17] 1
[FILE SOME ASPECTS ON RATIONAL EXPRESSIONS(Pages 621-653)
Sometimes we will encounter mathematical expressions containing fractions and we must be able to simplify them. There are several techniques that can be used to simplify such an expression. Now let us examine some of the concepts under the above heading.
I. REDUCING RATIONAL EXPRESSIONS
Note: To reduce rational expressions is to divide; both, the numerator and the denominator by the same non-zero number an even number of times. Do not convert an improper fraction to a mixed number unless you are told to do so. Sometimes rational expressions involve polynomial expressions in which case you would factor, if possible, then reduce by cancellation. You MUST know the rules on FACTORING. Examples: Reduce 1) 24 = 2 36 3 2) x(x+2) = x y(x+2) y Hint: To reduce, divide, both, the numerator and the denominator by 12. Hint: Since the given expressions are already in factored form, reduce by canceling like quantities, if possible.
3) x2+5x+2 = (x+3)(x+2) = x+3 x2-4 (x+2)(x-2) x-2
Hint: When factoring, always look for a common monomial factor, first. If not, factor, and apply the cancellation process to reduce. 4) 4x2(6x2-13xy+6y2) = 4xx(3x-2y)(2x-3y) = 2x(3x-2y) 6x(2x2+5xy-12y2) 6x(2x-3y)(x+4y) 3(x+4y) TRY # 5 AND # 6 ON YOUR OWN: 5) 6) 6x2-54 2x2+8x+6 =
2x3+20x2+50x = 4x3-100x
ANSWERS: 5) 3(x-3) 6) x+5 x+1 2(x-5) NOTE: Leave the result(s) in factored form as in examples 5 and 6 above. Exercise: Do problems 3,11,17,21,27,31,36,41,47,55,61, (Pages 630-632) II. Multiplication/Division with Rational Expressions See pages 633-638.
2 NOTES: When multiplying or dividing rational expressions, i. factor where possible ii. if dividing, invert the divisor and multiply iii. use the cancellation process to reduce and simplify Examples: Perform the indicated operation. 1) 2 18 7 9 35 4 4) x2+3x+2 x+1 5) 3 11 = 3 7 14 7 =1 5
2
2)
9b 7c
28c = 4 81b 9
3)
4-x2 14
7 2-x
= (2+x)(2-x) 14
7 = 2+x (2-x) 2
x+3 = (x+2)(x+1) x +5x+6 (x+1) 14 = 6 11 11
(x+3) =1 (x+3)(x+2) d =c c 8) 2x+4 4x+8 = 2(x+2) 5(x+3) = 5 3x+9 5x+15 3(x+3) 4(x+2) 6
6) c2 c = (c)(c) dd d = x+y xy (x-5) =1
7) x2-y (x-y) = (x+y)(x-y) 1 xy xy (x-y) 9) 10) x2+4x+3 x+3 = (x+3)(x+1) x2-4x-5 x-5 (x-5)(x+1) x2-y2 x2
x+y x2+2x+y2 = (x+y)(x-y) x-y xy xx
(x+y) (x-y)
xy =y (x+y)(x+y) x
Exercise: Do problems 3,11,17,21,27,31,37 ; Pages 636-637 III. SOME Aspects on the LCM (Least Common Multiply) [Pages 640-641] Sometimes we may be asked to find the sum or difference (that is to say, add or subtract) of rational expressions with unlike denominators. If that is the case, then we need to find the least common denominator, some of these denominators will contain numerical numbers and some will contain polynomial expressions. Now, let us take a look at those denominators with only the numerical numbers. There are two methods that may be used to find the LCM, the factoring method or an alternate method. Examples Find the LCM for the following set of numbers:
1) {18, 24, 36}
3 SOLUTION: FACTORING METHOD i) Factor each number into a set of prime factors. Use the tree diagram. 18 29 33 24 36 2) 18 - 24 - 36 2) 9 - 12 - 18 2) 9 - 6 - 9 3) 9 - 3 - 9 3) 3 - 1 - 3 ) 1 - 1 - 1 stop ii) Multiply all the numbers on the left side of the ladder system to determine the LCM. LCM =(3)(3)(2)(2)(2)= 72 ALTERNATE METHOD i) use the prime numbers to divide until you have a result of ones(1's) across the bottom of the ladder system.
2 12 2 18 26 29 23 33 ii) Express as a product all of the different factors for the numbers. 18= (2)(3)2; 24= (2)3(3); 36= (2)2(3)2 Note: The set of factors for each number must be all prime factors. LCM = (23)(32) = (8)(9)= 72 2) {9,27,63} Solution: 9 33 3 27 63 9 3 21 33 37 2 3 2 9= 3 ; 27= 3 ; 63= (3) (7) LCM= (3)3(7) = 189
2) {9,27,63} Solution: 3) 9 - 27 - 63 3) 3 - 9 - 21 3) 1 - 3 - 7 7) 1 - 1 - 7 )1-1 -1 LCM= (7)(3)(3)(3)= 189
Now, let us take a look at those denominators containing polynomial expressions. The following procedures may be used to determine the LCM: i) Factor all the given expressions (in the denominators) completely, ii) Find the LCM for all numerical coefficient factors, iii) Express as a product all of the different factors in the given expressions. Be sure to select those factors with the highest exponents. Examples Find the LCM for the following: 1) 18a2, 48a Solution: 2) 30ab, 75a2b Solution: 3) 2x2+7x-4, x2+6x+8 Solution:
4 i) Find Lcm for 18 and 48. Hence, 144 ii) Select the a the with highest exponent: a2 iii) Lcm =144a2 Lcm for 30 and 75 is 150. Lcm for a, a2,b is a2b Hence, Lcm=150a2b 2x +7x-4=(2x-1)(x+4) x2+6x+8= (x+4)(x+2)
2
Lcm= (2x-1)(x+4)(x+2)
4) x2-16, 2x-8 5) x2-25, x2-10x+25, x2-x-20 Solution: Solution: i) x2-16=(x+4)(x-4) i) x2-25=(x+5)(x-5) 2x-8=2(x-4) x2-10x+25=(x-5)2 x2-x-20=(x+4)(x-5) LCM=2(x+4)(x-4) LCM=(x+4)(x+5)(x-5)2
6) x2-49, x2-14x+49, x2-5x-1 Solution: i) x2-49=(x+7)(x-7) x2-14x+49=(x-7)2 x2-5x-14=( x+2)(x-7) LCM= ?
ASSIGNMENT: IV. ADDITION AND SUBTRACTION OF POLYNOMIALS[ PP 639-646] A. To add or subtract polynomials with like denominators, combine the numerators, write the result over one common denominator, and reduce whenever possible. EXAMPLES Perform the indicated operations. 1) 4 + 1 = 4+1 = 5 2) 5 + x+4 =5+x+4 =x+9 3x 3x 3x 3x x x x x 3) x2-3x + 3x-4 = x2-3x+3x-4 = x2-4 = (x+2)(x-2) = x+2 x-2 x-2 x-2 x-2 (x-2) 4) 7x - 4x+3 = 7x-(4x+3) = 7x-4x-3 = 3x-3 = 3(x-1) 5x+5 5x+5 5x+5 5(x+1) 5(x+1) 5(x+1)
B. To add or subtract rational expressions with unlike denominators, factor all given denominators, where possible, then find the LCM. Find the equivalent terms for the numerator and simplify. EXAMPLES: Perform the indicated operations. 1) 5 + 7 = 15+14 = 29 2) 3x + 4 = 3x(x-1)+4(x+3) = 3x2+x+12 12 18 36 36 x+3 x-1 (x+3)(x-1) (x+3)(x-1) NOTE: Be sure to reduce fractions in the end result, where possible. In the above problem, the numerator can not be factored; therefore, the fraction can not be reduced. Also, leave the results in factored form. 3) x+5 - 3x = x+5 3x = (x+5)(x+2) - 3x = x2+7x+10-3x = x2+4x+10 2 x-2 x -4 x-2 (x+2)(x-2) (x+2)(x-2) (x+2)(x-2) (x+2)(x-2) 5x-10 10 = (5x-10)(3)-10(x) = 15x-30-10x = 5x-30 =5(x-6) = 5 x(x-6) 3(x-6) 3x(x-6) 3x(x-6) 3x(x-6) 3x(x-6) 3x
4) 5x-10 - 10 = x2-6x 3x-18
5 NOTE: Do 5-12 on your own. 5) 3y-2 + 2 7y 14y2 9) 12) 6 3x-2
2
6) 6x-1 - 5 12x 18x2 10)
2
7)
x+3 - 8 x+2 x-2 11)
2
8)
5x+1 - 3 x-4 x-3 3y+5 3y2-7y+2
+
7x 6x-4
5y 2 2 y -6y+9 y -9
4y 3y +5y-2
3x-2 + x -3x+2
5x-1 x2+x-6
ANSWERS 5) 9) 6y2-4y+2 14y2 7x+12 2(3x-2) 6) 10) 18x2-3x-10 36x2 5y2_13y+6 (y+3)(y-3)2 7) x2-7x-22 (x+2)(x-2) 11) 8) 5x2-17x+9 (x-4)(x-3) 8x2+x-5 (x-2)(x-1)(x+3)
y2-19y-10 12) (3y-1)(y+2)(y-2)
ASSIGNMENT: Math -0096/0097: 7,17,19,23,27,33,41,45,51; pages 645-646 SOME ASPECTS ON COMPLEX RATIONAL EXPRESSIONS TO BE DISCUSSED pp 647-652 Earlier, we discussed complex rational expressions containing numerical numbers and we were asked to simplify them using a system that we applied. Now we will simplify complex rational expressions containing polynomial expressions. To simplify complex rational expressions, apply the LCM method and that is: i) Find the LCM for all fractions in the given expression ii) Multiply each term, both the numerator and the denominator, by the LCM iii) Factor when possible iv) apply the cancellation rule when possible to reduce
Examples: 1) 1+7 y 2) a2-b2 ab
6 1 -5 y Solution: LCM= y 1+7 y y = 1+7y 1 -5 y 1-5y y 3) 5+3 x+3 3-x 14 3+x x-3 a-b b Solution: LCM= ab a2-b2 ab a-b b ab = a2-b2 = (a+b)(a-b) = a+b ab a(a-b) a(a-b) a
Solution: Transform the expression (3+x) so that the x term is first and positive. Hence, 3+x = x+3 by the commutative property for addition. Transform the expression (3-x) so that the x term is first and positive. Hence, 3-x = -(x-3) by factoring out a negative one (-1). So, the expression 1 will become 1 and + 3 will become - 3 3+x x+3 3-x x-3 Now, the original problem becomes
53 x+3 x -3 14 x+3 x-3
(x+3)(x-3) = 5(x-3)-3(x+3) = 5x-15-3x-9 = 2x-24 = -2(x-12) (x+3)(x-3) x-3-4(x+3) x-3-4x-12 -3x-15 3(x+5)
Note that the LCM is (x+3)(x-3) ASSIGNMENT: Solving Rational Equations [Pages 654-662] NOTE: To solve rational equations, 1) for those rational equations containing algebraic expressions in the denominators, factor them 2) find the LCM for all fractions in the given equation, 3) multiply each term in the given equation by the LCM to clear all fractions 4) solve the equation for the unknown variable
7 Examples: Solve. 1) 2x + 3 = x -4 3 2 6 Solution: LCM = 6 6 2x + 6 3 = 6 x - 6 4 13 12 161 4x + 9 = x 24 -x -9 -x -9 3x = -33 x = -11 { -11 } 2) ___4__ - ___2__ = __6__ x-2 x+2 x2 4
Solution: ___4__ - ___2__ = __6_____ x-2 x+2 (x+2)(x-2) LCM = (x+2)(x-2) (x+2)(x-2) ___4__ - (x+2)(x-2) ___2__ = (x+2)(x-2) __6_____ 1 (x 2) 1 (x + 2) 1 (x+2)(x-2) 4(x+2) - 2(x-2) = 1(6) 4x + 8 -2x + 4 = 6 2x + 12 = 6 -12 = -12 2x = -6 x = -3 {-3} ASSIGNMENT
8 Math-0096/0097: Pages 661-662; Problems 3,11,17,19,23,25,29,33,35