Lecture 5.Addition and subtraction of rational expressionsTwo rational expressions in general have different denominators, therefore if you want toadd or subtract them you need to equate the denominators first. The common denominatorwith the smallest possible degree is the Least Common Multiple of the original ones. On thislecture we will consider the procedure of finding the Least Common Multiple and solve someexamples on addition and subtraction of rational expressions.5.1. Addition and subtraction of rational expressionsThe rules for adding and subtracting rational expressions are the same as rules foradding and subtracting fractions.ab+cb=a + cb,ab−cb=a − cb.In this example both the denominators are equal and are not zero.Example 5.1. (3.4 ex.13 )x2x2+ 4−x2+ 1x2+ 4=x2− x2− 1x2+ 4=1x2+ 4.(3.4 ex.16 )4x + 5x − 4−2x + 44 − x=4x + 5x − 4−2x + 4−(x − 4)=4x + 5x − 4+2x + 4x − 4=6x + 9x − 4.If denominators of two rational expressions are not equal, we can use following generalrule for adding and subtracting quotientsab+cd=ad + bcbd, b 6= 0, d 6= 0.Let us prove this formulaab+cd=ab·dd+cd·bb=adbd+cbdb=ad + cbbd.Lecture 5. Addition and subtraction of rational expressions 23Similarly,ab−cd=ad − bcbd.Example 5.2. (3.4 ex.22 )x23+xx + 1=x2(x + 1) + 3x3(x + 1)=x3+ x2+ 3x3(x + 1)= x(x2+ x + 33(x + 1).(3.4 ex.36 )2x − 1x − 1−2x + 1x + 1=(2x − 1)(x + 1) − (x − 1)(2x + 1)(x − 1)(x + 1)=2x2− x + 2x − 1 − 2x2+ 2x − x + 1(x − 1)(x + 1)=2x(x − 1)(x + 1).(3.4 ex.41 )xx + 1+(x − 2)x − 1+x + 1x − 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)(x2− 4x + 4) + (x2+ 2x + 1)(x − 1)(x + 1)(x − 1)(x − 2)=x3− 3x2+ 3x + x3− 4x2+ 4x + x2− 4x + 4 + x3+ 2x2+ x − x2− 2x − 1(x + 1)(x − 1)(x − 2)=3x3− 5x2+ 2x + 3(x + 1)(x − 1)(x − 2)=(x − 1)x(3x − 2) + 3(x + 1)(x − 1)(x − 2).One can check that there is no extra canceling by trying long division of the numeratorby the factors (x + 1), (x −1), and (x −2) of the denominator.5.2. Least common multipleThe general formula for summing up two rational expressions will work in every situa-tion. But not in every situation the general formula is the best. Let us consider followingexampleExample 5.3. (3.4 ex.69 )x + 4x2− x − 2−2x + 3x2+ 2x − 8=generalformula=(x + 4)(x2+ 2x − 8) − (x2− x − 2)(2x + 3)(x2− x − 2)(x2+ 2x − 8).Lecture 5. Addition and subtraction of rational expressions 24A little bit smarter way to do it is to factor the denominators, first:(x + 4)(x − 2)(x + 1)−(2x + 3)(x + 4)(x − 2).Then, obviously, we can equate the denominators by multiplying the first rational expres-sion by(x + 4)(x + 4)and the second one by(x + 1)(x + 1):(x + 4)(x + 4) − (2x + 3)(x + 1)(x − 2)(x + 1)(x + 4)=x2+ 8x + 16 − 2x2− 3x − 2x − 3(x − 2)(x + 1)(x + 4)=−x2+ 3x − 13(x − 2)(x + 1)(x + 4).Let us summarize what we did. We wanted to equate the denominators and we wantedto keep our work simple. The work will be kept simple if a common denominator has thesmallest possible degree. In this example a polynomial (x −2)(x + 1)(x+4) is the one andit is called the Least Common Multiple of polynomials (9x −2)(x −1) and (x −2)(x + 4).Definition 5.1. The polynomial P (x) is the least common multiple of polynomialsP1(x), ···Pn(x) ifa) each of piis a factor of P (x);b) there is no polynomial q(x) with degree less than degree of P (x) such that a) issatisfied.Remark 5.1. Part b) says that P (x) has the least possible degree for property a).Question 5.1. Let P (x) be the least common multiple of polynomials P1(X), P2(X),P3(x).a) ConsiderP (x)P1(x), will P1(x) - cancel?b) ConsiderP (x)P2(x), will P2(x) - cancel?c) ConsiderP (x)P1(x)P2(x), will P1(x)P2(x) - cancel?Let us discuss the process of finding the least common multiple of two or more poly-nomials. The first step is to factor each polynomial completely. Then to construct theleast common multiple we successively combine prime factors of the original polynomials.For each original polynomial we add only those of it’s factor which are missing in theLecture 5. Addition and subtraction of rational expressions 25least common multiple. In particular, let polynomials M(x), N (x) , P (x) have factoring,correspondinglyM(x) = a(x)b(x)c(x), N (x) = b(x)c(x)d(x), P (x) = b(x)d(x)e(x),then the least common multiple for M(x), N(x), and P (x) will beLCM = a(x)b(x)c(x)d(x)b(x)e(x).Here initially we added to LCM all factors from M(x) since LCM should have all factorsof M(x) and had nothing initially. On the second step we added d(x) from N(x) sinceother two multipliers already were LCM. On the third step we added e(x) and b(x)from P (x). Although LCM already had one b(x), still P (x) had two of them so one wasmissing, thus, we added it.Example 5.4. (3.4 ex.51 ) Find the LCM to x3−x, x3−2x2+ x, and x3−1. First,we will factor three polynomialsx3− x = x(x + 1)(x −1),x3− 2x2+ x = x(x −1)(x −1),x3− 1 = (x − 1)(x2+ x + 1).The LCM is then given byLCM = x(x + 1)(x − 1)(x − 1)(x2+ x + 1).(3.4 ex.72 )x(x − 1)2+2x−x + 1x3− x2=x(x − 1)2+2x−x + 1x2(x − 1)=x · x2+ 2 · x · (x − 1)2− (x + 1)(x − 1)(x − 1)2x2= x3+ 2x3− 4x2+ 2x − x2+ 1 =3x3− 5x2+ 2x + 1(x − 1)2x2.By doing long division of the numerator over the (x − 1) one can see that (x − 1) doesnot cancels.Lecture 6.Mixed quotientsWhen the numerator or the denominator of a quotient contains combinations of rational func-tions we call it a mixed quotient. On this lecture we will consider how to simplify mixedquotients.6.1. Mixed quotientsDefinition 6.1. When sums and/or differences of rational expressions appear as thenumerator and/or denominator of a quotient, the quotient is called a mixed quotient.To simplify a mixed quotient means to write it as a rational expression reduced tolowest terms.Example of Mixed Quotientab+cdef+kl, where a, b, c, d, e, f, k, l are polynomials.There are two favorite methods for simplifying mixed quotients. The first methods is tosimplify mixed quotient step by step by considering rational expressions is the numerator,then in the denominator and, finally, the whole thing.In the second approach you will need to compute the least common multiplied of thedenominators of all rational expressions entering the mixed quotient. Then the mixedquotient is simplified in one step by multiplying both numerator and