(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 n th -degree polynomial in x , written in descending powers of x, has the following general form : a n x n + a n 1 x n 1 + ... + a 1 x + a 0 , a n 0 ( ) The coefficients , denoted by a 1 , a 2 , …, a n , are typically assumed to be real numbers, though some theorems will require integers or rational numbers. a n , the leading coefficient , must be nonzero, although any of the other coefficients could be zero (i.e., their corresponding terms could be “missing”). a n x n is the leading term . a 0 is the constant term . It can be thought of as a 0 x 0 , where x 0 = 1 . • 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.
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(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 n th -degree polynomial in x , written in descending powers of x, has the
following general form:
anxn+ an 1x
n 1+ ... + a1x + a0 , an 0( )
The coefficients, denoted by a
1, a
2,…, a
n, are typically assumed to be real
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”).
anxn is the leading term.
a0 is the constant term. It can be thought of as a
0x
0, where x
0= 1.
• 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)
4x35
2x2 +1 is a 3
rd-degree polynomial in x with leading coefficient 4,
leading term 4x3, and constant term 1. The same would be true even if the
terms were reordered:
15
2x2 + 4x3 .
The polynomial 4x35
2x2 +1 fits the form
anxn+ an 1x
n 1+ ... + a1x + a0 , with degree n = 3.
It can be rewritten as:
4x35
2x2 + 0x + 1 , which fits the form
a3x3+ a2x
2+ a1x + a0 , where the coefficients are:
a3= 4 leading coefficient( )
a2=
5
2
a1= 0
a0= 1 constant term( )
§
Example Set 2 (Constant Polynomials)
7 is a 0th-degree polynomial. It can be thought of as 7x
0.
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
of Terms Type Examples
1 Monomial x5
2 Binomial x3+ 4x
3 Trinomial 5x2
x +1
PART E: SQUARING BINOMIALS
Formulas for Squaring Binomials
a + b( )
2
= a2+ 2ab + b
2
a b( )
2
= a2 2ab + b
2
WARNING 1: When squaring binomials, don’t forget the “middle term” of the
resulting Perfect Square Trinomial (PST).
For example, x + 3( )2= x2 + 6x + 9 . Observe that 6x is twice the product of the
terms x and 3: 6x = 2 x( ) 3( ) .
The figure below implies that
x + y( )2
= x2+ 2xy + y
2 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:
a) 1
x.
b) 5x3 1
x2 + 7x 2. Irrational coefficients such as 2 are permissible
as coefficients of either polynomial.
c) x7+ x which equals
x7+ x
1. In fact, all polynomials are rational
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 .
b) x3 + 7x5/7
x x + 53+
.
• (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 2 4x + 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)
Factor 8x5 6x
3 .
§ Solution
8x5 6x
3= 8x
5+ 6x
3( )= 2x
3 4x2+ 3( )
WARNING 1: Sometimes, people confuse signs if they try to
factor out 2x3 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)
x3 + x( )5 is not equivalent to x x2 +1( )
5.
The following is correct:
x3 + x( )5= x x2 +1( )
5
= x5 x2 +1( )5
Each factor of the base must be raised to the exponent, 5.
See Section 0.5, Law 4:
xy( )n
= xny
n . §
(Section 0.7: Factoring Polynomials) 0.7.3
PART C: FACTORING FORMULAS
Factoring Formulas
Factoring a … Formula
Perfect Square Trinomial
(PST)
a2+ 2ab + b
2= a + b( )
2
a
2 2ab + b2= a b( )
2
Sum of Two Squares (A rule will be provided for a
2+ b
2
when we discuss imaginary numbers in
Section 2.1. As is, it is prime for now.)
Difference of Two Squares a
2b
2= a + b( ) a b( )
Sum of Two Cubes a3+ b
3= a + b( ) a
2ab + b
2( )
Difference of Two Cubes a
3b
3= a b( ) a
2+ ab + b
2( )
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+ b
3, and a b( ) for a
3b
3. 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
discriminant 0 and can be factored as x 3( )2
.
• 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
g) 3x2+ 6x 3
h) x4 16
i) a3 3a + 2a
2b 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.)
j) 4x2+ 9y
2
k) x3+125y
3
l) x3 125y
3
(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( ) .
b)
x2
x( )2
20x + 100
10( )2
Guess that this is aPST for now.
= x 10( )
Check:2 x( ) 10( )=
20x
2, or x 10( ) x 10( ) = x 10( )
2
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)
Think: What? • What?? = –12.
Sum = b = 4?
( a = 1 case )
12, +1 No
6, +2 Yes Can stop
4, +3 No
d) F + O + I( ) + L = 3x2 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 = 3x2 ; 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) 4x
2+11x + 6 = 4x + 3( ) x + 2( )
Method 1: Trial-and-Error ("Guess") Method
( )( )
4x x
2x 2xF = 4x2
+ 1 + 6
+ 6 + 1
+ 2 + 3
+ 3 + 2
L = 6; need both " + " because of + 11x
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.
4x2+ 11x + 6 = 4x2
+ 8x + 3x
OK to switch
+ 6
= 4x2+ 8x( ) + 3x + 6( ) Group terms
= 4x x + 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 = 2x2 ; 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
4 2( ) 5( ) = 100 40 = 60 , which is not a perfect square, and
the GCF = 1, so the polynomial is prime.
g)
3x2+ 6x 3 = 3
GCF
x2 2x + 1( )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
x2( )
2
16
4( )2
= x2+ 4( )
prime
x2
x( )2
4
2( )2
= x2+ 4( ) x + 2( ) x 2( )
i) Use Factoring by Grouping:
a3 3a + 2a
2b 6b = a
3 3a( ) + 2a2b 6b( )
= a a2 3( ) + 2b a
2 3( )= a + 2b( ) a
2 3( )
j) 4x2+ 9y
2 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).
k) Apply the Sum of Two Cubes formula
a3+ b3
= a + b( )
"Expectedfactor"
a2 ab
NOT2ab
+ b2
The visible signs follow the pattern:same, different, "+"
:
x3
x( )3
+ 125y3
5y( )3
= x + 5y( ) x2 5xy + 25y2( )
l) Apply the Difference of Two Cubes formula
a3 b3= a b( )
"Expectedfactor"
a2+ ab
NOT+2ab
+ b2
The visible signs follow the pattern:same, different, "+"
:
x3
x( )3
125y3
5y( )3
= x 5y( ) x2 + 5xy + 25y2( ) . §
(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.
Then, u
2= x
3( )2
= 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.
2x3 +1( ) x3 1( ) = 2x3 +1( ) x 1( ) x2 + 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 x
3 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 2x 1
over the integers.
§ Solution
Observe that 7 is the least exponent on x. Our GCF is x7, so we will
factor it out and subtract 7 from each of the exponents.
x 7+ x 4 2x 1
= x 7 1+ x 4 7( ) 2x 1 7( )( )= x 7 1+ x 4+ 7 2x 1+ 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)
Factor x3 + 2( )1/3
+ x3 + 2( )5/3
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.
Observe that 5
3 is the least exponent on x3 + 2( ) . Our GCF is x3 + 2( )
5/3,
so we will factor it out and subtract
5
3 from each of the exponents on
x3 + 2( ) .
(Section 0.8: Factoring Rational and Algebraic Expressions) 0.8.4