(Section 0.6: Polynomial, Rational, and Algebraic Expressions) 0.6.1
SECTION 0.6: POLYNOMIAL, RATIONAL, AND ALGEBRAIC EXPRESSIONS
LEARNING OBJECTIVES
• Be able to identify polynomial, rational, and algebraic expressions. • Understand terminology and notation for polynomials.
PART A: DISCUSSION
• In Chapters 1 and 2, we will discuss polynomial, rational, and algebraic functions, as well as their graphs.
PART B: POLYNOMIALS
Let n be a nonnegative integer.
An nth -degree polynomial in x , written in descending powers of x, has the following general form:
n n 1 an x + an 1x + ... + a1x + a0 , ()an 0
The coefficients, denoted by a , a ,…, a , are typically assumed to be real 1 2 n numbers, though some theorems will require integers or rational numbers.
an , the leading coefficient, must be nonzero, although any of the other coefficients could be zero (i.e., their corresponding terms could be “missing”).
n an x is the leading term.
a is the constant term. It can be thought of as a x0 , where x0 = 1. 0 0
• Because n is a nonnegative integer, all of the exponents on x indicated above must be nonnegative integers, as well. Each exponent is the degree of its corresponding term. (Section 0.6: Polynomial, Rational, and Algebraic Expressions) 0.6.2
Example 1 (A Polynomial)
5 4x3 x2 + 1 is a 3rd-degree polynomial in x with leading coefficient 4, 2 leading term 4x3 , and constant term 1. The same would be true even if the terms were reordered: 5 1 x2 + 4x3 . 2
5 The polynomial 4x3 x2 + 1 fits the form 2 n n 1 an x + an 1x + ... + a1x + a0 , with degree n = 3. It can be rewritten as:
5 4x3 x2 + 0x + 1, which fits the form 2
3 2 a3 x + a2 x + a1x + a0 , where the coefficients are:
a 4 leading coefficient 3 = () 5 a = 2 2 a = 0 1 a = 1 constant term 0 () §
Example Set 2 (Constant Polynomials)
7 is a 0th-degree polynomial. It can be thought of as 7x0 .
0 is a polynomial with no degree. § (Section 0.6: Polynomial, Rational, and Algebraic Expressions) 0.6.3
PART C: CATEGORIZING POLYNOMIALS BY DEGREE
Degree Type Examples 0 [Nonzero] Constant 7 1 Linear 3x + 4 2 Quadratic 5x2 x + 1 3 Cubic x3 + 4x 4 Quartic x4 5 Quintic x5
PART D: CATEGORIZING POLYNOMIALS BY NUMBER OF TERMS
Number Type Examples of Terms 1 Monomial x5 2 Binomial x3 + 4x 3 Trinomial 5x2 x + 1
PART E: SQUARING BINOMIALS
Formulas for Squaring Binomials
2 a + b = a2 + 2ab + b2 () 2 a b = a2 2ab + b2 ()
WARNING 1: When squaring binomials, don’t forget the “middle term” of the resulting Perfect Square Trinomial (PST).
2 For example, ()x + 3 = x2 + 6x + 9. Observe that 6x is twice the product of the terms x and 3: 6x = 2 x 3 . ()() 2 The figure below implies that x + y = x2 + 2xy + y2 for x > 0 and y > 0 . ()
(Section 0.6: Polynomial, Rational, and Algebraic Expressions) 0.6.4
PART F: RATIONAL AND ALGEBRAIC EXPRESSIONS
A rational expression in x can be expressed in the form:
polynomial in x
nonzero polynomial in x
Example Set 3 (Rational Expressions)
Examples of rational expressions include:
1 a) . x
5x3 1 b) . Irrational coefficients such as 2 are permissible 2 x + 7x 2 as coefficients of either polynomial.
x7 + x c) x7 + x which equals . In fact, all polynomials are rational 1 expressions. §
An algebraic expression in x is also permitted to contain non-integer rational powers of variable expressions (and their equivalents in radical form).
Example Set 4 (Algebraic Expressions)
Examples of algebraic expressions include:
a) x1/2 , or x .
x3 + 7x5/7 b) . x 3 x + 5 +
• (See Footnote 1.)
All rational expressions are algebraic. § (Section 0.6: Polynomial, Rational, and Algebraic Expressions) 0.6.5
The Venn diagram below is for expressions in x that correspond to functions (see Chapter 1):
FOOTNOTES
1. Algebraic expressions. Some sources forbid the presence of in an algebraic expression, since is a transcendental (i.e., non-algebraic) number. That means that is not a zero of any polynomial with integer coefficients, as, say, 2 is. (Section 0.7: Factoring Polynomials) 0.7.1
SECTION 0.7: FACTORING POLYNOMIALS
LEARNING OBJECTIVES
• Know techniques and formulas for factoring polynomials. • Know the Test for Factorability for factoring quadratic trinomials. • Recognize polynomials in quadratic form and be able to factor them.
PART A: DISCUSSION
• Factoring is a very commonly used technique in precalculus and calculus. Factoring helps us simplify expressions, find zeros, solve equations and inequalities, and find partial fraction decompositions (see Section 7.3).
• Rewriting a sum of terms as a product of factors helps us perform sign analyses, as we will see in Sections 2.4 and 2.10.
PART B: FACTORING OUT GCFs
For now, when we factor a polynomial, we factor it completely over the integers ( ), meaning that the factors cannot be broken down further using only integer coefficients. That is, the factors must be prime (or irreducible) over the integers.
• In Chapter 2, we will factor over other sets, such as , , or .
TIP 1: The Greatest Common Factor (GCF), if it is not 1, should typically be factored out first, although it can be factored out piece-by-piece for more complicated expressions. (Unfortunately, there is no simple, standard definition for the GCF.)
Example 1 (Factoring out a GCF)
We factor 8x + 6 as 24()x + 3 , because 2 is the GCF. 2 is the greatest common divisor of 8 and 6. §
Example 2 (Factoring out a GCF)
We factor x5 + x3 as x3 ()x2 + 1 , because x3 is the GCF. x3 is the power of x with the least exponent. § (Section 0.7: Factoring Polynomials) 0.7.2
TIP 2: Sometimes, it is helpful to factor out 1, particularly when a polynomial has a negative leading coefficient.
Example 3 (Factoring out -1 First)
5 3 Factor 8x 6x .
§ Solution
8x5 6x3 = 8x5 + 6x3 () = 2x3 4x2 + 3 ()
WARNING 1: Sometimes, people confuse signs if they try to 3 factor out 2x immediately. §
WARNING 2: Be careful when factoring the base of a power. Make sure to apply the exponent to all factors.
Example 4 (Factoring out of a Power)
5 5 ()x3 + x is not equivalent to xx()2 + 1 . The following is correct:
5 5 ()x3 + x = xx()2 + 1 5 = x5 ()x2 + 1
Each factor of the base must be raised to the exponent, 5. n See Section 0.5, Law 4: xy = xn yn . § () (Section 0.7: Factoring Polynomials) 0.7.3
PART C: FACTORING FORMULAS
Factoring Formulas
Factoring a … Formula 2 2 2 Perfect Square Trinomial a + 2ab + b = ()a + b (PST) 2 2 2 a 2ab + b = a b () (A rule will be provided for a2 + b2 Sum of Two Squares when we discuss imaginary numbers in Section 2.1. As is, it is prime for now.) Difference of Two Squares a2 b2 = a + b a b ()() Sum of Two Cubes a3 + b3 = ()a + b ()a2 ab + b2 Difference of Two Cubes a3 b3 = a b a2 + ab + b2 ()()
WARNING 3: Many math students forgot or never learned the last two formulas.
WARNING 4: In the last two formulas, there is no “2” or “ 2” coefficient on the ab term of the trinomial factor. If a and b have no common factors (aside from 1 and 1), the trinomial factors are typically prime. Sometimes, people confuse these trinomials with Perfect Square Trinomials (PSTs), which we introduced in Section 0.6, Part E.
TIP 3: In the last two formulas, observe that the binomial factor is “as expected”: ()a + b for a3 + b3 , and ()a b for a3 b3 . The visible signs on the right-hand sides follow the pattern: “same,” “different,” and “+.” (Section 0.7: Factoring Polynomials) 0.7.4
PART D: TEST FOR FACTORABILITY and PRACTICE EXAMPLES
Test for Factorability
This test applies to any quadratic trinomial of the form ax2 + bx + c , where a, b, and c are nonzero, integer coefficients.
(Assume the GCF is 1 or 1; if it is not, factor it out.)
The discriminant of the trinomial is b2 4ac .
• If the discriminant is a perfect square (such as 0, 1, 4, 9, etc.; these are squares of integers), then the trinomial can be factored over the integers. For example, x2 + 3x + 2 has discriminant 1 and can be factored as x + 2 x + 1 . ()()
•• In fact, if the discriminant is 0, then the trinomial is a perfect square trinomial (PST) and can be factored as the square of a binomial with integer coefficients. For example, x2 6x + 9 has 2 discriminant 0 and can be factored as ()x 3 .
• If the discriminant is not a perfect square, then the trinomial is prime over the integers.
This test may be applied in Example Set 5, a) through g), which serve as review exercises for the reader.
• The discriminant is denoted by (uppercase delta), though that symbol is also used for other purposes. It is seen in the Quadratic Formula in Section 0.11. We will discuss a method for factoring quadratic trinomials using the Quadratic Formula in Chapter 2. (Section 0.7: Factoring Polynomials) 0.7.5
Example Set 5 (Factoring Polynomials)
Factor the following polynomials over the integers.
a) x2 + 9x + 20
b) x2 20x + 100 (Hint: This is a Perfect Square Trinomial (PST).)
c) x2 4x 12
d) 3x2 20x 7
e) 4x2 + 11x + 6
f) 2x2 + 10x + 5
2 g) 3x + 6x 3
h) x4 16
i) a3 3a + 2a2b 6b (Hint: Use Factoring by Grouping. This is when we group terms and factor each group “locally” before we factor the entire expression “globally” by factoring out the GCF.)
2 2 j) 4x + 9y
3 3 k) x + 125y
l) x3 125y3 (Section 0.7: Factoring Polynomials) 0.7.6
§ Solution
a) ax2 + bx + c = x2 + 9x + 20 = ()x + 5 ()x + 4
We want 5 and 4, because they have product = c = 20 and (since this is the a = 1 case) sum = b = 9. We can rearrange the factors: x + 4 x + 5 . ()()
2 2 b) x2 20x + 100 = ()x 10 ,or()x 10 ()x 10 = ()x 10 2 2 ()x ()10 Check: 2()x ()10 = Guess that this is a 20x PST for now.
c) x2 4x 12 = ()x 6 ()x + 2
How do we know we need 6 and + 2?
The constant term, c, is negative, so use opposite signs: one "+" and one " ."
The middle coefficient, b, is negative, so the negative number must be higher in absolute value than the positive number; it “carries more weight.”
2-factorizations of –12 (which is c) Sum = b = 4? Think: What? • What?? = –12. ( a = 1 case) 12, +1 No 6, +2 Yes Can stop 4, +3 No
d) F + ()O + I + L = 3x 2 20x 7 = ()3x + 1 ()x 7
F = First product (product of the First terms) O = Outer product (product of the Outer terms) I = Inner product (product of the Inner terms) L = Last product (product of the Last terms)
()3x ()x F =3x 2 ; factors must be 3x and x Need L = 7 + 7 1 1 + 7 Makes O + I =20x. We need O + I to be 20x, which is the middle term of the trinomial. We're only off by a sign, so we change both signs. + 1 7 Makes O + I = 20x. This works. 7 + 1
Also, b = 20 , a "very negative" coefficient, so we are inclined to pair up the 3x and the 7 to form the outer product, since they form 21x . (Section 0.7: Factoring Polynomials) 0.7.7 e) 4x2 + 11x + 6 = 4x + 3 x + 2 ()()
Method 1: Trial-and-Error ("Guess") Method
()() 4xx 2 F =4x 2x 2x + 1 + 6 + 6 + 1 L = 6; need both " + " because of + 11x + 2 + 3 + 3 + 2
Method 2: Factoring by Grouping
4 and 6 are neither prime nor “1,” so we may prefer this method. We want two integers whose product is ac = (4)(6) = 24 and whose sum is b = 11. We want 8 and 3; split the middle term accordingly.
4x 2 + 11x + 6 = 4x 2 + 8x + 3x + 6 OK to switch 2 = ()4x + 8x + ()3x + 6 Group terms = 4xx()+ 2 + 3()x + 2 "Local factoring" = ()4x + 3 ()x + 2 "Global factoring" f) 2x2 + 10x + 5 is prime or irreducible over the integers (i.e., it cannot be broken down further using integer coefficients). None of these combinations work: ()2x ()x F =2x 2 ; factors must be 2x and x Need L = 5; need both " + " because of + 10x
+ 1 + 5 + 5 + 1 We could also apply the Test for Factorability. The discriminant b2 4ac = ()10 2 42()()5 = 100 40 = 60 , which is not a perfect square, and the GCF = 1, so the polynomial is prime. g) 3x 2 + 6x 3 = 3 x 2 2x + 1 () GCF a PST You should usually factor out the GCF first. = 3()x 1 2 (Section 0.7: Factoring Polynomials) 0.7.8
h) Apply the Difference of Two Squares formula []a2 b2 = ()a + b ()a b twice:
x4 16 x2 4 x2 4 = ()+ 2 2 2 2 ()x2 ()4 prime ()x ()2 = x2 + 4 x + 2 x 2 ()()()
i) Use Factoring by Grouping:
a3 3a + 2a2b 6b = ()a3 3a + ()2a2b 6b = aa()2 3 + 2ba()2 3 = ()a + 2b ()a2 3
2 2 j) 4x + 9y is prime. The GCF = 1, and we have no formula for the Sum of Two Squares (for now…; this will change when we discuss imaginary numbers in Section 2.1).
3 3 2 2 k) Apply the Sum of Two Cubes formula a + b = ()a + b a ab + b : "Expected NOT 2ab factor" The visible signs follow the pattern: same, different, "+"
x 3 125y3 x 5y x 2 5xy 25y2 + = ()+ () + 3 3 ()x ()5y
3 3 2 2 l) Apply the Difference of Two Cubes formula a b = ()a b a + ab + b : "Expected NOT +2ab factor" The visible signs follow the pattern: same, different, "+"
x 3 125y3 x 5y x 2 5xy 25y2 . § = () ()+ + 3 3 ()x ()5y (Section 0.7: Factoring Polynomials) 0.7.9
PART E: FACTORING EXPRESSIONS IN QUADRATIC FORM
An expression is in quadratic form
It can be expressed as au2 + bu + c after performing a u substitution, where a 0, and a, b, and c are real coefficients.
• The term “quadratic form” is defined differently in higher math; that definition requires each term to have degree 2.
Example 6 (Factoring an Expression in Quadratic Form)
Factor 2x6 x3 1 over the integers.
§ Solution
The trinomial 2x6 x3 1 is in quadratic form, because the exponent on x in the first term is twice that in the second term (6 is twice 3), and the third term is a constant.
We will use the substitution u = x3 , the power of x in the “middle” term. 2 Then, u2 = x3 = x6 . ()
2x6 x3 1 = 2u2 u 1 Now, factor as usual. = ()2u + 1 ()u 1 Substitute back. Replace u with x3. = ()2x3 + 1 ()x3 1
With practice, the substitution process can be avoided. Either way, we are not done yet! It is true that ()2x3 + 1 is prime over the integers; Chapter 2 will help us verify that. However, ()x3 1 is not prime, because we can apply the Difference of Two Cubes formula.
3 3 3 2 ()2x + 1 ()x 1 = ()2x + 1 ()x 1 ()x + x + 1
This is factored completely over the integers. The Test for Factorability can be used to show that the trinomial factor ()x2 + x + 1 is prime, as expected. § (Section 0.8: Factoring Rational and Algebraic Expressions) 0.8.1
SECTION 0.8: FACTORING RATIONAL AND ALGEBRAIC EXPRESSIONS
LEARNING OBJECTIVES
• Know techniques for factoring rational and algebraic expressions.
PART A: DISCUSSION
• In this section, we will extend techniques for factoring polynomials to other rational and algebraic expressions, including those with negative and fractional exponents. Prior to a precalculus course, most students have no experience factoring such expressions.
PART B: FACTORING OUT GCFs
When we factor x5 + x3 as x3 ()x2 + 1 , we factor out x3 , the power of x with the least exponent; x3 is the GCF. We then divide each term of x5 + x3 by x3 to obtain the other factor, ()x2 + 1 . When we divide x5 by x3 , we subtract the exponents in that order and get x2 .
TIP 1: Think: “We’re factoring x3 out of x5 . 5 takeaway 3 is 2.”
These techniques apply even when the exponents involved are negative and/or fractional.
(Section 0.8: Factoring Rational and Algebraic Expressions) 0.8.2
Example 1 (Factoring with Negative Exponents)
Factor x 7 + x 4 2x1 over the integers.
§ Solution
7 Observe that 7 is the least exponent on x. Our GCF is x , so we will factor it out and subtract 7 from each of the exponents.
x 7 + x 4 2x1 = x 7 ()1+ x 4 ()7 2x1 ()7 = x 7 ()1+ x 4 + 7 2x1+ 7 = x 7 ()1+ x3 2x6
TIP 2: Observe that this last trinomial has no negative exponents on x. This is a sign that we have factored out the GCF correctly.
WARNING 1: We usually try to avoid negative exponents in final answers, so we will rewrite the expression as a fraction.
1+ x3 2x6 = x7
We are not done yet! We can factor the numerator further over the integers.
TIP 3: We will first factor out 1 so that the new leading coefficient is positive. This tends to make factoring easier.
()1 x3 + 2x6 = x7
The indicated 1 factor can be moved in front of the fraction with no other sign changes.
WARNING 2: This is because it is a factor of the entire numerator.
1 x3 + 2x6 = x7
(Section 0.8: Factoring Rational and Algebraic Expressions) 0.8.3
TIP 4: We will now rewrite the numerator in descending powers of x. This tends to make factoring easier.
2x6 x3 1 = x7
Fortunately, we have already factored the numerator in Section 0.7, Example 6.
()2x3 + 1 ()x3 1 = x7
()2x3 + 1 ()x 1 ()x2 + x + 1 = x7 §
Example 2 (Factoring with Negative and Fractional Exponents)
1/3 5/3 Factor ()x3 + 2 + ()x3 + 2 over the integers.
WARNING 3: Exponents do not typically distribute over sums.
WARNING 4: Likewise, the root of a sum is not typically equal to the sum of the roots.
§ Solution
WARNING 5: All negative exponents are less than all positive exponents.
5 5/3 Observe that is the least exponent on x3 + 2 . Our GCF is x3 + 2 , 3 () () 5 so we will factor it out and subtract from each of the exponents on 3 ()x3 + 2 .
(Section 0.8: Factoring Rational and Algebraic Expressions) 0.8.4
1 5 1/3 5/3 5/3 3 3 3 3 ()x + 2 + ()x + 2 = ()x + 2 ()x + 2 3 3 + 1 1 5 5/3 + 3 3 3 3 = ()x + 2 ()x + 2 + 1 5/3 2 = x3 + 2 x3 + 2 + 1 ()() 3 5/3 6 3 = ()x + 2 x + 4x + 4 + 1 3 5/3 6 3 = ()x + 2 x + 4x + 5 x6 + 4x3 + 5 = 5/3 ()x3 + 2
Although x6 + 4x3 + 5 is in quadratic form, it is prime over the integers. §
(Section 0.9: Simplifying Algebraic Expressions) 0.9.1
SECTION 0.9: SIMPLIFYING ALGEBRAIC EXPRESSIONS
LEARNING OBJECTIVES
• Be able to use factoring and canceling (i.e., division) to simplify algebraic expressions. • Be able to rationalize the numerator or the denominator of a fraction containing radical(s). • Be aware of common errors when rewriting and simplifying expressions.
PART A: DISCUSSION
• The factoring techniques from Sections 0.7 and 0.8 will help us simplify fractions by canceling (i.e., dividing out) common factors.
• We can also re-express or simplify a fraction by rationalizing the numerator or the denominator.
PART B: CANCELING COMMON FACTORS IN A FRACTION
We are typically required to provide answers in simplified form. While there are different opinions as to what that means, there is common agreement that factors (other than 1 and 1) that are common to both the numerator and the denominator of a fraction must be canceled (i.e., divided out).
WARNING 1: Some instructors object to the term “canceling,” because students often abuse the idea by inappropriately and carelessly deleting matching expressions. Remember to rely on mathematical rules, not merely wishful thinking!
Canceling Rule for Fractions
As we simplify a fraction, we can “cancel”:
a nonzero factor of the entire numerator with
an equivalent factor of the entire denominator.
• That is, we can divide the entire numerator and the entire denominator by equivalent nonzero factors. (Section 0.9: Simplifying Algebraic Expressions) 0.9.2
x For example, we can divide the numerator and the denominator of x2 x + 4 () 1 by x and obtain . We have canceled a pair of x factors: xx+ 4 () ()1 x 1 = x2 ()x + 4 xx()+ 4 ()x
x WARNING 2: We cannot cancel a pair of x factors in , because x is x2 + 4 not shown as a factor of the entire denominator.
Canceling and Restrictions
When canceling, if a factor is eliminated from the denominator of a fraction, check to see if any restrictions have been “hidden.” Such restrictions must be written out separately.
• It is usually acceptable to wait until the end to write such restrictions.
For example,
()x x2 x2 = ()x 0 xx()+ 4 x ()x + 4 ()1 x = ()x 0 x + 4
(Section 0.9: Simplifying Algebraic Expressions) 0.9.3
PART C: CANCELING OPPOSITE FACTORS
()a b and ()b a are opposites.
Canceling Opposite Factors
The quotient of nonzero, opposite factors is 1.
Example 1 (Canceling Opposite Factors)
9 x2 Simplify . x 3
§ Solution Method 1
9 x2 ()3+ x ()3 x = x 3 x 3 ()
WARNING 3: It is incorrect to factor 9 x2 as x + 3 x 3 , ()() which is the factorization of x2 9.
3 x and x 3 are opposites, so their quotient in either () () order is 1.
()1 ()3+ x ()3 x = ()x 3 ()x 3 ()1
TIP 1: The parentheses around the “ 1” remind us that it is a factor of the numerator, not a term.
= ()3+ x ()x 3 , or
3 xx 3 ()
The choice between 3+ x and 3 x may depend on context. § ()
(Section 0.9: Simplifying Algebraic Expressions) 0.9.4
§ Solution Method 2
2 9 x2 ()x 9 = x 3 x 3 x2 9 = x 3 ()1 ()x + 3 ()x 3 = ()x 3 ()x 3 ()1 = ()x + 3 ()x 3 , or x 3 x 3 () §
PART D: SIMPLIFYING ALGEBRAIC EXPRESSIONS
Example 2 (Simplifying an Algebraic Expression)
1/3 2 1 2/3 ()4x + 7 ()2x ()x ()4x + 7 ()4 3 Simplify 2/3 . ()4x + 7
Do not leave nonpositive exponents in the final expression.
• In calculus, the given expression is obtained by applying the Quotient Rule for x2 Differentiation to . 3 4x + 7
§ Solution
We begin by “cleaning up” the numerator.
1/3 2 1 2/3 ()4x + 7 ()2x ()x ()4x + 7 ()4 3 2/3 4x + 7 () 1/3 4 2/3 2x()4x + 7 x2 ()4x + 7 3 = 2/3 ()4x + 7
(Section 0.9: Simplifying Algebraic Expressions) 0.9.5
Method 1 (Factor the Numerator)
2/3 We will factor out the GCF of the numerator, which is x()4x + 7 .
TIP 2: It may be easier to factor out x in one step and then 2/3 ()4x + 7 in another step. The GCF does not have to be factored out in one step.
TIP 3: It may be easier to substitute u = 4x + 7 and obtain 4 2xu1/3 x2u 2/3 3 . Make sure to substitute back later. u2/3
1/3 4 2/3 4 2x()4x + 7 x2 ()4x + 7 2xu1/3 x2u 2/3 3 3 2/3 or ()4x + 7 u2/3 1 2 1 2 2/3 4 2/3 4 3 3 x()4x + 7 24()x + 7 3 3 x xu 2u x 3 3 = 2/3 or ()4x + 7 u2/3
2/3 4 2/3 4 x()4x + 7 24()x + 7 x xu 2u x 3 3 = 2/3 or ()4x + 7 u2/3
2/3 2/3 We now divide ()4x + 7 by ()4x + 7 . If we let u = 4x + 7 :
2/3 2/3 2 2 4 ()4x + 7 u 3 3 3 1 1 2/3 = = u = u = = 4/3 ()4x + 7 u2/3 u 4/3 ()4x + 7
4/3 The division yields ()4x + 7 in the denominator.
4 x 24()x + 7 x 3 = 4/3 ()4x + 7
4 x 8x + 14 x 3 = 4/3 ()4x + 7 (Section 0.9: Simplifying Algebraic Expressions) 0.9.6
WARNING 4: Compound fractions are typically unacceptable in a simplified expression.
We will multiply the numerator and the denominator by 3.
4 3x 8x + 14 x 3 = 4/3 34()x + 7 24x2 + 42x 4x2 = 4/3 34()x + 7 20x2 + 42x 2x()10x + 21 = 4/3 , or 4/3 34()x + 7 34()x + 7
WARNING 5: We should factor the numerator and see if there are any common factors with the denominator, aside from 1 and 1. Such factors must be canceled (divided out).
WARNING 6: Do not distribute the 3 in the denominator. The 4/3 exponent forbids that.
• No restrictions need to be written. The only restriction on the given 7 expression was x , and it is implied by the final expression. 4
Method 2 (Rewrite as a Compound Fraction First)
2 Because of the negative exponent , the expressions in blue below can be 3 rewritten as fractions containing positive exponents.
Again, the substitution u = 4x + 7 may help.
1/3 4 2/3 4 2x()4x + 7 x2 ()4x + 7 2xu1/3 x2u 2/3 3 3 2/3 or ()4x + 7 u2/3 2 1/3 4x 2 2x 4x 7 1/3 4x ()+ 2/3 2xu 34()x + 7 3u2/3 = 2/3 or ()4x + 7 u2/3
(Section 0.9: Simplifying Algebraic Expressions) 0.9.7
We will multiply the numerator and the denominator by the Least 2/3 Common Denominator (LCD), 34()x + 7 , or 3u2/3.
2 2/3 1/3 4x 34x + 7 2x 4x + 7 () () 2/3 34()x + 7 = 2/3 2/3 34()x + 7 ()4x + 7 2 2/3 1/3 4x 3u 2xu 2/3 3u or 2/3 2/3 3u u
WARNING 7: Do not cancel the expressions above in red. We must apply the Distributive Property. If it helps, write the following step; with experience, you can skip it.
2 2/3 1/3 2/3 4x 34x + 7 2x 4x + 7 34x + 7 () () () 2/3 34()x + 7 = 34x 7 2/3 4x 7 2/3 ()+ ()+ 4x2 3u2/3 2xu1/3 3u2/3 2/3 3u or 2/3 2/3 3u u
6xu 4x2 = 3u 4/3
6x()4x + 7 4x2 = 4/3 34()x + 7 24x2 + 42x 4x2 = 4/3 34()x + 7 20x2 + 42x 2x()10x + 21 = 4/3 , or 4/3 34()x + 7 34()x + 7 §
(Section 0.9: Simplifying Algebraic Expressions) 0.9.8
PART E: RATIONALIZING DENOMINATORS AND NUMERATORS
When we rationalize the denominator of a fraction, we eliminate radicals and non-integer exponents in the denominator but possibly introduce them in the numerator.
1 For example, is unacceptable in a simplified expression, so we 2 1 1 2 2 rewrite it: = = . For more complicated expressions, there 2 2 2 2 are different opinions as to when denominators need to be rationalized. (See Footnote 1)
When we rationalize the numerator of a fraction, we eliminate radicals and non-integer exponents in the numerator but possibly introduce them in the denominator.
Example 3 (Rationalizing a Numerator)
x + h x Re-express by rationalizing the numerator. h
• This expression is an example of a difference quotient (see Section 1.10). Rationalizing the numerator will help us find a derivative (see Section 1.11).
§ Solution
We will multiply the numerator and the denominator by x + h + x , the conjugate of x + h x .
x + h x x + h x x + h + x = () () h h x + h + x ()
To multiply out the numerator, we use the rule: a b a + b = a2 b2 . ()()
(Section 0.9: Simplifying Algebraic Expressions) 0.9.9
2 2 x + h x = ()() hx()+ h + x
WARNING 8: Do not forget the radical expression in the denominator.
()x + h x = hx()+ h + x x + h x = hx()+ h + x h = hx()+ h + x
We now cancel the h factors and note the restriction h 0 . ()
()1 h = ()h 0 h x + h + x ()1 () 1 = ()h 0 x + h + x §
(Section 0.9: Simplifying Algebraic Expressions) 0.9.10
PART F: COMMON ERRORS
Beware of the following errors when rewriting or simplifying an expression.
Errors involving Radicals and Exponents
Error (crossed out) Related Correct Formulas Comments a2 = a 2 3 3 The formula a = a is 2 a = a a = a, if a < 0 2 often forgotten. a = a, if a 0 () Do not apply radicals or a + b = a + b exponents term-by-term. ab = a b , if a 0 and b 0. a b = a b There is no general
formula for the square 1/2 1/2 1/2 a a ()a + b = a + b = , root of a sum (or a b b difference), as there is 1/2 1/2 1/2 ()a b = a b if a 0 and b > 0. for a product or a quotient. ab + ac = ab+ c () When factoring a ab + ac = ab+ c = a b + c, radicand or the base of a if a 0 and b + c 0. power, make sure to
1/2 apply the appropriate 1/2 ()ab + ac = ab()+ c radical or exponent to all 1/2 1/2 1/2 factors. ab + ac = ab+ c = a1/2 b + c , ()() () if a 0 and b + c 0.
(Section 0.9: Simplifying Algebraic Expressions) 0.9.11
Errors involving Fractions
Error (crossed out) Related Correct Formulas Comments ()x 2 2 ()1 x x = x 0 We can cancel (nonzero, x 1 xx 4 ()equivalent) factors of the = ()+ x ()x + 4 x2 + 4 x + 4 ()1 entire numerator and the ()x x entire denominator. = ()x 0 x + 4 If we add like fractions, a b a + b we add the numerators + = . 1 1 1 and keep the common denominator. 1 1 1 + = 1 1 1 b 1 a a b a + b + = + If we add unlike a b a b b a fractions, we do not add b + a a + b = , or . the denominators. ab ab A fraction groups a b + c a b + c a b + c a ()b + c together its numerator = = d d d d d d (and groups together its denominator). 1 1 1 Reciprocals cannot be = x1 + x 2 + = x1 + x 2 x + x2 x x2 taken term-by-term. 1 1 1 The negative exponent = 7x4/3 = x4/3 ()x 0 7x 4/3 7x 4/3 7 only applies to x.
FOOTNOTES
2x()10x + 21 1. Rationalizing denominators. The expression 4/3 from Example 2 is often 34()x + 7 considered simplified. If we were to rationalize the denominator, we would have (nontrivial) powers of 4x + 7 as factors of the numerator and the denominator: ()
2/3 2/3 2x()10x + 21 2x()10x + 21 ()4x + 7 2x()10x + 21 ()4x + 7 4/3 = 4/3 2/3 = 2 34()x + 7 34()x + 7 ()4x + 7 34()x + 7
2x()10x + 21 We may be inclined to divide those powers and go back to 4/3 . 34()x + 7
(Section 0.10: More Algebraic Manipulations) 0.10.1
SECTION 0.10: MORE ALGEBRAIC MANIPULATIONS
LEARNING OBJECTIVES
• Learn additional ways to write equivalent expressions.
PART A: DISCUSSION
• The techniques of this section will help us rewrite expressions so that they are easier to manipulate.
• They will also help us fit expressions to particular forms and templates.
PART B: SPLITTING A FRACTION THROUGH ITS NUMERATOR
Example 1 (Splitting a Fraction Through its Numerator)
Fill in the boxes below with appropriate, simplified numbers:
7x + 3 = x + x 4 x
§ Solution
4 1/4 We begin by rewriting x as x , because the template on the right-hand side suggests a preference for exponential form over radical form.
7x + 3 7x + 3 = 4 x x1/4
We will rewrite this as a sum by splitting it through its numerator.
7x1 3 = + x1/4 x1/4 = 7x3/4 + 3x1/4
(Section 0.10: More Algebraic Manipulations) 0.10.2
3 1 7x + 3 Answer: = 7 x 4 + 3 x 4 4 x
• In calculus, the right-hand side is much easier to differentiate and integrate. §
WARNING 1: Do not split a fraction in this way through its denominator. 1 1 1 For example, is not equivalent to + . x + y x y
PART C: MAKING COMPOUND FRACTIONS
When we divide by a nonzero number, we multiply by its reciprocal. When we multiply by a nonzero number, we divide by its reciprocal.
Example 2 (Making Compound Fractions)
Fill in the boxes below with appropriate, simplified numbers:
3y2 x2 y2 9x2 + = + 4
§ Solution
3y2 9 3 9x2 + = x2 + y2 4 1 4
3 3 WARNING 2: is not equivalent to y2 . 4y2 4
x2 y2 = + 1 4 9 3
3y2 x2 y2 Answer: 9x2 + = + 4 1 4 9 3 §
• This technique will be used in Chapter 10 to set up standard forms for equations of conics. (Section 0.10: More Algebraic Manipulations) 0.10.3
PART D: COMPENSATION
Sometimes, we can alter an expression if we compensate for the change and maintain equivalence.
Example 3 (Compensation Using Addition and Subtraction)
Fill in the box below with an appropriate, simplified expression:
x = 1 x + 1
§ Solution
x + 1 We know that = 1 x 1 , so we would like to add 1 to the x + 1 () x numerator of . To compensate for that, we must also subtract 1 from x + 1 the numerator.
WARNING 3: If we add 1 to the numerator of a fraction, we cannot 1 2 compensate by adding 1 to the denominator. For example, . 2 3
x x + 1 1 = x + 1 x + 1 x + 1 1 = x + 1 x + 1 1 = 1 x + 1
x 1 Answer: = 1 x + 1 x + 1 §
• A similar compensation technique is used when Completing the Square (CTS) in Sections 0.11, 0.13, and 2.2 and Chapter 10.
(Section 0.10: More Algebraic Manipulations) 0.10.4
Example 4 (Compensation Using Multiplication and Division)
Fill in the box below with an appropriate, simplified number:
5 5 ()4x + 1 = ()4x + 1 ()4
§ Solution
The expression on the left-hand side has been multiplied by 4. 1 We compensate by dividing by 4, or by multiplying by its reciprocal, . 4
5 1 5 Answer: () 4x + 1 = ()4x + 1 ()4 4
• In calculus, this technique helps us integrate using the u-substitution method.