Chapter 1D Rational Expressions - Texas A&M Universitybrlynch/150fall17/Misc/... · Chapter 1D-Rational Expressions Definition of a Rational Expression A rational expression is the

© : Pre-Calculus - Chapter 1D

Chapter 1D - Rational Expressions

Definition of a Rational Expression

A rational expression is the quotient of two polynomials. (Recall: A function p�x is a polynomial in xof degree n, if there are constants a0, a1,C, an, with an ≠ 0 such that p�x � a0 � a1x �C � anxn.)

More formally, a rational expression is an expression of the form pq , where p and q are polynomials,

and q�x cannot be the zero polynomial. The denominator of a rational expression can be a constantpolynomial though. For example, r�x � x is a polynomial of degree 1, and it is also a rationalexpression, for r�x � x � x

1 .

Example 1: 4x − 76x − 16 ,

2x − 9 ,

x3 − 282 , and x � 5

x2 � 25are examples of rational expressions.

Example 2: x2 � 12x � 7 is not a rational expression.

A rational expression involves a quotient, and since division by 0 is not defined for real numbers, thereare sometimes restrictions on the variable x. In particular, the restrictions occur wherever thedenominator is zero.Example 3: For the rational expression 4x − 7

6x − 16 , the values of x which make the denominator zerocan be found by solving the equation

6x − 16 � 06x � 16

x � 166 � 8

3Therefore, the variable x cannot equal 83 in the rational expression

4x − 76x − 16 . We also say that the

domain of this expression is all x not equal to 83 .

Example 4: For the rational expression 2x − 9 , the values of x which make the denominator zero

can be found by solving the equationx − 9 � 0

x � 9Therefore, the variable x cannot equal 9 in the rational expression 2

x − 9 . The domain of thisexpression is all x not equal to 9.

Example 5: For the rational expression x � 5x2 � 25

, there are no values of x which make thedenominator zero since x2 � 25 is always positive. Therefore, there are no restrictions on the variable x.The domain of this expression is all x.

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Example 6: For the rational expression x2 − 9x2 − 1

, the values of x which make the denominator zerocan be found by solving the equation

x2 − 1 � 0x2 � 1x � o1

Therefore, the variable x cannot equal o1 in the rational expression x2 − 9x2 − 1

. The domain of thisexpression is all x not equal to o1.

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Simplifying Rational ExpressionsTo simplify a rational expression means to reduce it to lowest terms. From working with fractions, youmay recall that simplifying is done by cancelling common factors. Therefore, the key to simplifyingrational expressions (and to most problems involving rational expressions) is to factor the polynomialswhenever possible.

Example 1: Write 8x412x6

in reduced form.Solution: Factor the numbers and cancel common factors. Use properties of exponents to helpcancel the x’s:

8x412x6

� 4 � 2 � x44 � 3 � x4 � x2

� 23x2

Example 2: Write y3 � 5y2 � 6yy2 − 4

in reduced form.

Solution:y3 � 5y2 � 6yy2 − 4

�y�y2 � 5y � 6

y2 − 4

�y�y � 2 �y � 3 �y � 2 �y − 2

�y�y � 3 y − 2 � y2 � 3y

y − 2 ; y ≠ o2

Question: Why is y ≠ o2?Answer: In order for our answer to be equivalent to the original fraction, the variable must havethe same restrictions. Since y cannot equal o2 in the original expression (the denominator would thenbe zero), we must restrict the domain of our answer in order for these fractions to be equivalent.

Example 3: Write 4x − x3x2 − x − 2

in reduced form.

Solution:4x − x3x2 − x − 2

�x�4 − x2 x2 − x − 2

�x�2 � x �2 − x �x − 2 �x � 1

At this point, it doesn’t look like anything will cancel. However, if we factor −1 from the last term inthe numerator, we obtain the following:

�−x�2 � x �x − 2 �x − 2 �x � 1

�−x�x � 2 x � 1 � −x

2 − 2xx � 1 ; x ≠ 2 or − 1

Question: What property of real numbers tells us that −x�2 � x � −x�x � 2 ?Answer: The commutative property of addition.

x � 2 � 2 � x .

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Operations with Rational ExpressionsThe operations of addition, subtraction, multiplication, and division with rational expressions followthe same rules that are used with common fractions. The key, as with simplifying rational expressions,is to factor the polynomials.

I. Multiplication of Rational ExpressionsRecall with fractions ab and

cd that

ab � cd � ac

bd . Also recall when mulitplying fractions that youmay cross-cancel; that is, reduce common factors of any numerator with any denominator.Example 1:

29 � 310

� 23 � 3 � 3

2 � 5� 13 � 15

� 115

The same is true of multiplying rational expressions. As before, the key is to factor the polynomialsfirst.Example 2: Simplify 5

x − 1 � x2 − 125x − 50

Solution: First factor the polynomials. Don’t forget to cancel common factors!5x − 1 � x2 − 1

25x − 50 � 5x − 1 �

�x � 1 �x − 1 25�x − 2

� 11 � x � 1

5�x − 2

� x � 15x − 10 ; x ≠ 1 or 2

Before looking at the next example, recall our strategies for factoring:1. Factor out any common factors2. If more than 3 terms, try factoring by grouping3. Recognize special products4. Factor trinomials using product/sum strategies:a) x2 � bx � c: factors into �x � m �x � n where

mn � c and m � n � bb) ax2 � bx � c: split bx into mx � nx, wheremn � ac and m � n � b, then factor by grouping.

Example 3: Simplify 2x2 � x − 6

x2 � 4x − 5� x

3 − 3x2 � 2x4x2 − 6x

� x � 5x3 − 8

Solution:2x2 � x − 6x2 � 4x − 5

� x3 − 3x2 � 2x4x2 − 6x

� x � 5x3 − 8

� �2x − 3 �x � 2 �x � 5 �x − 1 �

x�x − 2 �x − 1 2x�2x − 3 � x � 5

�x − 2 �x2 � 2x � 4

� x � 22�x2 � 2x � 4

� x � 22x2 � 4x � 8

; x ≠ −5,1,0, 32 ,2

Example 4: Simplify 2x − 4x2 − 4

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Solution: 2x − 4x2 − 4

�2�x − 2

�x − 2 �x � 2 � 2x � 2 , x ≠ o2.

II. Division of Rational ExpressionsRecall with fractions ab and

cd that

ab µ cd � a/bc/d � ab � dc . In other words, dividing by a fraction

is the same as multiplying by its reciprocal. The same is true for rational expressions. As before, thekey to making the work easier is to factor the polynomials first.Example 5: x2 � 2x − 15

x � 2 µ �x2 � 7x � 10

Solution: Begin by writing the second term as x2 � 7x � 10

1 :

x2 � 2x − 15x � 2 µ x

2 � 7x � 101 � x2 � 2x − 15

x � 2 � 1x2 � 7x � 10

� �x � 5 �x − 3 x � 2 � 1

�x � 5 �x � 2

� x − 3�x � 2 2

� x − 3x2 � 4x � 4

; x ≠ −5, − 2

Notice that the x � 2 terms cannot be cancelled since both are in the denominator.

Multiplication and Division can also appear together. Only take the reciprocal of the fractions you aredividing.

Example 6: y2 − 7y � 12y2 � 3y − 18

µ y2 � 3y − 28y2 � 12y � 36

� 3y � 214y � 24Solution: We only take the reciprocal of the second fraction:

y2 − 7y � 12y2 � 3y − 18

µ y2 � 3y − 28y2 � 12y � 36

� 3y � 214y � 24

� y2 − 7y � 12y2 � 3y − 18

� y2 � 12y � 36y2 � 3y − 28

� 3y � 214y � 24

� �y − 3 �y − 4 �y � 6 �y − 3 �

�y � 6 2

�y � 7 �y − 4 �3�y � 7 4�y � 6

� 3; y ≠ −7, − 6, 3, 4Question: Explain why the values −7, −6, 3, and 4 are excluded in the previous example.Answer: The easiest way to understand why these values have been excluded is to writey2 − 7y � 12y2 � 3y − 18

µ y2 � 3y − 28y2 � 12y � 36

� 3y � 214y � 24 as a fraction.

y2 − 7y � 12y2 � 3y − 18y2 � 3y − 28y2 � 12y � 36

3y � 214y � 24

y ≠ −6 follows since 4y � 24 cannot equal 0.We also get y ≠ 3 because the term

y2 − 7y � 12y2 � 3y − 18

� y2 − 7y � 12�y − 6 �y � 3

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The next observation is that the denominator ofy2 − 7y � 12y2 � 3y − 18y2 � 3y − 28y2 � 12y � 36

that is, y2 � 3y − 28y2 � 12y � 36

cannot equal zero. Factoring the numerator and denominator of this rational

expression we havey2 � 3y − 28 � �y � 7 �y − 4

The numerator is zero if y � −7 or if y � 4. This accounts for excluding the numbers −7 and 4. Thedenominator of

y2 � 3y − 28y2 � 12y � 36

factors into �y � 6 2 which means we have to exclude −6, but we have already done that.

Question: What values of x are not allowed in the rational expression x − 1x − 6 µ x − 3x − 4 ?

Answer: 3, 4, 6 (The value 3 is not allowed because x − 3x − 4 is zero at x � 3 and we cannotdivide by 0. x � 4 is not allowed because we have the term x − 4 in the denominator. Similarly x � 6 isnot allowed because the term x − 6 is the denominator of x − 1x − 6 .)

III. Addition and Subtraction of Rational ExpressionsRecall that addition and subtraction is done by first finding a common denominator. The least commondenominator (LCD) of several fractions is the product of all prime factors of the denominators. Afactor only occurs more than once in the LCD if it occurs more than once in any one fraction. Onceyou have the LCD, convert each fraction to an equivalent fraction with the LCD, then add or subtractthe numerators.Example 7: 1

x −1x − 1

Solution: The LCD here is x�x − 1 . We convert each fraction to an equivalent one with thisdenominator: That is, 1x � x − 1

x�x − 1 and1x − 1 � x

x�x − 1 1x −

1x − 1 �

1�x − 1 x�x − 1 −

1�x �x − 1 �x

� x − 1x�x − 1 −

xx�x − 1

� �x − 1 − xx�x − 1 � −1

x�x − 1

Example 8: y − 4y � 6 −

y2 � 3y − 28y2 � 12y � 36

Solution: We must factor the denominators first to find the LCD:

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y − 4y � 6 −

y2 � 3y − 28�y � 6 2

The LCD is �y � 6 2. We now convert each fraction to an equivalent fraction using this denominator:y − 4y � 6 � �y − 4 �y � 6

�y � 6 2

y2 � 3y − 28�y � 6 2

� y2 � 3y − 28�y � 6 2

Next subtract the two expressionsy − 4y � 6 −

y2 � 3y − 28y2 � 12y � 36

� �y − 4 �y � 6 �y � 6 2

− y2 � 3y − 28�y � 6 2

� y2 � 2y − 24�y � 6 2

− y2 � 3y − 28�y � 6 2

��y2 � 2y − 24 − �y2 � 3y − 28

�y � 6 2

� y2 � 2y − 24 − y2 − 3y � 28�y � 6 2

� −y � 4�y � 6 2

Note that the minus sign in the numerator was distributed across the expression−�y2 � 3y − 28 � −y2 − 3y � 28.

Question: What is the common denominator in x − 5�x − 7 �x � 1 −

1x2 − 4

?

Answer: �x − 7 �x � 1 �x2 − 4

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Compound FractionsA compound fraction (also called a complex fraction) is an expression which contains nested fractions.In general, there is one ”main fraction” which will have one or more fractions in the numerator and/ordenominator. There are two strategies which may be used to simplify compound fractions. First,simplify both the numerator and the denominator individually, then divide the numerator by thedenominator by multiplying with the reciprocal of the denominator.

Example 1:6y −

52y � 1

6y � 4

Solution: First simplify the numerator using the common denominator y�2y � 1 :Example

6y −

52y � 1

6y � 4

�

6�2y � 1 y�2y � 1 −

5�y �2y � 1 �y

6y � 4

�

�12y � 6 − �5y y�2y � 1 6y � 4

�

7y � 6y�2y � 1 6y � 4

Now simplify the denominator using the common denominator y (remember that 4 � 41 :

6y −

52y � 1

6y � 4

�

7y � 6y�2y � 1 6y �

4�y 1�y

�

7y � 6y�2y � 1 6 � 4yy

We now divide the two fractions 7y � 6y�2y � 1 and

6 � 4yy :

6y −

52y � 1

6y � 4

� 7y � 6y�2y � 1 µ

6 � 4yy

� 7y � 6y�2y � 1 �

y6 � 4y

� 7y � 6�2y � 1 �6 � 4y ; y ≠

−32 ,

−12 , 0 .

As you can see, this is often a long, tedious process. A second technique is to multiply thenumerator and the denominator of the ”main fraction” by the LCD of all the smaller fractions andsimplify the result.

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Example 2:6y −

52y � 1

6y � 4

Solution: The LCD of all the smaller fractions is y�2y � 1 . We multiply the numerator anddenominator by this LCD:

6y −

52y � 1

6y � 4

�

6y −

52y � 1 �y �2y � 1

6y � 4 �y �2y � 1

�

6y �y �2y � 1 − 5

2y � 1 �y �2y � 1

6y �y �2y � 1 � 4�y �2y � 1

�6�2y � 1 − 5y

6�2y � 1 � 4y�2y � 1

� 12y � 6 − 5y�2y � 1 �6 � 4y

� 7y � 6�2y � 1 �6 � 4y

The following example is a common one in many calculus classes:

Example 3: Simplify the expression 2�x � h � 1 −1 − 2�x � 1 −1

h .

Solution: First rewrite the expression without negative exponents. (Recall x−1 � 1x )

2�x � h � 1 −1 − 2�x � 1 −1

h �2

x � h � 1 −2x � 1

hMethod I Solution: Now simplify the numerator using the common denominator �x � h � 1 �x � 1 :

2�x � h � 1 −1 − 2�x � 1 −1

h �

2�x � 1 �x � h � 1 �x � 1 −

2�x � h � 1 �x � 1 �x � h � 1

h

�

�2x � 2 − �2x � 2h � 2 �x � h � 1 �x � 1

h

�

2x � 2 − 2x − 2h − 2�x � h � 1 �x � 1

h

�

−2h�x � h � 1 �x � 1

hThe denominator is h � h

1 which is already simplified. We now divide the two fractions−2h

�x � h � 1 �x � 1 andh1 :

2�x � h � 1 −1 − 2�x � 1 −1

h � −2h�x � h � 1 �x � 1 �

1h

� −2�x � h � 1 �x � 1 ; h ≠ 0

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Method II Solution: The LCD of all the smaller fractions is �x � h � 1 �x � 1 . We mulitply thenumerator and denominator by this LCD:

2x � h � 1 −

2x � 1 �x � h � 1 �x � 1

h�x � h � 1 �x � 1

�2

x � h � 1 �x � h � 1 �x � 1 − 2x � 1 �x � h � 1 �x � 1

h�x � h � 1 �x � 1

�2�x � 1 − 2�x � h � 1 h�x � h � 1 �x � 1

� 2x � 2 − 2x − 2h − 2h�x � h � 1 �x � 1

� −2hh�x � h � 1 �x � 1 � −2

�x � h � 1 �x � 1 ; h ≠ 0

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Exercises for Chapter 1D - Rational Expressions

For problems 1-5, determine if the given expression is a rational expression. If it is arational expression, state any restrictions on the variable.

1. 3x2 − 6x � 72. 5x − 10

5x � 10

3. x � 8x2 � 4x � 6

4. 5x � 74

5. 2x − 4

For problems 6-8, reduce the rational expression if possible.

6. x3yx2y2

7. x2 � 4x − 125x − 10

8. x3 − 4xx2 � x − 2

For problems 9-20, perform the indicated operations and simplify.9. x3 − 8

x2 − 4� x3 � 8x2 � 2x � 4

10. z2 − 5z − 24z2 − 9

� 9z2 − 12z � 4

z2 − 10z � 16� z

2 − z − 69z − 12

11. 3x − 1 µ 6x � 9

x2 − 9x � 812. y2 � 8y − 20

y2 � 11y � 10� y

3 � 1y3 − 8

µ y2 − y � 1y2 � 2y � 4

13. y3 − 272y3

� 4yy2 − 5y � 6

µ 2y − 6y2 − 2y

14. xx − 3 � 2

3x � 415. 1

x − 2 −3x � 3

16. 1x � 3 � 1

x − 3 −10x2 − 9

17. 72x � 1 −

8x2x − 1 � 4

18. 3x − 1 −

2x � x � 3

x2 − 119. y2 − 7y � 12

y2 � 3y − 18� y2 − 25y2 − 8y � 15

(HINT: Simplify first)

20. y − 7y2 � 2y � 1

− 5y � 1

For problems 21-25, simplify the expression completely.

21.3�x � y 4x � y2

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22.5x � 1 � x

x − 12

x2 − 1� 1

23. 11a � 1b

24.1x −

12

x − 2

25. �x � h −2 � x−2h

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Answers to Exercises for Chapter 1D - RationalExpressions

1. This is a rational function and there are no restrictions on the variable x.2. Is a rational function

Restriction:5x � 10 ≠ 0

5x ≠ −10x ≠ −2

3. Not a rational function because of the radical in the numerator.4. This is a rational function and there are no restrictions on the variable x.5. Is a rational function.

Restriction:x − 4 ≠ 0

x ≠ 46.

x3yx2y2

� x2 � x � yx2 � y � y

� xy ; x ≠ 0 and y ≠ 0

7.x2 � 4x − 125x − 10 � �x � 6 �x − 2

5�x − 2 � x � 65 ; x ≠ 2

8.x3 − 4xx2 � x − 2

�x�x � 2 �x − 2 �x � 2 �x − 1 �

x�x − 2 x − 1 ; x ≠ −2 and x ≠ 1

9.x3 − 8x2 − 4

� x3 � 8x2 � 2x � 4

��x − 2 �x2 � 2x � 4 �x � 2 �x − 2 �

�x � 2 �x2 − 2x � 4 x2 � 2x � 4

� x2 − 2x � 4; x ≠ 2,−2

10.z2 − 5z − 24z2 − 9

� 9z2 − 12z � 4

z2 − 10z � 16� z

2 − z − 69z − 12

� �z − 8 �z � 3 �z � 3 �z − 3 �

�3z − 2 2

�z − 2 �z − 8 ��z − 3 �z � 2 3�3z − 4

� 13 �3z − 2

2 z � 2�z − 2 �3z − 4 z ≠ −3,3,8,−2,

43

11.3x − 1 µ 6x � 9

x2 − 9x � 8� 3x − 1 � x

2 − 9x � 86x � 9 � 3

x − 1 ��x − 8 �x − 1 3�2x � 3 � x − 8

2x � 3 ; x ≠ 1,8

12.

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y2 � 8y − 20y2 � 11y � 10

� y3 � 1y3 − 8

µ y2 − y � 1y2 � 2y � 4

� �y � 10 �y − 2 �y � 10 �y � 1 �

�y � 1 �y2 − y � 1 �y − 2 �y2 � 2y � 4

� y2 � 2y � 4y2 − y � 1

� 1; y ≠ −10,−1,213.

y3 − 272y3

� 4yy2 − 5y � 6

µ 2y − 6y2 − 2y

��y − 3 �y2 � 3y � 9

2y3� 4y�y − 2 �y − 3 �

y�y − 2 2�y − 3

� y2 � 3y � 9y�y − 3 ; y ≠ 2

14. LCD: �x − 3 �3x � 4 xx − 3 � 2

3x � 4 �x�3x � 4

�x − 3 �3x � 4 �2�x − 3

�3x � 4 �x − 3 ��3x2 � 4x � �2x − 6 �x − 3 �3x � 4 � 3x2 � 6x − 6

�x − 3 �3x � 4 15. LCD: �x − 2 �x � 3

1x − 2 −

3x � 3 �

1�x � 3 �x − 2 �x � 3 −

3�x − 2 �x � 3 �x − 2 � �x � 3 − �3x − 6

�x − 2 �x � 3 � −2x � 9�x − 2 �x � 3

16. LCD: �x � 3 �x − 3

�1�x − 3

�x � 3 �x − 3 �1�x � 3

�x − 3 �x � 3 −10

�x � 3 �x − 3

� �x − 3 � �x � 3 − 10�x � 3 �x − 3

� 2x − 10�x � 3 �x − 3

17.7

2x � 1 −8x2x − 1 � 41

LCD: �2x � 1 �2x − 1

�7�2x − 1

�2x � 1 �2x − 1 −8x�2x � 1

�2x − 1 �2x � 1 �4�2x − 1 �2x � 1 �2x − 1 �2x � 1

��14x − 7 − �16x2 � 8x � �16x2 − 4

�2x � 1 �2x − 1

� 6x − 11�2x � 1 �2x − 1

18.3x − 1 −

2x � x � 3

x2 − 1� 3x − 1 −

2x � x � 3

�x � 1 �x − 1 LCD: x�x � 1 �x − 1

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�3x�x � 1

�x − 1 �x �x � 1 −2�x � 1 �x − 1 x�x � 1 �x − 1 � �x � 3 �x

�x � 1 �x − 1 �x

��3x2 � 3x − �2x2 − 2 � �x2 � 3x

x�x − 1 �x � 1

� 2x2 � 6x � 2x�x − 1 �x � 1

19.y2 − 7y � 12y2 � 3y − 18

� y2 − 25y2 − 8y � 15

� �y − 4 �y − 3 �y � 6 �y − 3 �

�y � 5 �y − 5 �y − 5 �y − 3

� y − 4y � 6 � y � 5y − 3 ; y ≠ 5

So LCD is �y � 6 �y − 3

� �y − 4 �y − 3 �y � 6 �y − 3 �

�y � 5 �y � 6 �y − 3 �y � 6

��y2 − 7y � 12 � �y2 � 11y � 30

�y � 6 �y − 3

� 2y2 � 4y � 42�y � 6 �y − 3 ; y ≠ 5

20.y − 7

y2 � 2y � 1− 5y � 1

� y − 7�y � 1 2

− 5y � 1

So LCD � �y � 1 2

� y − 7�y � 1 2

− 5�y � 1 �y � 1 2

� �y − 7 − �5y � 5 �y � 1 2

� −4y − 12�y � 1 2

21.3�x � y 4 � 4x � y2 � 4

�3�x � y 2�x � y � 3

2

22.

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5x � 1 � x

x − 12

x2 − 1� 1

�5x � 1 � x

x − 1 �x � 1 �x − 1

2�x � 1 �x − 1 � 1 �x � 1 �x − 1

�5�x − 1 � x�x � 1 2 � �x � 1 �x − 1

� 5x − 5 � x2 � x2 � x2 − 1

� x2 � 6x − 5x2 � 1

; x ≠ 1,−1

23.1�ab

1a � 1b �ab

� abb � a ; a ≠ 0;b ≠ 0

24.1x −

12

x − 2

�1x −

12 �2x

x − 21 �2x

� 2 − x2x�x − 2

�−�x − 2 2x�x − 2

� −12x25.

�x � h −2 − x−2h

�

1�x � h 2

− 1x2

�x � h 2�x2

h�x � h 2�x2

�x2 − �x � h 2

hx2�x � h 2

� x2 − �x2 � 2xh � h2 hx2�x � h 2

� x2 − x2 − 2xh − h2hx2�x � h 2

� −2xh − h2hx2�x � h 2

�h�−2x − h hx2�x � h 2

� −2x − hx2�x � h 2

; h ≠ 0

© : Pre-Calculus

Chapter 1D Rational Expressions - Texas A&M Universitybrlynch/150fall17/Misc/... · Chapter 1D-Rational Expressions Definition of a Rational Expression A rational expression is the

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