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Unformatted text preview: (Preliminaries: Basic Algebra) P.30
TOPIC 5: RATIONAL AND ALGEBRAIC EXPRESSIONS
(See Section A.4.)
A rational expression in x can be expressed in the form Examples: polynomial in x
.
nonzero polynomial in x ⎛
1
5x3 − 1
x7 + x ⎞
,2
, x 7 + x ⎜ which equals
x x + 7x − 2
1⎟
⎝
⎠ Observe in the second example that irrational numbers such as
permissible. 2 are The last example correctly suggests that all polynomials are rational
expressions.
An algebraic expression in x looks like a rational expression, except that radicals and
⎛
5⎞
exponents that are noninteger rational numbers ⎜ such as ⎟ are allowed even when x
7⎠
⎝
appears in a radicand or in a base (but not in an exponent).
Examples: x, x 3 + 7 x5/7
x− 3 x+5+4 All rational expressions are algebraic. (Preliminaries: Basic Algebra) P.31
Venn diagram for standard symbolic mathematical expressions: (Preliminaries: Basic Algebra) P.32
TOPIC 6: FACTORING WITH “WEIRD” EXPONENTS:
NEGATIVE AND FRACTIONAL EXPONENTS;
QUADRATIC FORMS AND SUBSTITUTIONS
(See Section A.4: p.A41, and Section A.7: pp.A72A73.)
These manipulations come up throughout Calculus, especially Calculus I
(Math 150 at Mesa).
Basic Factoring Example
We factor 8 x + 6 as 2 ( 4 x + 3) .
Observe that, when we factor out a 2, we divide both terms in 8 x + 6 by 2.
Basic Factoring Example ( ) We factor x 5 + x 3 as x 3 x 2 + 1 .
Observe that we factor out the power of x with the lowest exponent.
Warning: All negative exponents are lesser than all positive exponents.
See the Example below.
When we divide x 5 by x 3 , we subtract the exponents in that order and get x 2 .
We can also think, “ x 3 times what gives us x 5 ?”
We can also think, “We’re factoring x 3 out of x 5 . 5 takeaway 3 is 2.”
Example
Factor x − 7 + x − 4 − 2 x − 1 completely over the integers.
Observe that − 7 is the lowest exponent on x.
We will factor out x − 7 and subtract − 7 from each of the exponents. ( x − 7 + x − 4 − 2 x − 1 = x − 7 1 + x − 4 − (− 7) − 2 x − 1 − (− 7) (
(1 + x = x − 7 1 + x − 4 + 7 − 2 x −1+ 7
= x− 7 3 − 2x6 ) ) ) (Preliminaries: Basic Algebra) P.33 ( ) Observe that 1 + x 3 − 2 x 6 has no negative exponents. ( ) We can actually factor x − 7 1 + x 3 − 2 x 6 further over the integers. We will
first factor out a − 1 factor so that the leading coefficient of our trinomial
factor is positive. Then, we will rewrite the trinomial in descending powers.
The trinomial will then be easier to factor. (
(2 x = − x− 7 −1 − x3 + 2x6
= − x− 7 6 ) ) − x3 −1 The trinomial 2 x 6 − x 3 − 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 nonzero constant. This means that the techniques for factoring
quadratic trinomials (see Topic 4) may be useful here.
If it helps, you may want to use the substitution u = x 3 , the power of x in the () “middle” term. Then, u 2 = x 6 , because u 2 = x 3 2 = x 6 . The quadratic form of the trinomial becomes more evident, and it may be easier to factor this
way: ( ) = − x − 7 2u 2 − u − 1 = − x − 7 ( 2u + 1) ( u − 1)
Now, substitute back by replacing u with x 3 . If you would rather avoid the
whole substitution process, you could have factored 2 x 6 − x 3 − 1 directly.
Either way, we obtain: ( ( ) x −1
) () = − x− 7 2x3 +1 ( ) 3 Difference
of Two
[Nice] Cubes ( ) = − x − 7 2 x 3 + 1 ( x − 1) x 2 + x + 1 This is factored completely over the integers.
2 x 3 + 1 ( x − 1) x 2 + x + 1
Note that this is equivalent to: −
.
x7 ( ) ( ) (Preliminaries: Basic Algebra) P.34
Example ( ) ( 1/ 3 Factor x 3 + 1 ) + x3 + 1 − 5/3 . Warning: Exponents do not distribute over sums. We cannot use the fact () 1/ 3 that x 3
= x . Likewise, 3 x 3 + 1 is not equivalent to x + 1 . The root of a
sum is not typically equal to the sum of the roots. (x ) ( ) − 5/3 ( 3 ⎡3
⎢ x +1
⎣ 1 ⎛ 5⎞
− − ⎟
3 ⎝ 3⎠ ⎡3
⎢ x +1
⎣ 15
+
33 ) − 5/3 )
+ 1)
+ 1) − 5/3 ⎡ x 3 + 1 2 + 1⎤
⎣
⎦ − 5/3 ⎡ x 6 + 2 x 3 + 1 + 1⎤
⎣
⎦ − 5/3 ⎡ x6 + 2x3 + 2⎤
⎣
⎦ = x +1 = x3 + 1
3 3 ( ) ⎤
+ 1⎥
⎦ ) (
= (x
= (x + x +1
3 3 ( +1 1/ 3 − 5/3 = x +1 3 ( ) ( ) ⎤
+ 1⎥
⎦ Although the trinomial x 6 + 2 x 3 + 2 is in quadratic form, it cannot be
factored (i.e., it is prime) over the integers. See the Test for Factorability in
Notes 1.48 and the solution to Problem f) in the Factoring Problems unit in
Topic 4.
Note that the last expression is equivalent to: x6 + 2x3 + 2 (x 3 ) +1 5/3 . (Preliminaries: Basic Algebra) P.35
TOPIC 7: SIMPLIFYING ALGEBRAIC EXPRESSIONS
(See Section A.4: pp.A40A42, and Section A.7: p.A74.)
We are typically expected to give our final answers in [completely] simplified form,
though there is some ambiguity about what that means. The “completely” goes without
saying.
For convenience, we will refer to the numerator (i.e., the top part) of a fraction as “N”
and the denominator (i.e., the bottom part) as “D.”
Example
Consider the fraction x
. Assume x ≠ 0 .
x2 We can divide the N and the D by x, and we obtain 1
.
x A “Canceling” Controversy
The spelling issue of one vs. two “l”s is the least of it!
Some would say that, in the above Example, we canceled a pair of xfactors
from the N and the D. However, some instructors object to the idea of
“canceling,” because students often abuse it. Instead of “canceling,” we may
prefer to use “dividing out” or “striking,” which has a more typographical
connotation.
Warning (“Too Much Canceling”): Students often cancel inappropriately.
x2
2
For example, we cannot cancel the “ x ”s in 2
, because x 2 is not a
x +4
factor of the entire denominator. (Preliminaries: Basic Algebra) P.36
Canceling / Dividing / Striking Rule for Fractions
As we simplify a fraction, we can “cancel”:
a nonzero factor of the entire numerator (“N”) with
an equivalent factor of the entire denominator (“D”).
More precisely, we can divide the entire N and the entire D by equivalent nonzero
factors.
Warning (“Too Little Canceling”): Sometimes, insufficient canceling can be a
problem.
If the entire N and the entire D of a fraction have any factors in common aside
from ±1 , then those factors must be canceled (or “divided out”) if we are to
simplify the fraction.
“Canceling” and Domain Issues
Warning: You may need to keep legality (i.e., domain) issues in mind when you
are canceling (dividing out).
Example x2
Consider the fraction
. At the outset, we assume x ≠ 0 .
x
x
We can divide the N and the D by x, and we obtain , or simply x.
1
It is still assumed that x ≠ 0 , even though the expression x in and of
1
itself does not imply that restriction the way that, say does.
x
We should explicitly state that restriction on our second expression:
x2
=x
x ( x ≠ 0) (Preliminaries: Basic Algebra) P.37
“Canceling” and the “Switch Rule for Subtraction”
Example 9 − x2
Simplify
. (It is typically automatically assumed that x ≠ 3 here.)
x−3
Solution ( )( ) Warning: 9 − x 2 factors as 3 + x 3 − x , which is the opposite of ( x + 3) ( x − 3) because of the “Switch Rule for Subtraction.”
It is true that ( 3 + x ) is equivalent to ( x + 3) , but ( 3 − x ) and ( x − 3)
are opposites: ( 3 − x ) = − ( x − 3) . ( )( 3+ x 3− x
9 − x2
=
x−3
x−3 ( ) ) The parentheses in the denominator may help. ( ) ( ) Because 3 − x and x − 3 are opposites, their
quotient in either order is − 1 : (3 + x ) (3 − x )
=
( x ≠ 3; see the previous Example)
( x − 3)
= − ( 3 + x ) or − 3 − x ( x ≠ 3)
−1 ( ) The choice between − 3 + x and − 3 − x may depend on context. (Preliminaries: Basic Algebra) P.38
Example Simplify ( 4 x + 7 )1/ 3 ( 2 x ) − ( x 2 ) ⎡ ( 4 x + 7 )− 2 / 3 ( 4 )⎤
⎢
⎥
1
⎣3
( 4 x + 7 )2 / 3 ⎦ (Given). You may not have nonpositive exponents in your final answer.
In Calculus: We do see monsters like these! (Given) is obtained by applying
x2
the Quotient Rule for Differentiation to 3
.
4x + 7
Solution
Let’s begin by “cleaning up” the N. 42
− 2/3
x ( 4 x + 7)
3
( 4 x + 7 )2 / 3 2 x ( 4 x + 7) 1/ 3 (Given ) = − Method 1: Factor first. (Factoring out the x in the N turns out to be unnecessary.)
1 ⎛ 2⎞
4⎤
− 2/3 ⎡
− ⎜− ⎟
42
− 2/3
x ( 4 x + 7 ) ⎢2 ( 4 x + 7 ) 3 ⎝ 3⎠ − x ⎥
2 x ( 4 x + 7) − x ( 4 x + 7)
3⎦
⎣
3
=
2/3
2/3
( 4 x + 7)
( 4 x + 7)
1/ 3 1 ⎛ 2⎞ 1 2 3
⎡
⎤
Fortunately, − ⎜ − ⎟ = + = = 1.⎥
⎢
3 ⎝ 3⎠ 3 3 3
⎣
⎦ = x ( 4 x + 7) 4⎤
⎡
⎢2 ( 4 x + 7) − 3 x ⎥
⎣
⎦
2/3
( 4 x + 7) − 2/3 ⎡
22
4
⎢
(blah )− 2 / 3
1
−−
−
Observe, with ( blah ) = ( 4 x + 7 ) :
= ( blah ) 3 3 = ( blah ) 3 =
⎢
(blah )2 / 3 Byetitule
(blah )4 / 3
⎢
th Quo en R
⎢
for Exponents
⎣ ⎤
⎥
⎥
⎥
⎥
⎦ (Preliminaries: Basic Algebra) P.39 4⎤
⎡
x ⎢2 ( 4 x + 7) − x ⎥
3⎦
=⎣
4 /3
( 4 x + 7)
4⎤
⎡
x ⎢ 8 x + 14 − x ⎥
3⎦
=⎣
4 /3
( 4 x + 7)
⎡ We will multiply the N and the D by 3 so that we may avoid a complex ⎤
⎢(or compound) fraction, which is unacceptable in a simplified answer. ⎥
⎣
⎦
4⎤
⎡
3x ⎢ 8 x + 14 − x ⎥
3⎦
⎣
=
4 /3
3( 4 x + 7 )
=
= 24 x 2 + 42 x − 4 x 2
3( 4 x + 7 ) 20 x 2 + 42 x 3( 4 x + 7 ) 4 /3 4 /3 or 2 x (10 x + 21)
3( 4 x + 7 ) 4 /3 Note: The benefit of factoring the N as our final step is to ensure that there
are, in fact, no more common factors in the N and the D, aside from ±1 .
Note: It is assumed throughout that x ≠ − 7
.
4 Method 2: First, rewrite using only positive exponents. 4 x2
1/ 3
42
− 2/3
2 x ( 4 x + 7) −
2/3
2 x ( 4 x + 7) − x ( 4 x + 7)
3( 4 x + 7 )
3
=
( 4 x + 7 )2 / 3
( 4 x + 7 )2 / 3
1/ 3 ⎡ We will multiply the N and the D of the overall fraction by the ⎤
⎢
⎥
2/3
⎢ LCD (Least Common Denominator), 3 ( 4 x + 7 ) , so that we ⎥
⎢ may avoid a complex (or compound) fraction.
⎥
⎣
⎦ (Preliminaries: Basic Algebra) P.40 ⎡
⎤
4 x2
⎡ 3 ( 4 x + 7 )2 / 3 ⎤ ⎢ 2 x ( 4 x + 7 )1/ 3 −
2/3 ⎥
⎣
⎦
3( 4 x + 7 ) ⎦
⎢
⎥
⎣
=
2/3
2/3
⎡ 3 ( 4 x + 7 ) ⎤ ⎡( 4 x + 7 ) ⎤
⎣
⎦⎣
⎦
Warning: Do not strike the expressions above in red until the Distributive
Property is applied first! Then, we may do some striking. If it helps, write
out the step below; with more experience, you may be able to skip it. ⎡
4 x2
⎡ 3 ( 4 x + 7 )2 / 3 ⎤ ⎡ 2 x ( 4 x + 7 )1/ 3 ⎤ − ⎡ 3 ( 4 x + 7 )2 / 3 ⎤ ⎢
⎣
⎦⎣
⎦⎣
⎦⎢
2/3
3( 4 x + 7 )
⎣
=
2/3
2/3
⎡ 3 ( 4 x + 7 ) ⎤ ⎡( 4 x + 7 ) ⎤
⎣
⎦⎣
⎦ ⎤
⎥
⎥
⎦ 213
⎡
⎤
Fortunately, when multiplying the powers of ( 4 x + 7 ) in the upper left, + = = 1.⎥
⎢
333
⎣
⎦
=
=
= 6 x ( 4 x + 7) − 4 x2
3( 4 x + 7 ) 4 /3 24 x 2 + 42 x − 4 x 2
3( 4 x + 7 ) 20 x 2 + 42 x 3( 4 x + 7 ) 4 /3 4 /3 or 2 x (10 x + 21)
3( 4 x + 7 ) 4 /3 Warning: Do not distribute the 3 in the D. The 4/3 exponent forbids that.
Rationalizing a Numerator in an Algebraic Expression
Read Example 11 on p.A42. (Preliminaries: Basic Algebra) P.41
TOPIC 8: COMMON ERRORS IN ALGEBRA
(See Section A.7: pp.A70A71.)
See all of my Warnings and Larson’s Study Tips.
Here’s another entry in “Errors Involving Fractions”:
Potential Error
a b+c
a−b+c
−
=
d
d
d Correct Form
a − (b + c )
a b+c
−
=
d
d
d Comment
A fraction groups together its numerator. (Preliminaries: Basic Algebra) P.42
TOPIC 9: MORE ALGEBRA TRICKS!
(See Section A.7: pp.A72A74.)
Splitting a Fraction Through its Numerator
Example 7x + 3
as a sum of two simplified terms, neither of which contains an
4
x
addition or a subtraction.
Write 7x + 3 7x + 3
= 1/ 4
4
x
x
7 x1
3
= 1/ 4 + 1/ 4
x
x
= 7x 1− 1
4 + 3 x − 1/ 4 = 7 x 3 / 4 + 3 x − 1/ 4 or 7 x 3/ 4 + 3
x 1/ 4 Warning: Do not split a fraction in this way through its denominator.
1
11
For example,
is not equivalent to + .
x+y
xy
1
11
= + .)
(In fact, there are no pairs of real values for x and y that make
x+y x y
For nonzero x and y: 11
+
xy = 1y
1x
⋅+⋅
xy
yx The LCD is xy. = y+x
x+y
or
xy
xy . (Preliminaries: Basic Algebra) P.43
Making Complex (or Compound) Fractions
We will do this in Chapter 10 on standard forms for equations of conics.
The key is that multiplying by a fraction (or any other nonzero quantity) is
equivalent to dividing by its reciprocal. (In your Algebra classes, you divided by a
fraction by multiplying by its reciprocal. Here, we do the reverse.)
Example 3y 2 9 2 3 2
9x +
= x+ y
4
1
4
⎡Observe that the y 2 can "pop out" like this ⎤
⎢
⎥
⎣ out of the N but not the D ( Warning!).
⎦
x2
y2
=
+
1/ 9 4 / 3
2 Compensation (Addition and Subtraction)
The idea behind compensation is that we have to “pay” for (nonzero) quantities we
insert into an expression in order to maintain equivalence.
Example
Divide x by x + 1 without using long division.
Solution
We would like to insert a “ + 1 ” term into the N, but we must then
“pay” for it with a “ − 1 ” term in the N. x
x +1−1
=
x +1
x +1
x +1
1
=
−
x +1 x +1
1
=1−
x +1
You also use this kind of trick when Completing the Square (CTS) in Section 2.1
and Chapter 10 on conics. (Preliminaries: Basic Algebra) P.44
Compensation (Multiplication and Division)
In Calculus: The usubstitution technique of integration (in Section 5.2 of the
Calculus I – Math 150 text at Mesa) uses this trick.
Example
Fill in the box to make an equivalent expression: ( 4 x + 1)5 = ( 4 x + 1)5 ( 4 ) Solution ( 4 x + 1)5 = 1
4 ( 4 x + 1)5 ( 4 ) If we insert a nonzero factor (here, 4) into the overall expression, we
must “pay” for it by inserting its reciprocal as a factor (i.e., dividing
the 4 out). ...
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This note was uploaded on 09/08/2011 for the course MATH 141 taught by Professor Staff during the Fall '11 term at Mesa CC.
 Fall '11
 staff
 Calculus, Algebra

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