Addition and subtraction of rational expressions


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1 Lecture 5. Addition nd subtrction of rtionl expressions Two rtionl expressions in generl hve different denomintors, therefore if you wnt to dd or subtrct them you need to equte the denomintors first. The common denomintor with the smllest possible degree is the Lest Common Multiple of the originl ones. On this lecture we will consider the procedure of finding the Lest Common Multiple nd solve some exmples on ddition nd subtrction of rtionl expressions Addition nd subtrction of rtionl expressions The rules for dding nd subtrcting rtionl expressions re the sme s rules for dding nd subtrcting frctions. b + c b + c, b b c b c. b In this exmple both the denomintors re equl nd re not zero. Exmple 5.1. (3.4 ex.13 ) (3.4 ex.16 ) x 2 x x2 + 1 x x2 x 2 1 x x x + 5 x 4 2x x 4x + 5 x 4 2x + 4 (x 4) 4x + 5 x 4 + 2x + 4 x 4 6x + 9 x 4. If denomintors of two rtionl expressions re not equl, we cn use following generl rule for dding nd subtrcting quotients Let us prove this formul b + c d d + bc, b 0, d 0. bd b + c d b d d + c d b b d bd + cb db d + cb. bd
2 Lecture 5. Addition nd subtrction of rtionl expressions 23 Similrly, b c d d bc. bd Exmple 5.2. (3.4 ex.22 ) x x x + 1 x2 (x + 1) + 3x 3(x + 1) x3 + x 2 + 3x 3(x + 1) x (x2 + x + 3 3(x + 1). (3.4 ex.36 ) 2x 1 x 1 2x + 1 (2x 1)(x + 1) (x 1)(2x + 1) x + 1 (x 1)(x + 1) 2x2 x + 2x 1 2x 2 + 2x x + 1 (x 1)(x + 1) 2x (x 1)(x + 1). (3.4 ex.41 ) x (x 2) + x + 1 x 1 + x + 1 x 2 x(x 1)(x 2) + (x 2)(x 2)(x 1) + (x + 1)(x + 1)(x 1) (x + 1)(x 1) x(x2 3x + 3) + (x + 1)(x 2 4x + 4) + (x 2 + 2x + 1)(x 1) (x + 1)(x 1)(x 2) x3 3x 2 + 3x + x 3 4x 2 + 4x + x 2 4x x 3 + 2x 2 + x x 2 2x 1 (x + 1)(x 1)(x 2) 3x3 5x 2 + 2x + 3 (x + 1)(x 1)(x 2) (x 1)x(3x 2) + 3 (x + 1)(x 1)(x 2). One cn check tht there is no extr cnceling by trying long division of the numertor by the fctors (x + 1), (x 1), nd (x 2) of the denomintor Lest common multiple The generl formul for summing up two rtionl expressions will work in every sitution. But not in every sitution the generl formul is the best. Let us consider following exmple Exmple 5.3. (3.4 ex.69 ) x + 4 x 2 x 2 2x + 3 x 2 + 2x 8 generl formul (x + 4)(x2 + 2x 8) (x 2 x 2)(2x + 3). (x 2 x 2)(x 2 + 2x 8)
3 Lecture 5. Addition nd subtrction of rtionl expressions 24 A little bit smrter wy to do it is to fctor the denomintors, first: (x + 4) (x 2)(x + 1) (2x + 3) (x + 4)(x 2). Then, obviously, we cn equte the denomintors by multiplying the first rtionl expression by nd the second one by (x + 4) (x + 1) (x + 4) (x + 1) : (x + 4)(x + 4) (2x + 3)(x + 1) (x 2)(x + 1)(x + 4) x2 + 8x x 2 3x 2x 3 (x 2)(x + 1)(x + 4) x 2 + 3x 13 (x 2)(x + 1)(x + 4). Let us summrize wht we did. We wnted to equte the denomintors nd we wnted to keep our work simple. The work will be kept simple if common denomintor hs the smllest possible degree. In this exmple polynomil (x 2)(x+1)(x+4) is the one nd it is clled the Lest Common Multiple of polynomils (9x 2)(x 1) nd (x 2)(x + 4). Definition 5.1. The polynomil P (x) is the lest common multiple of polynomils P 1 (x), P n (x) if ) ech of p i is fctor of P (x); b) there is no polynomil q(x) with degree less thn degree of P (x) such tht ) is stisfied. Remrk 5.1. Prt b) sys tht P (x) hs the lest possible degree for property ). Question 5.1. Let P (x) be the lest common multiple of polynomils P 1 (X), P 2 (X), P 3 (x). ) Consider P (x) P 1 (x), will P 1(x)  cncel? b) Consider P (x) P 2 (x), will P 2(x)  cncel? c) Consider P (x) P 1 (x)p 2 (x), will P 1(x)P 2 (x)  cncel? Let us discuss the process of finding the lest common multiple of two or more polynomils. The first step is to fctor ech polynomil completely. Then to construct the lest common multiple we successively combine prime fctors of the originl polynomils. For ech originl polynomil we dd only those of it s fctor which re missing in the
4 Lecture 5. Addition nd subtrction of rtionl expressions 25 lest common multiple. In prticulr, let polynomils M(x), N(x), P (x) hve fctoring, correspondingly M(x) (x)b(x)c(x), N(x) b(x)c(x)d(x), P (x) b(x)d(x)e(x), then the lest common multiple for M (x), N (x), nd P (x) will be LCM (x)b(x)c(x)d(x)b(x)e(x). Here initilly we dded to LCM ll fctors from M(x) since LCM should hve ll fctors of M(x) nd hd nothing initilly. On the second step we dded d(x) from N(x) since other two multipliers lredy were LCM. On the third step we dded e(x) nd b(x) from P (x). Although LCM lredy hd one b(x), still P (x) hd two of them so one ws missing, thus, we dded it. Exmple 5.4. (3.4 ex.51 ) Find the LCM to x 3 x, x 3 2x 2 + x, nd x 3 1. First, we will fctor three polynomils The LCM is then given by (3.4 ex.72 ) x 3 x x(x + 1)(x 1), x 3 2x 2 + x x(x 1)(x 1), x 3 1 (x 1)(x 2 + x + 1). LCM x(x + 1)(x 1)(x 1)(x 2 + x + 1). x (x 1) x x + 1 x 3 x 2 x (x 1) x x + 1 x 2 (x 1) x x2 + 2 x (x 1) 2 (x + 1)(x 1) (x 1) 2 x 2 x 3 + 2x 3 4x 2 + 2x x x3 5x 2 + 2x + 1 (x 1) 2 x 2. By doing long division of the numertor over the (x 1) one cn see tht (x 1) does not cncels.
5 Lecture 6. Mixed quotients When the numertor or the denomintor of quotient contins combintions of rtionl functions we cll it mixed quotient. On this lecture we will consider how to simplify mixed quotients Mixed quotients Definition 6.1. When sums nd/or differences of rtionl expressions pper s the numertor nd/or denomintor of quotient, the quotient is clled mixed quotient. To simplify mixed quotient mens to write it s rtionl expression reduced to lowest terms. Exmple of Mixed Quotient b + c d e f + k, where, b, c, d, e, f, k, l re polynomils. l There re two fvorite methods for simplifying mixed quotients. The first methods is to simplify mixed quotient step by step by considering rtionl expressions is the numertor, then in the denomintor nd, finlly, the whole thing. In the second pproch you will need to compute the lest common multiplied of the denomintors of ll rtionl expressions entering the mixed quotient. Then the mixed quotient is simplified in one step by multiplying both numertor nd denomintor with the lest common multiple. We will consider some exmples for ech of two methods Simplifying by prts Method Summry: Consider numertor, simplify; consider denomintor, simplify, etc. b + c d e f + k l d + bc bd el + kf fl (d + bc) fl (el + fk) bd
6 Lecture 6. Mixed quotients 27 Exmple 6.1. (3.5 ex.1 ) x x x + 1 x x + 4 x + 1 x (x + 4) (x + 1)(x + 4) 2 x Simplifying by one strike! Method Summry: Tke LCM of ll little denomintors (b d f e) nd multiply both the numertor nd the denomintor of the mixed quotient by LCM. b + c d e f + k l LCM LCM b LCM + c d LCM e f LCM + k. e LCM Exmple 6.2. (3.5 ex.16 ) 1 x x x 1 x LCM (x + 1)x, Multiplying both the numertor nd the denomintor by LCM 1 x x x 1 x (3.5 ex.22 ) (x + 1)x (x + 1)x (x + 1)x x 2 2(x + 1)x 1(x 1)(x + 1) x 2 + x x 2 2x 2 + 2x x (2x + 5) x x x 3 x 2 (x + 1)2 x 3 x + 3 (2x + 5)(x 3) x 2 x(x 3) x 2 (x + 3) (x + 1) 2 x 3 (x 3)(x + 3) ((2x + 5)(x 3) x2 )(x 3)(x + 3) x(x 3)(x 2 (x + 3) (x + 1) 2 (x 3)) x (x + 1) 2. (2x2 + 5x 6x 15 x 2 )(x + 3) x(x 3 + 3x 2 ) (x 2 + 2x + 1)(x 3) (x 2 x 15)(x + 3) x(x 3 + 3x 2 x 3 2x 2 x + 3x 2 + 6x + 3) (x2 x 15)(x + 3). x(4x 2 + 5x + 3)
7 Lecture 6. Mixed quotients 28 (3.5 ex.27 ) x 1 1 x 1 x 1 x x 1 x 1 x x 1 1 x x. Solving the sme exmple by the Method 2 looks little more strightforwrd x x x x 1 x x 1 x 1 x x x.
8 Lecture 7. Negtive exponents. Scientific nottion. Squre root On this lecture we will discuss the negtive exponents. Properties of negtive exponent turn out to be the sme s tht of usul positive exponent. Therefore, both negtive nd positive exponents represent the sme object. We will lso consider the scientific nottions nd how to convert numbers into scientific nottion nd will discuss properties of squre root. In the end of the lecture we will discuss rtionlizing of mthemticl expressions Negtive exponents Let us strt from some simple remrks. If n is counting number then rised to the power n is defined by n... }{{} n fctors. The number n is lso cn be referred s the exponents of. Let us derive simple properties for the exponents n+m n m {}}{ } {{... } } {{... } n+m ; n m (b) n (b)(b)... (b) }{{} n b n ; n times ( ) n ( ) ( ) ( )... n b b b b b ; n n m {}}{ ( n ) m (... ) (... )... (... ) }{{}}{{}}{{} n m ; n n n }{{} n m n {}}{ }{{} m m n m, m < n.
9 Lecture 7. Negtive exponents. Scientific nottion. Squre root 30 Let us tke better look t the lst formul: m < n is n obvious the limittion for it. The formul is not working for m n simply becuse it is undefined for tht cses. From the other side wht if we could extend existing definitions to new cses? If this is possible, we will improve the performnce of our old formul. The first bd cse is m n for which 0 is undefined. Obviously, the cse m n cn be cptured by defining 0 1. You cn check tht the whole set of five formuls will still fit together. Similrly, to mke the lst formul meningful for m > n we need to define 1 1, 2 1 2, 3 1 3, etc. Eventully, we rrive to the definition of the negtive exponent n 1 n. It is esy do check tht properties of exponents re the sme for negtive exponents nd we cn improve the lst formul: m n m n m+( n) m n for ny numbers m nd n. Exmple 7.1. (4.1 ex.11 ) (4.1 ex.13 ) As n exercise, let us derive one more property of the exponents (4.1 ex.44 ) ( b ) n 1 ( b ) n 1 n b n b n ( b ) n n 4x 2 (yz) 1 ( 5) 2 x 4 y 2 z 2 4 ( 5) 2 x 2 y 1 z 1 x 4 y 2 z 2 4( 5) 2 x 6 y 3 x 1 4z 25x 6 y 3. (4.1 ex.52 ) (3xy 1 ) 2 (2x 1 y) 3 32 x 2 y x 3 y 3 32 x 2 x y 3 y 2 32 x y 5.
10 Lecture 7. Negtive exponents. Scientific nottion. Squre root Scientific Nottion Numbers you meet in pplictions my very from very lrge to very smll. As n exmple let us tke or these numbers re tedious to write nd difficult to red. Therefore people use exponents to mke better representtion of such numbers. Recll if you hve some number in decimls. the multipliction by 10 will move the point to the right nd division by 10 will move comm to the left. Therefore,. 10 (. 1 ) (. 10) Definition 7.1. When number hs been written s product of it is sid to be written in scientific nottion. 10 S, 1 10 Exmple 7.2. (4.1 ex.69 ) Write in scientific nottion: (4.1 ex.81 ) Write s deciml Question 7.1. Try to put the following number in ny redble form: Squre roots A rel number is sid to be squred when it is rised to the power 2. Tking squre root is the opertion inverse to squring. Definition 7.2. The squre root of number is number b such tht b 2
11 Lecture 7. Negtive exponents. Scientific nottion. Squre root 32 Question 7.2. If b is squre root of, wht could be sid bout b? How mny roots hs if is negtive? Let us list some properties of squre root if > 0 there exist two squre roots; if 0 the squre root of 0 is 0; if < 0 there re no squre roots. The positive squre root is clled the Principl squre root nd the symbol clled rdicl sign is used to denote the principle (or nonnegtive) squre root. So, > 0; nother squre root is given by. Definition 7.3. In some cses n expression cn be presented s squre of something not ended with the rdicl sign. Such expression re clled perfect squres. Exmples: ( + x) 2, ( x + ) 2 re perfect squres, ( 3) 2 is not perfect squre. Remrk 7.1. Tking squre root or perfect squres is esy nd plesnt. Note nevertheless, tht some cre still should be tken. You should keep in mind tht 2. As n exercise, try to prove why bsolute vlue is need to be putted. Exmple 7.3. (4.2 ex.6 ) Evlute Some more properties of squre roots b b, b b to prove the bove identities we need to recll the definition of squre root. The definition sys tht b ( ) 2 b if nd only if b b. Let us check it ( b ) 2 ( ) 2 ( b ) 2 b. Exmple 7.4. (4.2 ex.16 )
12 Lecture 7. Negtive exponents. Scientific nottion. Squre root 33 Next we will consider some exmples on sums nd differences of squre roots. The ide is to combine the like terms. (4.2. ex.27 ) ( ) 3( ) 3(19 3 2). (4.2. ex.36 ) (3 2)(3 + 2) 3 2 ( 2) Rtionlizing There re severl conventions in mthemtics describing how to write mthemticl expressions. In prticulr, there is rule sying tht the denomintor of quotient should not contin rdicls. The process of removing the rdicls from the denomintor is clled rtionlizing. Exmple 7.5. Rtionlize the following expressions. (4.2. ex.43 ) ; (4.2. ex.56 ) (4.2. ex.62 ) 1 x + h 1 x h Solution 2: x + h x h ( 5 3) x x + h x + h x h h( x x + h) x + h x h(x x + h (x + h) x) (x + h)x ( x + h x ) h x + h ; x 1 x + h h 5 3; h x + h x( x x + h)h (x + h)x h (x x + h (x + h) x) (x + h)x 1 x h h h ( x + h h x + h x (x + h) Let us discuss the properties of the nth root. It will tke some dvnced nlysis to x x ).
13 Lecture 7. Negtive exponents. Scientific nottion. Squre root 34 First, let us summrize some fcts bout the exponents > 0, if b > 0 nd n is even b n > 0, if b < 0 nd n is even is > 0, if b > 0 nd n is odd < 0, if b < 0 nd n is odd
14 Lecture 8. Rdicls On this lecture we will define root of degree n nd discuss it s properties. In the end we will dd some review exmples on squre roots since techniques for working with nth roots nd with squre roots re very similr Rdicls Definition 8.1. An nth root of number is number b such tht b n. n b b n. Note, tht the squre root is specil cse of this definition. Let us discuss the properties of the nth root. By doing some not very dvnced nlysis one cn prove tht there exists t lest one nth root of if > 0. Here we will ccept this fct without proof nd will summrize other informtion which is vilble for rdicls. First, if the number n is even then with ny root b there will be root b since ( b) n (b) n, if n is even. Second, if the number n is odd then if is negtive then its nth root will be negtive nd if is positive then its nth root will be positive. We summrize this informtion in the next tble two, > 0, < 0 if > 0 nd n is even none, if < 0 nd n is even the number of nth roots of is exctly one, > 0 if > 0 nd n is odd one, < 0 if < 0 nd n is odd Definition 8.2. The principl nth root of number denoted by the rdicl sign is n {positive nth root if n is even, nth root if n is odd.
15 Lecture 8. Rdicls 36 Exmple 8.1. (4.3 ex.1 ) (4.3 ex.2 ) (4.3 ex.4 ) (4.3 ex.10 ) x (x 2 ) 3 3 (3x 2 ) 3 3x Properties of rdicls The first property follows from the definition of rdicls nd the definition of the principl root { n if n is odd n if n is even Exmple ( 3)3 3, 2 ( 2)2 2 2 Another properties follow from the definition of rdicls nd properties of exponents n b n n b, n n b n, b ( ) n m n m, m n mn. Remrk 8.1. Note tht these identities re only welldefined when both nd b re positive. For exmple, the first one is meningless for < 0, b < 0, nd n even. Question 8.1. How to prove these identities? Let us prove the lst one. When we rise both sides to degree nm. Using the properties of integer exponents nd the definition of rdicls we rewrite the left s ( ) m n m n (( ) m n m ) n ( ) n n. At the sme time, ccording to the definition of the nth root the right side is ( nm ) mn. The identity is proved.
16 Lecture 8. Rdicls 37 Exmple 8.3. (4.3 ex.27 ) 3x 2 12x 3 x x x 6 x x (4.3 ex.31 ) (4.3 ex.30 ) 4 9x 2 y 4 3 x2 y 3 125x 3 3 8x3 y x 2 y x y x 5 y 23 x 3 y 5x 3 x 2 3 y 4 3 2xy 3 y 5 3 x 2 2y. (4.3 ex.47 ) 8x3 3 50x + 2x 5 2 x 2x 3 5 2x x 2 2x (2 x 15 x 2 ) 2x. (4.3 ex.51 ) (4.3 ex.54 ) (3 3 6)(2 3 9) 3( 3 3 2)2( 3 3 3) ( 4 4 2)( ) ( 4 4) 2 (2) Review exmples Rtionlize the denomintor in ech expression (4.2 ex.55 ) ( 3 + 2) (4.2 ex.57 ) ( 3 2)( ) ( 3 + 2) Rtionlize the numertor: (4.2 ex.59 ) (2 5)(2 + 5) (3 2 5)(2 + 5) Solve eqution x + 3 2x 5 0, x + 3 2x 5, x + 3 2x 5, x 8.
17 Lecture 9. Rtionl exponents On this lecture we will give the definition of rtionl exponent nd will discuss its properties which re, in fct, similr to the properties of integer exponent Rtionl exponents Let us compre two identifies, one for exponents nd one for rdicls ( m ) n mn m n mn Since the similrity is obvious it give us hope tht both the nturl exponents nd rdicls represent the sme but more generl object. The reltion is even more obvious if we consider the following identity for rdicls ( n ) n ( n 1 n) (n 1n) ( ) n n Finlly, the gp between integer exponents nd rdicls will be eliminted if we define 1 n n, or, more generl, for ny rtionl number m n, n > 0 define m n ( n ) m Exmple 9.1. (4.4 ex.6 ) (64) 3 2 (4 16) 3 2 (2 6 ) (2 6 ) (4.4 ex.8 ) (25) 5 2 (5 2 ) (5 2 ) 5 1 (52 )
18 Lecture 9. Rtionl exponents 39 Let us list bsic identifies for the rtionl exponents. These identifies will combine together properties of integer exponents nd rdicls. r s r+s, ( r ) s rs, (b) r r b r, ( b ) r r b r, r 1 r, r s r s. As n exercise let us check the second one r s m n p ( ) q n m ( ) q p ( n ( q ) m ( ) q q ( n ) p ) n ( ) n q qm ( ) q n pn ( nq ) qm ( nq ) pn ( nq ) qm+pn qm+pn nq r+s. Agin, the identifies re welldefined only for positive nd b. When nd b re negtive some identifies will fil. Exmple 9.2. (4.4 ex.32 ) (4.4 ex.16 ) (4.4 ex.28 ) ( x 3 2 y 5 2 ) 4 3 ( x 1 3 y 1 3 ( 1 9 ) 3 ( 3 x y ) 1.5 ( 1 9)15 10 ( ) 4x 1 y )( 1 x 3 ( 3) ) x 2 x 1 y 1 3 ( 3) y 10 3 y 1 1 x 3 y 7 3 ( 1 9) (x 1 ) (1 9 ( ) y ) 3 ( 1 3 ) x 3 2 y y x 3 1 x 3 3 y 7. Question 9.1. Wht is equl to? Answer: (x 2 ) 3 2 (x 2 ) 3 2 x 3
19 Lecture 10. Geometry topics On this lecture we will consider some re formuls, prove the Pythgoren theorem, nd stte some extr formuls which will be used lter Are formuls For rectngle of length l nd width w re is defined to be Are lw. For the right tringle with the bse l nd ltitude the re ccording to the picture will be Are 1 2 lh. For n rbitrry tringle: Are 1 2 l 1h l 2h 1 2 lh. Let us use re formuls to prove the Pythgoren Theorem The Pythgoren theorem
20 Lecture 10. Geometry topics 41 Definition A right tringle is one tht contins right ngle  tht is, n ngle of 90. The side of the tringle opposite to 90 ngle is clled the hypotenuse; the remining two sides re clled legs. Theorem Pythgoren Theorem. In right tringle, the squre of the length of the hypotenuse is equl to the sum of the squres of the lengths of the legs. c b 2 Proof. The side of the big squre is + b. Note tht ll four tringles re right nd congruent. The inside figure is squre with the side c. Obviously, the re of the big squre equls to the sum of res of tringle nd the smll squre. the Pythgoren Theorem is proved. ( 1 ( + b) 2 4 b + c 2) b + b 2 2b + c b 2 c 2 Exmple (4.6 ex.34 ) The re of squre ABCD is S ABCD 100 squre feet, the re of squre BEF G is S BEF G 16 squre feet Wht is the re of the tringle CGF. Solution. Since S ABCD 100 then side BC 10. Since S BEF G 16 then side BG GF 4, therefore GC BC BG 6 nd S CGF (4.6 ex.8 ) Check if the tringle with the sides below is right one: Tke 6, b 8, nd c 10, then 6, 8, b c 2.
21 Lecture 10. Geometry topics 42 Answer: the tringle is right. (4.6 ex.14 ) Check if the tringle with the sides below is right one: Tke 5, b 4, nd c 7, then Answer: the tringle is not right. 5, 4, b c 2. Question Why the hypotenuse is lwys the longest side of the tringle. Prove using Pythgoren theorem Some more geometricl formuls Volume of rectngulr box of length l, width w, nd height h: Volume lwh For the circle with rdius r (dimeter d 2r) Are πr 2, Circumference 2πr πd Perimeter of rectngulr of length l nd width w is given by Perimeter 2l + 2w. Exmple (4.6 ex.38 ) How fr person cn see. R is the rdius of the Erth. h is the height of tower. d is the distnce the person cn see.
22 Lecture 10. Geometry topics 43 According to Pythgoren theorem d 2 + R 2 (R + h) 2 or d (R + h) 2 R 2.
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