Beginning and Intermediate Algebra Chapter 7: …

1 Beginning and Intermediate AlgebraChapter 7: rational ExpressionsAn open source (CC-BY) textbookbyTylerWallace1 Beginning and Intermediate Algebra by Tyler Wallace is licensed under a Creative CommonsAttribution Unported License. ( )You are free: to Share: to copy, distribute and transmit the work to Remix: to adapt the workUnder the following conditions: Attribution: You must attribute the work in the manner specified by the author orlicensor (but not in any way that suggests that they endorse you or your use of thework).With the understanding that: Waiver: Any of the above conditions can be waived if you get permission from the copy-right holder.

2 Public Domain: Where the work or any of its elements is in the public domain underapplicable law, that status is in no way affected by the license. Other Rights: In no way are any of the following rights affected by the license: Your fair dealing or fair use rights, or other applicable copyright exceptions andlimitations; The author s moral rights; Rights other persons may have either in the work itself or in how the work is usedsuch as publicity or privacy rights Notice: For any reuse or distribution, you must make clear toothers the license term ofthis work. The best way to do this is with a link to the following web page: is a human readable summary of the full legal code which can be read at the followingURL: 7: rational Expressions - ReducingRational expressions are expressions written as a quotientof of rational expressions include:x2 x 12x2 9 +20and3x 2anda bb aand32As rational expressions are a special type of fraction, it isimportant to rememberwith fractions we cannot have zero in the denominator of a fraction.

3 For thisreason, rational expressions may have one more excluded values, or values thatthe variable cannot be or the expression would be the excluded value(s):x2 13x2+ 5xDenominator can tbe zero3x2+ 5x 0 Factorx(3x+ 5) 0 Set each factor not equal to zerox 0or3x+ 5 0 Subtract5from second equation 5 53x 5 Divide by333x 53 Second equation is solvedx 0or 53 Our Solution3 This means we can use any value forxin the equation except for 0 and 53. Wecan however, evaluate any other value in the expression. rational expressions areeasily evaluated by simplify substituting the value for thevariable and using orderof 4x2+ 6x+ 8whenx= 6 Substitute 5in for each variable( 6)2 4( 6)2+ 6( 6) + 8 Exponents first36 436+ 6( 6) + 8 Multiply36 436 36+ 6 Add and subtract326 Reduce,dividing by2163 Our SolutionJust as we reduced the previous example, often a rational expression can bereduced, even without knowing the value of the variable.

4 When we reduce wedivide out common factors. We have already seen this with monomials when wediscussed properties of exponents. If the problem only has monomials we canreduce the coefficents, and subtract exponents on the ,subtract exponents move to denominator3x25y4 Our SolutionHowever, if there is more than just one term in either the numerator or denomi-nator, we can t divide out common factors unless we first factor the numerator4and 16 Denominator hasacommon factor of8288(x2 2)Reduce by dividing 24 and8by472(x2 2)Our SolutionExample 318x 6 Numerator hasacommon factor of3,denominator of63(3x 1)6(3x 1)Divide out common factor(3x 1)

5 And divide3and6by312 Our SolutionExample 25x2+ 8x+15 Numerator is difference of squares,denominator is factored using ac(x+ 5)(x 5)(x+ 3)(x+ 5)Divide out common factor(x+ 5)x 5x+ 3 Our SolutionIt is important to remember we cannot reduce terms, only factors. This means ifthere are any + or between the parts we want to reduce we cannot. In the pre-vious example we had the solutionx 5x+ 3, we cannot divide out thex s becausethey are terms (separated by + or ) not factors (separated by multiplication).5 Practice - Reduce rational ExpressionsState the excluded values for )3k2+30kk+103)15n210n+255)10m2+ 8m10m7)r2+ 3r+ 25r+109)b2+12b+32b2+ 4b 322)27p18p2 36p4)x+108x2+80x6)10x+166x+208)6n2 21n6n2+ 3n10)10v2+30v35v2 5vSimplify each )21x218x13)24a40a215)32x38x417)18m 246019)204p+ 221)x+ 1x2+ 8x+ 723)32x228x2+28x25)n2+ 4n 12n2 7n+1027)9v+54v2 4v 6029)12x2 42x30x2 42x31)6a 1010a+ 433)2n2+19n 109n+9035)9m+1620m 1237)2x2 10x+ 83x2 7x+ 439)7n2 32n+164n 1641)n2 2n+ 16n+ 643)7a2 26a 456a2 34a+2012)12n4n214)21k24k216)90x220x18)10 81n3+36n220)n 99n 8122)28m+123624)49r+5656r26)b2+14b+48b2+ 15b+5628)

6 30x 9050x+4030)k2 12k+32k2 6432)9p+18p2+ 4p+ 434)3x2 29x+405x2 30x 8036)9r2+81r5r2+50r+4538)50b 8050b+2040)35v+3521v+ 742)56x 4824x2+56x+3244)4k3 2k2 2k9k3 18k2+ Expressions - Multiply & DivideMultiplying and dividing rational expressions is very similar to the process we useto multiply and divide 1445 First reduce common factors from numerator and denominator(5and7)37 29 Multiply numerators across and denominators across663 Our SolutionThe process is identical for division with the extra first step of multiplying by thereciprocal. When multiplying with rational expressions wefollow the same pro-cess, first divide out common factors, then multiply straight 24y455x7 Reduce coefficients by dividing out common factors(3and5)

7 Reduce,subtracting exponents,negative exponents in denominator53y4 811x5 Multiply across4033x5y4 Our SolutionDivision is identical in process with the extra first step of multiplying by b44 Multiply by the reciprocala4b2a 4b4 Subtract exponents on variables,negative exponents in denominator7a31 4b2 Multiply across4a3b2 Our SolutionJust as with reducing rational expressions, before we reduce a multiplicationproblem, it must be factored 9x2+x 20 x2 8x+163x+ 9 Factor each numerator and denominator(x+ 3)(x 3)(x 4)(x+ 5) (x 4)(x 4)3(x+ 3)Divide out common factors(x+ 3)and(x 4)x 3x+ 5 x 43 Multiply across(x 3)(x 4)3(x+ 5)Our SolutionAgain we follow the same pattern with division with the extrafirst step of multi-plying by the x 12x2 2x 8 5x2+15xx2+x 2 Multiply by the reciprocalx2 x 12x2 2x 8 x2+x 25x2+15xFactor each numerator and denominator(x 4)(x+ 3)(x+ 2)(x 4) (x+ 2)(x 1)5x(x+ 3)Divide out common factors.

8 (x 4)and(x+ 3)and(x+ 2)11 x 15xMultiply acrossx 15xOur Solution8 Practice - Multiply / Divide rational ExpressionsSimplify each )8x29 923)9n2n 75n5)5x24 657)7 (m 6)m 6 5m(7m 5)7(7m 5)9)7r7r(r+10) r 6(r 6)211)25n+255 430n+3013)x 1035x+21 735x+2115)x2 6x 7x+ 5 x+ 5x 717)8k24k2 40k 115k 2519)(n 8) 610n 8021)4m+36n+ 9 m 55m223)3x 612x 24(x+ 3)25)b+ 240b2 24b(5b 3)27)n 76n 12 12 6nn2 13n+4229)27a+369a+63 6a+ 8231)x2 12x+32x2 6x 16 7x2+14x7x2+21x33)(10m2+100m) 18m3 36m220m2 40m35)7p2+25p+126p+48 3p 821p2 44p 3237)10b230b+20 30b+202b2+10b39)7r2 53r 247r+ 2 49r+2149r+142)8x3x 474)9m5m2 726)10p5 8108)710(n+ 3) n 2(n+ 3)(n 2)10)6x(x+ 4)x 3 (x 3)(x 6)6x(x 6)12)9b2 b 12 b 5b2 b 1214)v 14 4v2 11v+1016)1a 6 8a+80818)p 8p2 12p+32 1p 1020)x2 7x+10x 2 x+10x2 x 2022)2rr+ 6 2r7r+4224)2n2 12n 54n+ 7 (2n+ 6)26)21v2+16v 163v+ 4 35v 20v 928)x2+11x+246x3+18x2 6x3+ 6x2x2+ 5x 2430)k 7k2 k 12 7k2 28k8k2 56k32)9x3+54x2x2+ 5x 14 x2+ 5x 1410x234)n 7n2 2n 35 9n+5410n+5036)7x2 66x+8049x2+ 7x 72 7x2+39x 7049x2+ 7x 7238)35n2 12n 3249n2 91n+40 7n2+16n 155n+ 440)12x+2410x2+34x+28 15x+ Expressions - LCDAs with fractions, the least common denominator or LCD is very important toworking with rational expressions.

9 The process we use to findand LCD is basedon the process used to find the LCD of the LCD of8and6 Consider multiples of the larger number8,16,24 .24 is the first multiple of8that is also divisible by624 Our SolutionWhen finding the LCD of several monomials we first find the LCD ofthe coeffi-cients, then use all variables and attach the highest exponent on each the LCD of4x2y5and6x4y3z6 First find the LCD of coefficients4and612 12 is the LCD of4and6x4y5z6 Use all variable with highest exponents oneach variable12x4y5z6 Our SolutionThe same pattern can be used on polynomials that have more than one , we must first factor each polynomial so we can identify all the factors tobe used (attaching highest exponent if necessary).

10 Example the LCD ofx2+ 2x 3andx2 x+12 Factor each polynomial(x 1)(x+ 3)and(x 4)(x+ 3)LCD uses all unique factors(x 1)(x+ 3)(x 4)Our SolutionNotice we only used(x+ 3)once in our LCD. This is because it only appears as afactor once in either polynomial. The only time we need to repeate a factor or usean exponent on a factor is if there are exponents when one of the polynomials isfactoredExample the LCD ofx2 10x+25 andx2 14x+45 Factor each polynomial(x 5)2and(x 5)(x 9)LCD uses all unique factors with highest exponent(x 5)2(x 9)Our SolutionThe previous example could have also been done with factoring the first polyno-mial to(x 5)(x 5).