TO CHECK YOUR ANSWER<*~> TO GET A HINTHE CORRECT FACTORIZATION OF20XY + 3UV - 2XV - 3UY IS(#J2X - 3U)(Y - V)T IS:2ZX - 5XY - 3Y2W2iW=wZ (AX + BY)(CX + DY)WjHERE A, B, C, AND D ARE INTEGERS ANDTwHE PRODUCT OF A AND C IS 2, AND THEPRODUCT OF B AND D IS -3. OU MUST TRYVARIOUS POSSIBILITIES BY "TRIAL ANDERROR", CHECKING THE MIDDLE TERM, UNTILBF YOU GOT A DIFFERENT ANSWER, TYPE TuO GET SOME HINTS. ACTOR 2X - 5XY - 3Y .2F 2 #ET'S SEE IF WE CAN HELP. INCE THEREI0S NO COMMON FACTOR IN THE TERMS, THEO=NLY WAY THE TRINOMIAL WILL FACTOR IS ASTJHE PRODUCT OF TWO BINOMIALS. HAN N(hVIIIVhIII) (X - Y)(X + XY + Y ) = X - Y2xe2e3e eHAT IS THE FACTORED FORM OF20X - 5XY - 3Y ?2#-2w-<1d119d> TO CHECK YOUR ANSWER<1~> TO GET A HINT2:X - 5XY - 3Y272i7=w: (2X + Y)(X - 3Y)??..?? htHE SPECIAL PRODUCT TO USE IS:X* + Y = (X + Y)(X - XY + Y )313T22ddd:>C:ee^cg_NE OF THE FOLLOWING SPECIAL PRODUCTSI*S THE MODEL TO USE FOR THIS PROBLEM.(QVIIVQII) (X + Y)(X - XY + Y ) = X + Y2xN2N3XYYYssY ET'S TRY ANOTHER. ACTOR 8A + 27B3F 3w #OUR ANSWER: (b#2A + 3B)(2A - 3B)AECAAREFUL!gAHE POLYNOMIAL 8A + 27B3>>INS THE SUM OF TWO CUBES; THAT IS,8aA + 27B = (2A) + (3B) .3^3F^3~^3^TGE THEMIDDLE TERM OF THE SECOND FACTOR.'''ԁrrԁ OOD!:HE POLYNOMIAL 8A + 27B3w737I:S THESGUM OF TWO CUBES:8hA + 27B = (2A) + (3B)3e3?e3we3e=M (2A + 3B)(4A - 6AB + 9B )2||:EC:AREFUL!c:HE POLYNOMIAL 8A + 27B37 7IGS THE SUM OF TWO CUBES;G THAT IS,8TA + 27B = (2A) + (3B) .3Q3?Q3wQ3QTHESaPECIAL PRODUCT TO USE IS:X#{ + Y = (X + Y)(X - XY + Y )3*x3Mx2x2x OOK AT THE MODEL, THEN CHANHAT IS THE FACTORED FORM OF80A + 27B3#-3T-?[0<d11d> (2A + 3B)(4A - 12AB + 9B )2a2a<~22~> (2A + 3B)(4A - 6AB + 9B )2{2{<33> (2A + 3B)(2A - 3B) ACTOR 8A + 27B3F 3w #OUR ANSWER: (2A + 3B)(4A - 12AB + 9B )2 THIS MATERIAL<* > KEY TO SELECT THE MENU\2\AA2\2U) LOG2*(1X+3)=W5aqXx+3=25=32 AND X=29**{4{4{AA4{4gFgSsSsFgFyyyᣀ IHIS IS THE END OF b=OGARITHMSdOU MAY PRESS THE<*~> KEY TO REVIEWP) LOG5?U1HP25=`P3kP BECAUSE 53M=P1P254o) LOGA*t6t>f>`O`37nXAk2HhmAkmYBtZ=Wo 1eimdi3etLnnOGAsXn2k/nYZ =1m3(2LOGA3X:-DLNOGAcYlZ)=2m3LOGA-LOGAԌY݇-LOG 5299 @<- 5X + 3(@@2*LX28I ?L- X - 3(`3)6+`X29] @`- X - 126`X2] `+ 11X + 4)dddd`hdd2*pX28m ?p+ 15X - 272pX2m p+ 17X - 9(4)3+X + 2 X + 5 2X + 7)ShʄX1 + 1-[Xi + 1+X + 1OR CHOOSEA~NOTHER ANSWER.#OUR ANSWER:3~#X2 +#2#X+#4#}&&X0+020JE C6JAREFUL. IDN'T YOU MISSAW SIGN? UWOW SELECT THE CORRECTAdNSWER.IMPLIFY EACH OF THE FOLLOWING:(1) 24A X Y3@ 2 332*6A Y5G' 2a)(<2)2+3j}X2xz-}2}X+}4}3jX2x+2X+4jX}+2#OO#D!3G=X2U:2=X4=E@`@u@@@@XGJ+QJ2[J-gBXwJ+J2J+BXJ+J2J=Uo3ciX2qf-{i2iX+iOMMON DENOM-I\NATOR AND A NUMERATOR WHICH IS THESoUM OF THE NUMERATORS OF THEEXPRESSIONS CONSIDERED.ELL WHICH OF THE FOLLOWING IS THES#UM OF3*4X2812[4X44$8@8W8q888X&B+0B2:B-J:XXB+bB2lB+{;XB+B2B.83jV)2s2sOW SIMPLIFY THE FRACTION IN THE LARGEPARENTHESES AND YOU WILL BE WITHIN AFEW STEPS OF THE ANSWER.HE SUM OF TWO OR MORE RATIONALI#NTEGRAL EXPRESSIONS HAVING THE SAMED6ENOMINATOR IS A RATIONAL INTEGRALEIXPRESSION WITH THE COOK AT IT.(KX - Y )2H2?HOHTHRERK^HgHHHHEKHHHXiC + XYXCY - X2n@2@YiT + XY Y - XY2pQ2Q=n (X - Y )28k2[kkkrkohonzk(gX + XY)(Y - XY)2d2dkkk?v?v(vY + XY)(XY - X (X + Y)Y(Y - X)rrY{(Y + X)X(Y - X)r= (X - Y )282[kroo1z= X* - Y212TeeHE PROBLEM IS NOT AS BAD AS IT SEEMS.OU MUST SIMPLY USE THE RULEA"/B,300"C8"/D = AD/BC /ET'S TAKE A L^-g----*0---Xi( + XYX(Y - X2n%2%Yi9 + XY Y - XY2p626=S (X - Y )28P2[PkPrPoMoSzP(LX + XY)(Y - XY)2I2IPP([Y + XY)(XY - X )2X2XP=u (X - Y )28r2[rkrrrooouzrXnTHIS ONE:(#X - Y )2 2? O T RR#^ g #  Xi + XYXY - X2n2Yi, + XY Y - XY2p)2)<1]1> TO CHECK YOUR ANSWER<1w>AwTUwO GET A HINTERE IS THE SOLUTION:(0X - Y )2-2?-O-T-R*R0d Cbf/D = AC/BD. E WANT TOFsACTOR THE NUMERATOR AND DENOMINATOR ANDCANCEL ANY COMMON FACTORS. OTICE THATX$ - Y = (X - Y )(X + Y )4+4N2x222ڍ=\ (X - Y)(X + Y)(X + Y )2ڝ2VVVnIn99InIhBOW TRY j=-D|(DX + Y)Y2|XK - Y=w=w=w ET'S GET YOU STARTED. E KNOW THATX - Y44?(,X - Y)2C)!J!.P"X^, + Y2e)2)Yp2w\!Z!!=8M (IWX - Y)(X - Y)(X + Y )2T2TIJJ(pEX - Y )Y4~B4B2BBfECAUSE A/B .TX2 + Y2)OW, SIMPLIFY THE FRACTION.ORRECT.4L4Z4h44 2X24. &2- Y;14B.Yx02-(?X - Y)2E=Xh@2o= v@+ Y2<=-` D^^(HYX-YYYcY)(X+YYY)(X2V Y+ Y2V)YY2V(IjX - Y)(X - Y)(X2g j+ Y2g)e>HIS IS THE END OFV=ACTORINGAGBGB--GBGqOU MAY USE THE:# KEY TO REVIEW THE SECTION. # KEY TO RETURN TO THE MENU.2 *#+ 10X + 21 = (X + 7)(X + 3)(((2C) 1 - 81X2M@ TC= (1 + 9X)(1 - 9X)fHgHg88HgH3c) AB - 3A - 2B + 6=Fp A(B - 3) - 2(B - 3)=F} (B - 3)(A - 2)PQQttQ4) 125 - C = (5 - C)(25 + 5C + C )3M2ee> KEY TO REVIEW THE SECTION< > KEY TO RETURN TO THE MENU#) X + 10X + 2129 2=2=) 1 - 81X2c:3W3W) AB - 3A - 2B + 64q4q) 125 - C3cnON'T GO ON UNTIL YOU HAVE FACTOREDALL OF THE PROBLEMS.ERE ARE THE ANSWERS:1#) X2# E= ==S=E=R=GG--GGqOU MAY USE THE:#$ KEY TO REVIEW THE SECTION. # $ KEY TO RETURN TO THE ENU.ION. # $ KEY TO RETURN TO THE ENU.--GGqOU MAY USE THE:<> KEY TO REVIEW THE SECTION< > KEY TO RETURN TO THE MENU@} mm } }m}mEEK HELP: mmmmMHIS IS THE END OF *=N2=S:=TB=RJ=UR=CZ=Tb=Ij=Or=Nz=S= =T=O= =T=H=F YOU PRESS A LETTER, THAT FUNCTIONWILL BE PERFORMED.ccmcmpACKUP ONE FRAME:mpmppmpm}ETURN TO THE MENU FRAME:m}m } }m}mEEK HELP:mmmmMHIS IS THE END OF *=NSTRUCTIONS TO THE SERGGyy0EVIEW OF THE PTIONS"EE IF YOU CAN REMEMBER WHICH KEY WILLP/ERFORM THE FUNCTION. RESS THE SPACEB@0: TEP TO THE NEXER: WHE SSWPP%WAA-WCC5WEE=W BBLWAATWRR\W IS USED TO STEP TO THENdEXT FRAME.~ DIFFERENT LETTER IS USED TO ~ACKUPONE FRAME.HAT LETTER WOULD YOU TYPE IF YOUWANTED TO REVIEW THE FRAME JUST BEFORET#HIS E JUST BEFORET#HIS ONE?=OUR ANSWER: WHE LETTER NW IS USED TO RETURN TO THE dENU.~ DIFFERENT LETTER IS USED TO ~ACKUPONE FRAME.HAT LETTER WOULD YOU TYPE IF YOUWANTED TO REVIEW THE FRAME JUST BEFORET#HIS ONE?=OUR ANSW c=WHE LETTER  IS USED TO ADVANCE TO THENdEXT QUESTION.~ DIFFERENT LETTER IS USED TO ~ACKUPONE FRAME.0XCELLENT!qEXT, LET'S SUMMARIZE WHAT YOU HAVEL~EARNED.HAT LETTER WOULD YOU TYPE IF YOUWANTED TO REVIEW THE FRAM~DEMEMBER TOA~QNSWER BY NUMBER.y;y;yUU;y;;;(B.=/A)B1N2[3h4u5RY THE ELP OPTIONIF YOU WANT TO.{{HAT LETTER WOULD YOU TYPE IF YOUWANTED TO REVIEW THE FRAME JUST BEFORET#HIS ONE?=OUR ANSWER: FRAME WHERE YOUApSKED FOR HELP.RY THE 9 OPTION AT THE QUESTION IN THENEXT FRAME.HAT LETTER WOULD YOU TYPE IF YOU WANTED TO REVIEW THE FRAME JUST BEFORET*HIS ONE?<N1>N <[2>[ <h3>h <u4>u <5> u"ACH QUESTION ALSO HAS THE "ELP OPTIONW/HICH YOU MAY CHOOSE BY TYPING THE L KEY TO REVIEW THE SECTION< > KEY TO RETURN TO THE MENUE. THE SECTION.  KEY TO RETURN TO THE MENU.-SX2P S+ 11X + 4(WWWWS[WW2)cX27` >c+ 15X - 272cX2` c+ 17X - 9=y2sX - 12X + 1vvڄۄiiۄ(4)3*X + 2 X + 5 2X + 7(RgɞX0 + 1-ZXh + 1+X + 1=ס 4FK ALL PROBLEMS BEFORE GOING ON.( 1) 24A X Y3? 2 3316A Y5F 2`(=2X Y2 3A2####(/2)2*/X28, ?/- 5X + 3'3~32)?X27< >?- X - 3=6X0 - 1X= + 144@((@@(S3)6*SX28P ?S- X - 126ECTION DISCUSSES A TECH-NqIQUE FOR FINDING THE INVERSE FUNCTION;H~OWEVER, THIS DEMONSTRATION SECTIONENDS HERE.FHIS IS THE END OFCOMPOSITE AND` NVERSE`UNCTIONS M 3 3 M MjOU MAY USE THE:<> KEY TO REVIEW THE SECTION< > KY-THING OTHER THAN X, K CAN NOT BE THEINVERSE OF F.E HAVE DISCUSSED WHAT HAS TO BE TRUEI#N ORDER FOR A FUNCTION G TO BE THEI0NVERSE FUNCTION OF F. OW GIVEN AF=UNCTION F, HOW DO WE FIND ITS INVERSEFJUNCTION?dHE COMPLETE SHE REQUIREMENTSO0F THE INVERSE FUNCTION. OR EXAMPLE,I=F K(X) = 2X - 1 IS THE INVERSE OF FJ(X) = 2X + 1, THEN (KjF)(X) = X. OESIWT? ERE'S HOW YOU START:(jKjF)(X) = K(F(X)) = K(2X + 1)=Tw2(2X + 1) - 1OW FINISH THE ALGEBRA. F YOU GET ANAVE(WKjF)(X) = K(F(X)) = K(2X + 1)=Td 2(2X + 1) - 1=Tq 4X + 1HEREFORE, KL IS NOT THE INVERSE FUNCTIONOF F BECAUSE (KjF)(X) =/ X.OU DON'T HAVE TO FIND THE INVERSEFUNCTION, JUST TEST THE GIVEN FUNCTIONST#O SEE WHICH ONE MEETS T+ 1 = X - 1 + 1=T XkJkPiY DEFINITION, H IS THE INVERSE OF F.OUR ANSWER:Ka DEFINED BY K(X) = 2X - 1#O. '#F K IS THE INVERSE FUNCTION OF F,T0HEN (KjF)(X) = X AND (FjK)(X) = X.=INCE F IS DEFINED BY 2X + 1, WE HRQRECT.!INCE F IS DEFINED BY F(X) = 2X + 1,(3HjF)(X) = H(F(X)) = H(2X + 1)=TF(b@2X + 1) - 1aCC2M=F m@m2@X2L=F X7H7JJASND(fFjH)(X) = F(H(X)) = FcX` - 1cc2mc=T 2mXu} - 1t2AND (FjG)(X) = X.KINCE F IS DEFINED BY F(X) = 2X + 1,WXE HAVE(#eGjF)(X) = G(F(X)) = G(2X + 1)=bx (pr2X + 1) + 1puu2=b X + 1HEREFORE, G TIS NOT THE INVERSE FUNCTIONOF F BECAUSE (GjF)(X) =/ X. O U ARE CO(X) = X= + 1@@2J<d2> H?d DEFINED BY H(X) = X^ - 1aa2k<3> K? DEFINED BY K(X) = 2X - 1OUR ANSWER: Ga DEFINED BY G(X) = X + 1  21O.1 F G IS THE INVERSE FUNCTION OF F,T>HEN (GjF)(X) = X X bTO F(X), THEN G ҘBRINGSF(X) BACK TO XZ. LSO, IF G CARRIES XTOG(X), THEN F TBRINGS G(X) BACK TO X.HICH ONE OF THE FOLLOWING IS THEINVERSE FUNCTION OF F DEFINED BYF)(X) = 2X + 1?<C1> G?C DEFINED BY G . NVERSE FUNCTIONS CAN BE ILLUSTRATEDPICTORIALLY IN THE FOLLOWING MANNER:1ă*))X*C3<]/Fb2k0JDKJFP(X)G~XRWxW4D9I;C5DXuwG{G l(X)t TO GET SOME HELP INCE F AND G ARE DEFINED BY:F9(X) = X + 2 ANDG9#(X) = X - X + 1,#(8GjF)(X) = G(F(X)) = G(X + 2)=[G X + 2 - (X + 2) + 1?>>=[Z X + 2 - X + 3RRONO__NONmHE DOMAIN OF F IS THE SE HUS, WHEN WORKING WITHTHE COMPOSITE FUNCTION GjF WE MUST RESTRICT X TO THE INTERVAL x2,e). F F AND G ARE DEFINED BYF(X) = X + 2 AND G(X) = X - X + 1,F/IND (GjF)(X) AND DETERMINE THE DOMAINO TO CHECK YOUR WORKLL(L<  } }$%0Ʃ 8(L$0pijlk"ћћliA[8`hH`klmniȅ ӥȅklȥȑȑȑȑȑiȅ` ieȅ q L H ϐ { lhHHH hHL? hυυ Ljݥ ,L mݰ % LߩƉ 慝 憨80 ў*%L L ު `  }LޢٹHH ޥLLޥШh^^h_H r륡HHHHHl^h#d j݄hJhhhhhhE` LJ }d.U"i L=8,LLTҐL {ݩ),(,,ѸLLԠhhL ߅` ȱLL  ߅ ei` 꺽  ' 8 uvL׊i , {$8$0`LԦƹƸ$HH `ީ 8ϐ*IEʼna Lݦ,{iweLi^ e^hٲаg jH hV_F*ƹƸٲаH]"`.?r$6P,a(ŠĠԠŠ)a(̠Š͠ӠĮ)6a(Π))}d?ed@e`M`.?r$]]ƠŬĠ]2o a(ǠԠϠŠĩ)a(Ġìİ)]ŠԠŠq.?6~?r]ð_6]]ϠĠҠΧԠ1L +mLd+rLd+wLd+|Ld+Ld+Ld+Ld+Ld+Ld+Ld+Ld+Ld+Ld+Ld+Ld+Ld+Ld+Ld+Ld81L 81L 81L 8  181L 81L 81L 8*Ld*Ld*Ld*Ld*Ld*Ld+Ld+ Ld+Ld+Ld+Ld+Ld+"Ld+'Ld+,Ld+1Ld+6Ld+;Ld+@Ld+ELd+JLd+OLd+TLd+YLd+^Ld+cLd+hLdLd1cLd1hLd1mLd1rLd1wLd1|Ld1Ld1Ld1Ld1Ld1Ld1Ld*Ld*Ld*Ld*Ld*Ld*Ld*Ld*Ld0Ld1Ld1 Ld1Ld1Ld1Ld1Ld1"Ld1,Ld11Ld16Ld1;Ld1@Ld1ELd1JLd1OLd1TLd1YLd1^EVIEW THE SECTION.  $ KEY TO RETURN TO THE MENU. THETqHE NEXT FRAME.HIS IS THE END OF THE DEMONSTRATIONPORTION OF00 0(00080 @0H0P0`X0`0h0p0`x0 0000000`000000 0000 : : :WOU MAY USE THE:q $q KEY TO R THEINVERSE OF F.#E HAVE DISCUSSED WHAT HAS TO BE TRUEI0N ORDER FOR A FUNCTION G TO BE THEI=NVERSE FUNCTION OF F. OW GIVEN AFJUNCTION F, HOW DO WE FIND ITS INVERSEFWUNCTION?  TECHNIQUE FOR FINDING THEIdNVERSE FUNCTION IS ILLUSTRATED INTION. OR EXAMPLE,I=F K(X) = 2X - 1 IS THE INVERSE OF FJ(X) = 2X + 1, THEN (KjF)(X) = X. OESIWT? ERE'S HOW YOU START:(jKjF)(X) = K(F(X)) = K(2X + 1)=Tw2(2X + 1) - 1OW FINISH THE ALGEBRA. F YOU GET ANY-THING OTHER THAN X, K CAN NOT BEEY TO RETURN TO THE MENUHEREFORE, K IS NOT THE INVERSE FUNCTIONOF F BECAUSE (KjF)(X) =/ X.OU DON'T HAVE TO FI N D THE INVERSEFUNCTION, JUST TEST THE GIVEN FUNCTIONST#O SEE WHICH ONE MEETS THE REQUIREMENTSO0F THE INVERSE FUNCƙ.șI)$Ъ0.ȩ+.8嚪-E/8 i:`  cd pLP袊 +륪 #쩊  UH A驊  hJ I`8;)q4X>Vt~w/z*|cYX ~u q80L>nk'nk(X 1vu8 4L:۠-$șȩ0LW     9ƙ U i 0 i8.ș0șylykyjyi00ڊIi i/ȄȪ) 芢 脤`)F 鄤`ɠ 넬I* L)腞` -+ [..E0 - +f \$8Lf$På8噅 U 9ƚ0`LH$ 9h80 LaH ch 륪EL祚 d$0L e_^^^^ ^`, r^_^^^ %^^`` r` L襝 *` 녞I*L)F``a`Ȫı`E0!` Şȱ`şȱ`Š ȩŬ`(IL륝J8$E ^`e0,iLR襫`I0hhLNL ci ` cP Li v r8坅 Ğğ Ġġ* e24(&&&0⨥塅堅埅垅L@ (LꢅLԥbcdeL.^vvvvvj`^Vy dv88; 54541r LᥝH- 2 f  \7 h < L bcde 饡 饠 饟 饞 LLJ eeededcecbebfbfcfdfefJ`^_^^^^W8Ie Ji `eeeeeLi&&&&8坰IiBfffff`IIIIII `ELԢai0ᦠL LN覸^e^`_a`H` Jh`` g R L楝ɑ 률PQ`PHQH RPhQhPL F犠P` F熅  L熆PE%`dL 饢IEL < LS리8$Ii0ǨV $` lݥ^_ 5^Hȱ^ȱ^h(po Heooph^_`T SRS` H hhhL* ьHH 㥌 hhe^^_ L* IL` ) ʊH񌰸Iš hhhhhhhHH` L ` ^LL g ^ ie^^__Ŕ #^05ȱ^0ȱ^+ȱ^ȱ^poŜ䛐^_e^^__`)Jeiop Ӥȥ敥ȑL䥡HH ` lhhqL 奌  奫  *LݠHȱȱh^_ H^qheqqr "e膮 㦫 R^LԥSR`FHI8eopnm opqrh`M0 䩀hЦstopmnU^_R #ij^_lk 󅔆nmLb^_^ȱ^ȱ^eȱ^e(ӊ0ȱL$8vС,L A  ީ j ީ HHHHH L  ߅Lj A㥋HH jhhȱ𙅄ȱH +륹HHȱHH ghh Lhhhhȑhȑhȑhȑ` jݠ 6hh R䆞`" ȱ ↭^e]eR Ӆmn殤ƕƮ8m因n圑bȱhhћȊћLLȥ e^eʅʥʆd ee`^de *& edeeƙ` 8ompn` gݥ0 ɐ ~LGHHHHH hhhHH ,҄ hh)klnm?Łћȱeȱeעk,5LԢx8` ћLK*L өȥʆȑ $PhihiȑȊ\DN`aFb+cdefg(hiijlm:oapqstvwuxzQ{}<~rx  5(kSS?wP)P +  ?HHSS?wP?PkSSXiP - 2XP - 2=M]5[]X - 10 + X - 2UUpHE DOMAIN  TpOF F wpIS THE SET OF ALL REALN}UMBERS; HOWEVER, THE LAST EQUALITY IM-PLIES THAT (GjF)(X) IS A REAL NUMBER ONLY IF X ~ 2. OF G.HIS TECHNIQUE IS ILLUSTRATED IN THENEXT FRAME. F THE FUNCTIONS F AND G ARE DEFINED BYF(X) = X - 2 AND G(X) = 5X + X,T#HEN WE MAY FIND (GjF)(X) AS FOLLOWS:(6GjF)(X) = G(F(X))=MC G(kFF?wC)C?wCkFFXiC - 2=MPT ANSWER. E STATED EARLIER THAT FOR THE COMPOSITEFUNCTION GjF, THE DOMAIN OF G MUST CON-T&AIN THE RANGE OF F. @F THIS IS NOT TRUE, THEN WE MAY RMESTRICT THE DOMAIN OF GjF TO A SUBSETOZF THE DOMAIN OF F SO THAT F(X) IS INTgHE DOMAIN8s5E C5AREFUL.e5E HAVE F(X) = 1/(1 + X )2ABND G(X) = 3X + 5. HEREFORE,(#UGjF)(X) = G(F(X))=bh G(1/(1 + X ))2e=b 3 1 + X21|+Ç 5HEN YOU SIMPLIFY THIS EXPRESSION YOU'LLBE READY TO SELECT THE CORREC) AND G(X) = 3X + 5, WE HAVE,(#3GjF)(X) = G(F(X))=bE G(1/(1 + X ))2B=b\ 3zY1e + X2bYY1UY+\ 5=b3sw + 5(1 + X )2tp||1 + X2=b8s + 5X2q1x + X2VWWWOUR ANSWER:1b + X2_s 1c + X2a8tOUR ANSWER:1a9b X2p w!+ 30X + 26DHE FUNCTIONS ARE:F8W(X) = 1/(1 + X )2TAWNDG7d(X) = 3X + 5|IDN'T YOU FIND FjG INSTEAD OF GjF?RY AGAIN. ERY`GOO9 DA n INCE F(X) = 1/(1 + X2 NT IN THE FUNCTION F. ET'S TRY ANOTHER ONE. F FUNCTIONS FAND G ARE DEFINED BY F(X) = 1/(1 + X )2A#ND G(X) = 3X + 5, WHICH ONE OF THEF0OLLOWING IS GjF? 9cRX2qO+R 30X + 26bGG1C 8cg + 5X2d`jj1gv + X2+ 1=w 4X2 + 1.kkkkOUR ANSWER: 2X + 10OR0RY, YOU MADE A MISTAKE. E HAVEF@(X) = X28= ?@+ 1 AND G(X) = 2X. HEREFORE,(?ZFjG)(X) = F(G(X))=~g F(2X)OW COMPLETE THE CALCULATION BY USING2X AS A DOMAIN ELEMEREFORE,(?WFjG)(X) = F(G(X))=~d F(2X)~OW COMPLETE THE CALCULATION BY USING2X AS THE DOMAIN ELEMENT FOR THE FUNCTION F.IGHT.8E HAVEF80(X) = X2p- w0+ 1 AND G8=(X) = 2X; TUHEREFORE,(8oFjG)(X) = F(G(X))=w| F(2X)=w (2X)2 FUNCTIONS:F'(X) = X2M$+[' 1G'(X) = 2X.<1N1> (FjG)(X) = 2X2K+N 1<1h2> (FjG)(X) = 4X2e+h 1<13> (FjG)(X) = 2X + 1OUR`ANSWER: 2X2i p+ 10OR0RY, YOU MADE A MISTAKE. E HAVEF=(X) = X28: ?=+ 1 AND G(X) = 2X. HEG#(X) = 2X AND F(X) = X2 +# 1, WE HAVE(=GjF)(X) = G(F(X))=[[ G(b''wAAw__)[Xw[2~X+[ 1)=[o '>'hss2io(X2~l+o 1)=[ 2X2w+ 2.HE DOMAIN OF GjF IS THE SET OF ALLREAL NUMBERS.OW FIND FjG FOR THE SAME TWO VOwOTICE THAT THE DOMAIN OF G wMUST CONTAINTHE RANGE OF F. HAT IS, IF A IS IN AND F(A) DOES NOT BELONG IN THE DOMAINOF G, THEN WE CAN'T FIND G(F(A)). ET'S CONSIDER AN EXAMPLE. F THE FUNCTIONS F AND G ARE DEFINED BY~OMPOSITE FUNCTION OF G BY F. ;; HE IDEA CAN BE REPRESENTED PICTORIALLYIN THE FOLLOWING MANNER.#~##bb:bAHFt<(A)GI(F(A))FD3G7B?2M2p7i8j3q7778@@9? M=?<=`K i! 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OR AN EL8 L? IF PM ST E[ b WITH CENTER AT THE ORIGINTHE EQUATION YISANDTHE GRAPH ISX825m8mY>82E5m>8m+.>=X> 1AG2DB>G2ED?MA>->AB/OORDINATES OFTUHE VERTEX? mmmmm(TU7,2)mTUmmmmo7.+7<&iHICH WAY DOES THE PARABOLA OPEN? mmmm&LiEFTmimmm}HAT ARE THE X-COORDINATES OF THESEPOINTS OF THE GRAPH: (mm,0) AND (mm,4)?-1-1mmmm.!.5.4o-Mm-----------------~-~-~-~-~x-~j-~c-~\-~U-~NnbOOySVY]_begimyoOsOR X = -2Y2M T+ 8Y - 1'HAT IS THEET'QUATION OF THEA4XIS OF SYMMETRY?49 w4mmmmmY~4 = 2m~4mmmm2)*HHAT ARE THE C<o11o> TO CHECK YOUR SKETCH<> TO GET HELP WITH THIS PROBLEMHE GRAPH OF X = -2Y2 + 8Y - 1I S A PA1 R8 A? BF OM LT A[ b OPENING TO THE LE F T H0AVING VE?0RF0TM0ET0X[0 b0AT (7,2).HHoo(d7,2)nbR TO THE LEFT IF ARF0\F O`XM ` HH ?  */ 00>1""'-137=B"'-137=BX  : _ _eOU WILL BE ABLE TO SKETCH THEGpRAPH OF. A PARABOLA WITH AXIS PARALLELT*O THE X-AXIS          \DK`aDbcCdgf=ghi+ijxk<klnNpqSqtivwlxyz}  JF LOG21O(8JX-HJ3QJ) \J= 3, THEN X J- J3 J= J23G=J8JOdR X = 11.MgM[5[5gMg#OUR ANSWER: 6=O. MF LOG2*R(1MX-AM3JM) UM= `M3, THEN X M- 3M =M 2M3J=M8M.]OW WHAT IS X?wOLVE FOR X MwANDWHAT IS X?#8J1d16~#OUR ANSWER:8b#=OU DIDN'T COMPLETE THE PROBLEM. MF LMOG2(R(/MX-3) =UM 3`M, TrMHEN X -M 3M =M 2M3J=M8M.]OW WHAT IS X?wHOOSE ANOTHER ANSWER.0OR0REC.0T70.UATIONSI#NVOLVING LOGARITHMS.=OR EXAMPLE:JOLVE THE EQUATION LOG3O(JX-1)=J2J.dOLUTION:qINCE LOG3Tv([qX-1)=q2q, X-q1q=q3q2nO~R X-4~1>~=H~9R~.HEREFORE, THE SOLUTION IS X=Θ1ؘ0. ##F LOG2M((T#X-3)=z#3#, IT AGAIN.#OUR ANSWER:9b#=O. F LOG3GBUP==\= -2, THEN U = 3-:2.=WHAT IS 3-?T2?OW qOW TRY IT AGAIN.0IGH0T! JF LOG3*OU3J = 2, THENUJ = 3-G2 J=J 1CmC9P.I:TT::E CAN ALSO SOLVE CERTAIN EQV[] SUCH THAT 3VZ ]= 81.OW SELECT THE CORRECT ANSWER. ##F LOG3M(UW# = -2, WHAT IS U?# -=69W1lml9y#OUR ANSWER:-b#6=O. F LOG3GBUP==\= -2, THEN U = 3-:2.=WHAT IS 3-?T2?OW qOW TRY G3~R8M1 =/M M3. M OU WANTT]HE EXPONENT V[] SUCH THAT 3VZ ]= 81.OW SELECT THE CORRECT ANSWER.#OUR ANSWER:5b#=E = C=A"=REFUL.MINCE 332J 9M=/@M GM81, bMLOG3~R8M1 =/M M5. M OU WANTT]HE EXPONENT gHENCzHOOSE A DIFFERENT ANSWER.M#HAT IS LOG3(8#1?4J3d5~#IGH#T!=INCE 343: :== 81, LOG3B8=1 = 4.D2g2gDD#OUR ANSWER:3b#=E = C=A"=REFUL.MINCE 332J 9M=/@M GM81, bMLO!JINCE 522G 9J= 25, LOG5O2J5 = 2AQeQeAA JBY DEFINITION.OUR ANSWER:LbOG2w2~5 = 50OU DIDN'T USE THE DEFINITION PROPERLY.V#K=-KL7KOGALPUUK IF AND ONLY IF AVH=KUK gDENTIFY A, U, AND V IN 52d g= 25;T3>LcOG2x25 = 5#OUR ANSWER:Lb#OG2w(5~# = 25=OU DIDN'T USE THE DEFINITION PROPERLY.V#T=-TL7TOGALYUUT IF AND ONLY IF AVQ=TUT pDENTIFY A,UUp,\p AND V IN 52m=p2p5; THENCHOOSE A DIFFERENT ANSWER.0I 0GH0TATION LOG4b1]6 = 2 IS CALLEDLpOGARITHMIC FORM AND 42m p= 16 ISCALLED EXPONENTIAL FORM.#HICH ONE OF THE FOLLOWING IS THEL0OGARITHMICFr0ORM EQUIVALENT TO 5@2==*@ 25? LcZOG2x_5Z = 25LctOG5xy2t5 = 2 (2X + Y )(2X - Y )2Z2Z<1t229t> (4X + Y)(X - Y )3q<1339> 4X - Y IS PRIME2c4<1449> (2X - Y )(2X - Y )22şO OD!5HE POLYNOMIAL 4X - Y2w242 5IS THEDBIFFEREcII)(1cX + Y) = X + 2XY + Y2b`2`2`(yIVIyV)(1yX - Y) = X - 2XY + Y2bv2v2v(VV)(1X + Y) = X + 3X Y + 3XY + Y3b322 (VIVI)(1X - Y) = X - 3X Y + 3XY - Y3b322 OW LET'S TRY ONE.#HAT IS THE D BECOME FAMILIAR WITH THESwPECIAL PRODUCTS AS YOU STUDY THE EX-AMPLES OF FACTORING GIVEN IN THE TEXT.# PECIAL RODUCTS - ART 1(*II*)(1*X + Y)(X - Y) = X - Y2'2'(@III@I)(1@AX + B)(CX + D) =ApMCX + (AD + BC)X + BD2J(cIIIID THEREFORE, X2p~ w+ X - 2 IS NŃOOԁTT܁ PRIME.xfxʅ )N FACTORING A GREAT MANY POLYNOMIALS,W6E USE THE SPECIAL PRODUCTS GIVEN INTCHE NEXT FRAME. SE THEM AS MMCOOCDDCEECLLCSSC FORYPOUR FACTORING PROBLEMS.jOU SHOUL BE WRITTEN AS A PRODUCT OF TWOP#OLYNOMIALS OF POSITIVE DEGREE, USINGT0HE RULE IN THE PREVIOUS FRAME CONCERN-I=NG THE COEFFICIENTS.ZINCE X - 1 AND X + 2 ARE POLYNOMIALSOgF DEGREE 1, (X - 1)(X + 2) IS THEFtACTORIZATION`OF X + X - 2;2qANS, THEN THEFJACTORS SHOULD HAVE RATIONAL COEFFI-CWIENTS. HAT IS:*xHE FACTORS SHOULD HAVE COEFFI-C*IENTS FROM THE SAME SUBSET OFR*EAL NUMBERS AS THE POLYNOMIAL.xh hhE WILL CALL A POLYNOMIAL P RIME ` IF ITCANNOTOLYNOMIAL ANDTHE FACTORS HAVE COEFFICIENTS THAT AREINTEGERS.ENERALLY, WE WILL USE THE RULE THAT IFTHE POLYNOMIAL HAS INTEGRAL COEFFI-C#IENTS, THEN THE FACTORS SHOULD HAVEI0NTEGRAL COEFFICIENTS. F THE POLYNO-M=IAL HAS RATIONAL COEFFICIENTFACTOR''  ''O0F THE ORIGINAL POLYNOMIAL.BOR EXAMPLE,X1O - 1 AND X + 2 ARE FACTORS OFX\2Y \+ X - 2 BECAUSEX1o28l ?o+ X - 2 = (X - 1)(X + 2)SSFFSSbbb[abSb[`cOTE THAT BOTH THE GIVEN P`?d& R####d*d*@  :SOU WILL BE ABLE TO:.`F#`ACTOR CERTAIN TYPES OF POLYNOMIALSdJ J d F A POLYNOMIAL IS WRITTEN AS A PRODUCTOF OTHER POLYNOMIALS, THEN EACH OF THEL#ATTER POLYNOMIALS IS CALLED A               \DN``b?ceCfhghijlMmnoqXr*sttCtu8w#wxRyz{! (X2b> iA- Y2>)A(X2> A+ Y2>)A=FQ (X - Y)(X + Y)(X2N Q+ Y2N)Q.hHUS,:grgggg 8s(X - Y)2pp ws X2p s+ Y2pX?c4F` Mc- Y4i`Yd2a= (#X-4Y>)(X+]Yg)(X2 + Y2)Y2(#X - Y)(X - Y)(a= (#X-4Y>)(X+]Yg)(X2 + Y2)Y2(#X - Y)(X - Y)(X2 + Y2)OW, SIMPLIFY THE FRACTION. OUR ANSWER: Y2i #OR#RY, 1#BUT YOU MADE A MISTAKE IN SIM-P0LIFYING. OTICE THATXA4> A- Y48> ?A= 1#BUT YOU MADE A MISTAKE IN SIM-P0LIFYING. OTICE THATXA4> A- Y48> ?A= (X2b> iA- Y2>)A(X2> A+ Y2>)A=FQ (X - Y)(X + Y)(X2N Q+ Y2N)Q.hHUS,:grgggg 8s(X - Y)2pp ws X2p s+ Y2pX?c4F` Mc- Y4i`Yd2 SIMPLIFY:O5 5A5.V6]55( AX - Y)2>> EA X2h> oA+ Y2>X14. 1- Y49.Yq22x/ (X - Y)Y2^ Y2u 8 <3> k s}({X+Y)Y2(X-Y) OUR ANSWER: (X - Y)Y2 #OR#RY,(\X + 2)(2X - 3X2Y \__(j2X - 3)(X - 4)2gTwHEN FACTOR:=i~ (w2X - 3)(X - 2)(X + 2)x{{?x?x(xX + 2)(2X - 3)XAND CANCEL COMMON FACTORS:=iy????y~qyqy~ XX~ - 2 ULTIPLY AND - 42Lx2N'X - 32'X - 3X2$.$4IRST, MULTIPLY:XA + 2 X - 42[> C9C@CGCNCCD@DF2NX - 3 2ONX - 3X2]K=F XA + 2 2AX - 3X2>.DCCCCC2NX - 3 ?N?NXN - 42K=aM55A\NDAsC8sms m mBD8.,{.,s=HyAVsDrsmVsmrsBWCs.ewAsDmsmBC=y(yB, C, D =/y 0).jjj NVERT ANDMULTIPLY.dAEICB}~ ET'S WORK THROUGH AN EXAMPLE.IMPLIFYXT + 2 X ON FACTOR IN THE NUMERATOR AND DENOMINATOR.>]]>> N ORDER TO MULTIPLY OR DIVIDE TWORATIONAL INTEGRAL EXPRESSIONS, WE USET#HE FOLLOWING RULES ESTABLISHED EARLIER:A< C ACm< m mmBI D BD.+@=FB(wBB, D =/B 0)5MTHE NUMERATOR. ET'S SEE IF WE CAN HELP. N ORDER TO SIMPLIFY THE FRACTION, YOUM)UST FACTOR THE NUMERATOR AND THED6ENOMINATOR:2IX2*F 1I- X - 6 (2X + 3)(X - 2)LmLLL=xOXX2#U-1X 7X + 10?X?X(XX - 5)(X - 2)OTE THE COMMm - 5kVkqqVkV ET'S MOVE ON TO MULTIPLICATION ANDDIVISION OF FRACTIONAL EXPRESSIONS.IMPLIFY:2JX2X _- X - 6HXI&2P# W&- 7X + 10SOUR ANSWER:XpM - 2nQQXpZ - 5ORRY, BUT YOU MADE A MISTAKE INFACTORING SE+QUAL TOF((2H$X - X - 62V!XF52M2 T5- 7X + 10:+ 2X + 3Xpi - 5i__ X - 2Xi - 5gIGHT!2=X2#: *=- X - 6 (2X + 3)(X - 2)@h@@@=sCXL2I #L- 7X + 10 (X - 5)(X - 2)=sf cc2`X + 3XAPND CANCELING:!*0;2]X + YXj + Ybb=weF COURSE, WE MUST ASSUME X - Y =/ 0.E SHALL ALWAYS TACITLY ASSUME SUCHRESTRICTIONS WHEN SIMPLIFYING RATIONALINTEGRAL ALGEBRAIC EXPRESSIONS.ETERMINE WHICH OF THE FOLLOWING IXPRESSIONS BY FACTORING THENUMERATOR AND DENOMINATOR AND CANCEL-ING ANY COMMON FACTORS.OR EXAMPLE, WE SIMPLIFY THIS FRACTIONBY FACTORING:2)X - XY - Y2#&2i&-n-X*9 - Y2162T6=w0 ?)--(9X + Y)(X - Y)?)()2X + Y)(X - Y)AVE A COMMON DENOMIN-A#TOR> > RATIONAL INTEGRAL ALGEBRAICE#XPRESSION IS A QUOTIENT OF TWO POLY-N0OMIALS. INCEAEJD AnMgMRMDMBFWD=[PBiW(PD =/P 0),;[;AA[;[WqE CAN SIMPLIFY RATIONAL INTEGRAL AL-G~EBRAIC E`1  ` i` 1!1!!7  :DOU WILL BE ABLE TO:.QS#QIMPLIFY FRACTIONAL EXPRESSIONS.^M#^ULTIPLY AND DIVIDE FRACTIONALE#kXPRESSIONS.xA#xDD AND SUBTRACT FRACTIONAL EXPRES-S#IONS WHICH H"" " " " " """""""""