Download M19500 Precalculus Chapter 1.4: Rational Expressions

Welcome
Algebra with numerical fractions
Algebra with polynomial fractions
Exercises
Quiz Review
Welcome to Math 19500 Video Lessons
Prof. Stanley Ocken
Department of Mathematics
The City College of New York
Fall 2013
An important feature of the following Beamer slide presentations is that you, the
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use the index dots at the top of this slide (or the index at the left, accessible from the
Adobe Acrobat Toolbar) to access the different sections of this document.
To prepare for the Chapter 1.4 Quiz (September 23rd at the start of class), please
Read all the following material carefully, especially the included Examples.
Memorize and understand all included Definitions and Procedures.
Work out the Exercises section, which explains how to check your answers.
Do the Quiz Review and check your answers by referring back to the Examples.
Stanley Ocken
M19500 Precalculus Chapter 1.4: Rational Expressions
Welcome
Algebra with numerical fractions
Algebra with polynomial fractions
Exercises
Quiz Review
What is a fraction?
A numerical fraction is a way of writing down a rational number. For example, 83 is the
length of the stick you get by pasting together 3 short sticks, each obtained by
splitting a length 1 (unit) stick into 8 equal parts. In a textbook or on a calculator
display, this fraction might be written as 3/8.
The ”unit stick” is the interval [0, 1] on the number line, shown below, with the blue
line segment showing the rational number 3/8.
0
1
8
2
8
3
8
4
8
5
8
6
8
7
8
1
Another example of a fraction is 17
7 . This sort of fraction (numerator ≥ denominator)
is sometimes called improper, but there’s absolutely nothing wrong with it!
There are at least six different ways to write a fraction, as follows:
3
1
1 1 1
1
= 3/8 = 3 ÷ 8 = 3 · = + + = · 3
8
8
8 8 8
8
Stanley Ocken
M19500 Precalculus Chapter 1.4: Rational Expressions
Welcome
Algebra with numerical fractions
Algebra with polynomial fractions
Exercises
Quiz Review
Every fraction is a real number. To find that number, divide the numerator by the
denominator. For example, 3/8 is 3 divided by 8. In this division we say that 3 is the
dividend, 8 is the divisor, and 3/8 is the quotient.
To express a fraction as a decimal number, use long division.
Sometimes you will get a repeating non-terminating decimal:
1/3 = .333333333..... or 1/7 = .142857 142857 142857....
You will get a terminating decimal if and only if the denominator is a power of 2 times
a power of 5. In such a case, you can do the calculation by building up the fraction’s
denominator to a power of 10. For example,
3
3
3 · 53
3 · 125
375
= 3 = 3 3 =
=
= .375
3
8
2
2 ·5
1000
10
Stanley Ocken
M19500 Precalculus Chapter 1.4: Rational Expressions
Welcome
Algebra with numerical fractions
Algebra with polynomial fractions
Exercises
Quiz Review
A plan for adding fractions
When you glue together two or more sticks, you want to find their total length. To do
so, you need to add fractions. You probably know that 53 + 47 can be computed as
41
3·7+5·4
= . However, this memorized procedure isn’t enough.
7
7
3
7
7
For example, can you add
+
+ ? Your answer should be a reduced fraction.
20 16 24
To do this problem in a sensible way, you need to rewrite the three fractions so that
they have the same denominator. You also want that denominator to be as small as
possible. In other words, you want to use the three fractions’ least common
denominator (LCD).
In this chapter we will need to find the LCD for addition problems involving both
numerical fractions (such as those listed above) as wells as polynomial fractions, such
b
c
x
x+7
2
a
as + + and 2
+
+
.
x y z
x + 4x + 3 (x + 1)2 (x + 3)2
Stanley Ocken
M19500 Precalculus Chapter 1.4: Rational Expressions
Welcome
Algebra with numerical fractions
Algebra with polynomial fractions
Exercises
Fraction identities
Definition
A numerical fraction is a fraction
P
where both P and Q are integers.
Q
The following fraction identities such be familiar:
Fraction identities
C
AC
D = BD
A/B
A D
AD
A
B ÷
C/D = B · C = BC
A
C
or subtraction: B
+D
= AD+BC
BD
Fraction multiplication:
Fraction division:
Fraction addition
A
B ·
C
D =
and
A
B
−
C
D
=
AD−BC
BD
When you substitute numbers for A, B, C, D , use parentheses:
A
C
AD + BC
1 3
(1)(4) + (2)(3)
+
=
becomes + =
B D
BD
2 4
(2)(4)
Stanley Ocken
M19500 Precalculus Chapter 1.4: Rational Expressions
Quiz Review
Welcome
Algebra with numerical fractions
Algebra with polynomial fractions
Exercises
Quiz Review
Definition
A reduced fraction is a fraction whose numerator and denominator have no common
factor other than 1 and −1.
Procedure
To reduce a fraction
Factor the top completely.
Factor the bottom completely.
A
C
Cancel common factors by using the identity B
=
C
A
B.
AC
A
C
A C
A
A
We can cancel B
because BC
=B
·C =B
·1= B
C
A
C
A
CA
A
or =B
. Here are some examples:
We usually write: B
=B
C
CB
12
6
2·6
14 = 2·7 = 7
100
25·
4
25
64 = 16·4 = 16
10
5
450
45·
45
= 180
= 5·9 = 20
= 14
1800 = 180·
10
20·
9
9
6
9−5
4
2 ·3
= 239−6 = 323 as discussed in Section
25 ·39
Stanley Ocken
1.2.
M19500 Precalculus Chapter 1.4: Rational Expressions
Welcome
Algebra with numerical fractions
Algebra with polynomial fractions
Exercises
Please be extremely careful when you apply the fraction reduction identity
Quiz Review
AC
BC
Warnings: It is A SERIOUS ERROR to write any of the following:
5+4
5+5
= 5+4 =
4
7
5+7
5·2
2
5·2
5+7 = 5+7 = 7
5+7
7
5+7
5·2 = 5·2 = 2
x
x
xy = xy = y
Rewrite as
xy
x
x
xy
=
= xy
x
y
xy
x
Rewrite as
=
Error: 5 is NOT a factor of either the top or bottom.
Error: 5 is NOT a factor of the bottom.
Error: 5 is NOT a factor of the bottom.
Error: The numerator is not a product.
x·1
xy
x1 =
= xy
1
y
Error: The denominator is not a product.
=
xy
x·1
= xy
=
x1
y
1
=y
Stanley Ocken
M19500 Precalculus Chapter 1.4: Rational Expressions
=
A
B
Welcome
Algebra with numerical fractions
Algebra with polynomial fractions
Exercises
Fractions are easy to multiply but difficult to add.
As stated above: to multiply fractions, multiply their numerators and multiply their
A C
AC
denominators: B
· D = BD
. What could be easier!
In the above formula, substitute C for D to get
A
A
A C
·C =B
·1= B
and this gives us two
But B
A
B
·
C
C
=
AC
BC
Procedures for rewriting fractions
Fraction cancellation:
A
C
B
C
=
A
B,
used to reduce fractions.
A
AC
Fraction building: B
= BC
, used to build up fractions so that they have a
common denominator when you want to add or subtract fractions.
Stanley Ocken
M19500 Precalculus Chapter 1.4: Rational Expressions
Quiz Review
Welcome
Algebra with numerical fractions
Algebra with polynomial fractions
Exercises
Quiz Review
A
AC
The fraction building rule B
= BC
says that you can rewrite a fraction by multiplying
both numerator and denominator by the same number C. When we do this, we say we
are building the fraction with building factor C.
Warning: The ONLY thing you can do to both numerator and denominator is to
multiply by the same number C. If you do anything else, you will change the fraction:
that’s something you never want to do. For example:
A
B
A+C
B+C
A
A2
If you square both the top and bottom you change the fraction: B
6= B
2
A
A+C
A
A−C
A
A2
In other words, never write B = B+C or B = B−C or B = B 2 .
If you add C to top and bottom, you change the fraction:
6=
These are INCORRECT identities. If you use them, you will receive zero partial
credit.
DON’T INVENT IDENTITIES.
Stanley Ocken
M19500 Precalculus Chapter 1.4: Rational Expressions
Welcome
Algebra with numerical fractions
Algebra with polynomial fractions
Exercises
Quiz Review
Adding fractions
The following very basic rule follows from the Distributive Law:
Procedure
Adding fractions with the same denominator:
A
C
+
B
C
=
A+B
C
Most of the time, addition problems involve different denominators. Fraction building
allows us to add fractions with different denominators by rewriting them as fractions
with the same denominator:
AD
C
BC
A
C
AD
BC
AD+BC
A
B = BD and D = BD and so B + D = BD + BD =
BD . This gives us a
Procedure
Adding fractions with different denominators :
A
B
+
C
D
=
AD+BC
BD
This procedure is often inefficient. We will improve it after doing some basic examples.
Example 1: Substituting 4 for A, 5 for B, 6 for C, and 7 for D gives
4 6
A
C
AD + BC
(4)(7) + (5)(6)
28 + 30
58
+ =
+
=
=
=
=
5 7
B D
BD
(5)(7)
35
35
Stanley Ocken
M19500 Precalculus Chapter 1.4: Rational Expressions
Welcome
Algebra with numerical fractions
Algebra with polynomial fractions
Exercises
Quiz Review
In the above example, we were careful to put parentheses around the substituted
numbers. However, in this case, using a dot for multiplication also works:
4 6
4·7+6·5
28 + 30
58
+ =
=
=
5 7
5·7
35
35
This answer is a reduced fraction, since 58 and 35 don’t have a common factor.
It’s fine to use a so-called improper fraction, with numerator larger than denominator.
In this course, do not rewrite improper fractions as mixed numbers.
C
A
+D
= AD+BC
introduces unpleasantly large
In many cases, however, the formula B
BD
numbers and also gives an answer that is not a reduced fraction.
7
11
7 · 3000 + 2000 · 11
21000 + 22000
43000
Example 2:
+
=
=
=
2000 3000
6000000
6000000
6000000
43000
43 · 1000
43
In this example, it is easy to reduce the sum:
=
= 6000 .
6000000
6000 · 1000
Multiplying denominators produces numbers that are (usually) much larger than
necessary. We want to rewrite fractions so that they have the same denominator, but
that denominator should be as small as possible.
Stanley Ocken
M19500 Precalculus Chapter 1.4: Rational Expressions
Welcome
Algebra with numerical fractions
Algebra with polynomial fractions
Exercises
Quiz Review
7
11
To find 2000
+ 3000
in a sensible way, we need to build both fractions so that they have
the same denominator. That denominator will be a multiple of both 2000 and 3000,
and should be as small as possible.
Definition
If a and b are positive integers, then a is a multiple of b if a = kb for some integer k.
The Least Common Multiple (LCM) of two or more positive integers is the smallest
positive integer that is a multiple of each of them.
One way to try to find the LCM of 2000 and 3000 ( which we abbreviate as
LCM(2000,3000) ) is to list multiples of 2000 until you get a multiple of 3000. That
method works nicely in this example, since the multiples of 2000 are 2000, 4000, 6000.
Stop here, since 6000 is a multiple of 3000. Thus LCM(2000,3000) = 6000.
11
7
+
, we use the LCM to build up both fractions as follows:
To find
2000 3000
7
7·3
21
11
11 · 2
22
=
=
and
=
=
.
2000
2000 · 3
6000
3000
3000 · 2
6000
7
11
21
22
43
Add these fractions to get
+
=
+
=
.
2000 3000
6000 6000
6000
Stanley Ocken
M19500 Precalculus Chapter 1.4: Rational Expressions
Welcome
Algebra with numerical fractions
Algebra with polynomial fractions
Exercises
Quiz Review
LCM and LCD
In the last example, we used the fact that LCM(2000,3000) = 6000 to add fractions
with denominators 2000 and 3000. To solve harder fraction addition problems, we need
to practice a technique called fraction building.
Procedure for fraction building
A
B as a fraction
AC
AC
BC = D
Assume B is a factor of D. To rewrite
solve BC = D for C and write
3
5
Example 3: Rewrite
Solution:
3
5
=
3C
5C
=
Solution:
3
120
=
3C
120C
First divide by 20
Next divide by 6
=
with denominator D,
as a fraction with denominator 50.
3C
50
3
120
Example 4: Rewrite
A
B
=
. Thus 5C = 50 ⇒ C = 10. and so
3
3 · 10
30
=
=
5
5 · 10
50
as a fraction with denominator 18000.
3C
18000
and so we need to solve 120C = 18000 .
3
3C
3(150)
450
6C = 900
Answer:
=
=
=
.
C = 150.
120
18000
18000
18000
Stanley Ocken
M19500 Precalculus Chapter 1.4: Rational Expressions
Welcome
Algebra with numerical fractions
Algebra with polynomial fractions
Exercises
Quiz Review
Exercise: Find w, x, and y in the following:
3
w
=
5
50
3
x
=
16
480
27
y
=
64
1600
3
as a fraction with denominator 50 · 28 · 711 .
5 · 22 · 73
Solution: We want to build the denominator 5 · 22 · 73 up to 50 · 28 · 711 .
In other words, solve 5 · 22 · 73 · X = 50 · 28 · 711 for X.
Find X by building its three factors separately:
Example 5: Rewrite
5 · ? = 50 22 · ? = 28 73 · ? = 711
The product of the blue building factors
5 · 10 = 50 22 · 26 = 28 73 · 78 = 711 .
on the last line is the total building factor 10 · 26 · 78 . Therefore:
3
3
10 · 26 · 78
30 · 26 · 78
=
·
=
5 · 22 · 73
5 · 22 · 73 10 · 26 · 78
50 · 28 · 711
Stanley Ocken
M19500 Precalculus Chapter 1.4: Rational Expressions
Welcome
Algebra with numerical fractions
Algebra with polynomial fractions
Exercises
Finding the least common denominator of two or more fractions
To add fractions, rewrite them so that their denominators are as small as possible.
Definition
The Least Common Denominator of Fractions is the LCM of their denominators.
Procedure for finding the Least Common Multiple (LCM) of whole numbers
Write the prime power factorization of each whole number.
For each prime that appears, choose the highest power of that prime.
The LCM is the product of those highest powers.
Example 6: Find the LCM of 2000 and 3000 by using the above method.
Solution: Prime power factorizations are 2000 = 24 · 53 and 3000 = 23 · 31 · 53 .
The highest power of 2 is 24 (appears in 2000 = 24 · 53 ).
The highest power of 3 is 31 = 3 (appears in 3000 = 23 · 31 · 53 ).
The highest power of 5 is 53 (appears in 3000 = 23 · 31 · 53 ).
Answer : LCM(2000,3000) is the product of highest powers:
Stanley Ocken
24 · 3 · 53 = 6000.
M19500 Precalculus Chapter 1.4: Rational Expressions
Quiz Review
Welcome
Algebra with numerical fractions
Algebra with polynomial fractions
Exercises
Quiz Review
The procedure just used was harder than the more basic method of listing the multiples
of the smaller number 2000. In general, however, our procedure is more efficient (as in
the example below) or absolute necessary (when we deal with polynomials).
Example 7: Find the LCM of 35 and 77.
Solution: Prime power factorizations are 35 = 5 · 7 and 77 = 7 · 11.
The highest power of 5 is 51 = 5 .
The highest power of 7 is 71 = 7 .
The highest power of 11 is 111 = 11.
Answer : LCM (35,77) is the product of highest powers: 5 · 7 · 11 = 385.
Try doing this problem by listing multiples of 35 until you get to a multiple of 77.
Procedure for Adding Fractions
The LCD of the fractions is the LCM of their denominators.
Find building factors: what multiple of each denominator gives the LCD?
Multiply numerator and denominator of each fraction by its building factor.
Add the built-up fractions by adding their numerators.
Reduce that answer to lowest terms.
Stanley Ocken
M19500 Precalculus Chapter 1.4: Rational Expressions
Welcome
Algebra with numerical fractions
Algebra with polynomial fractions
Exercises
Quiz Review
7
11
Example 8: Find
+
by using the above procedure for finding the LCD.
2000 3000
Solution:
From Example 6, LCM (2000, 3000) = 6000. This is the LCD of the fractions.
Find the building factors: what multiple of each denominator is the LCD 6000 ?
2000 · 3 = 6000
3000 · 2 = 6000
Rewrite the fractions by building both numerators and denominators.
7
11
7·3
21
11 · 2
22
=
=
=
=
2000
2000 · 3
6000
3000
3000 · 2
6000
21
22
43
Add the rewritten fractions:
+
=
6000 6000
6000
This answer can’t be reduced since 43 is a prime and is not a factor of 6000.
Answer:
7
11
43
+
=
.
2000 3000
6000
Stanley Ocken
M19500 Precalculus Chapter 1.4: Rational Expressions
Welcome
Algebra with numerical fractions
Algebra with polynomial fractions
Exercises
Quiz Review
2
5
Example 9: Find
+ .
35 77
Solution:
The LCD of the fractions is LCM(35,77). As in Example 7 above, prime power
factorizations are 35 = 5 · 7 and 77 = 7 · 11, and so the LCM is the product of
highest prime powers: 5 · 7 · 11. Don’t multiply out, since doing so will make some
of the following arithmetic more difficult.
Find the building factors needed to build the denominators 35 = 5 · 7 and
77 = 7 · 11 up to that LCD:
5 · 7 · 11 = 5 · 7 · 11
5 · 7 · 11 = 5 · 7 · 11
Rewrite the fractions by building both numerator and denominators:
2
2 · 11
22
5
5·5
25
=
=
=
=
35
5 · 7 · 11
5 · 7 · 11
77
5 · 7 · 11
5 · 7 · 11
22
25
22 + 25
47
Add the rewritten fractions:
+
=
=
.
5 · 7 · 11 5 · 7 · 11
5 · 7 · 11
5 · 7 · 11
2
5
47
Since 47 is prime, no cancelling is possible and
+
=
.
35 77
385
Stanley Ocken
M19500 Precalculus Chapter 1.4: Rational Expressions
Welcome
Algebra with numerical fractions
Algebra with polynomial fractions
Exercises
Quiz Review
Adding three fractions is more difficult.
3
7
7
Example 10: Find
+
+
20 16 24
Solution:
Find the LCM of 20 = 22 · 5; 16 = 24 ; and 24 = 23 · 3 as the product of highest
prime powers 24 · 3 · 5 = 240. This is the LCD of the fractions.
Find the building factors needed to build the denominators up to that LCD:
20 · 12 = 240
16 · 15 = 240
24 · 10 = 240
Rewrite the fractions by building both numerator and denominators:
3 · 12
36
7
7 · 15
105
7
7 · 10
70
3
=
=
=
=
=
=
20
20 · 12
240
16
7 · 15
240
24
24 · 10
240
36
105
70
36 + 105 + 70
211
Add the rewritten fractions
+
+
=
=
240 240 240
240
240
2
2
See if the fraction reduces. Since 13 < 211 < 19 , the possible prime factors of
211 are 2, 3, 5, 7, 11, and 13. Since 211 is not evenly divisible by any of these, 211
211
3
7
7
211
is prime, the numerator of
does not factor, and so
+
+
=
.
240
20 16 24
240
Challenge: try to do this problem without first finding the LCD.
Stanley Ocken
M19500 Precalculus Chapter 1.4: Rational Expressions
Welcome
Algebra with numerical fractions
Algebra with polynomial fractions
Exercises
Quiz Review
Complex Fractions
Definition
A complex fraction is an expression of the form E
F in which E and/or F contains a
fraction (which might be written as a negative power). These contained fractions are
sometimes called nested fractions.
Note: complex fractions have nothing to do with complex numbers!
1
1
1+
5 + 6 and
4
2
5
5
+
7−3·
4 23
4
1+
Examples of complex numerical fractions include
Procedure
To simplify a complex fraction:
Multiply both numerator and denominator by the LCD of all the nested fractions.
Rewrite numerator and denominator as numbers.
Reduce the resulting fraction.
Stanley Ocken
M19500 Precalculus Chapter 1.4: Rational Expressions
Welcome
Algebra with numerical fractions
Example 11: Rewrite
Solution Method 1:
Algebra with polynomial fractions
Exercises
Quiz Review
5 + 7−2
without negative exponents.
3
1
5+ 2
1
7 by 72 :
Rewrite 7−2 as 2 , then multiply top and bottom of
7
3
5 + 712
5 + 712 72
49 · 5 + 1
245 + 1
246
82 · 3
82
72 · 5 + 1
=
=
=
=
=
=
=
2
2
3
3·7
7 ·3
49 · 3
147
147
49 · 3
49
Solution Method 2:
Multiply top and bottom of the original fraction
5 + 7−2
by 72 :
3
5 + 7−2 72
5 + 7−2
246
82 · 3
82
5 · 72 + 7−2 · 72
5 · 72 + 1
=
= 2
=
=
=
=
2
2
2
3
3·7
3·7
7 ·3
7 ·3
49 · 3
49
Here we used: 7−2 · 72 = 7−2+2 = 70 = 1.
Stanley Ocken
M19500 Precalculus Chapter 1.4: Rational Expressions
Welcome
Algebra with numerical fractions
Algebra with polynomial fractions
Exercises
Quiz Review
Algebra with polynomial fractions
Definition
A rational expression is any expression involving one or more fractions, which may
have numerators and denominators involving both numbers and variables.
In earlier slides we discussed numerical fractions such as 37/44. The rules for working
with fractions stay the same when the fractions involve polynomials:
Fraction identities
Fraction multiplication:
Fraction division:
A
B
÷
A
B
C
D
·
=
C
D
=
A
B
C
D
Fraction addition or subtraction:
AC
BD
=
A
B
·
D
C
A
B
+
C
D
=
=
AD
BC
AD+BC
BD
and
A
B
−
C
D
=
AD−BC
BD
When you substitute polynomials such as x + 1, x + 2, x + 3, x + 4 for A, B, C, D in identities
(x+1)(x+4)+(x+2)(x+3)
x+1
A
C
such as B
+D
= AD+BC
, use parentheses: x+2
+ x+3
BD
x+4 =
(x+2)(x+4)
Stanley Ocken
M19500 Precalculus Chapter 1.4: Rational Expressions
Welcome
Algebra with numerical fractions
Algebra with polynomial fractions
Exercises
The procedure for reducing fractions with variables is the same as for reducing
numerical fractions.
Procedure
To reduce a fraction
Factor the top completely.
Factor the bottom completely.
Cancel common factors.
AC
A
C
A C
A
A
We can cancel B
because BC
=B
·C =B
·1= B
C
A
C
A
We usually write: B
=B
. Here are some examples:
C
x2 +4x
= x(x+4)
= x+4
x2 +5x
x(x+5) x+5
1
(x+2)
x+2
1
=
= x+3
(x+2)(x+3)
x2 +5x+6
2 2 −1)
2
(x+1)
x4 −x2
= xx3(x
= x (x−1)
x4 +x3
(x+1)
x3
(x+1
=
x−1
x
2
2
x +4x
+4x
Warning: It is A SERIOUS ERROR to write xx2 +5x
=
=
x2 +5x
2
Reason: x is NOT a factor of top or bottom.
Stanley Ocken
4x
5x
M19500 Precalculus Chapter 1.4: Rational Expressions
Quiz Review
Welcome
Algebra with numerical fractions
Algebra with polynomial fractions
Exercises
Fractions are easy to multiply but difficult to add.
As stated above: to multiply fractions, multiply their numerators and multiply their
A C
AC
denominators: B
· D = BD
. What could be easier!
In the above formula, substitute C for D to get
A
A
A C
·C =B
·1= B
and this gives us two
But B
A
B
·
C
C
=
AC
BC
Procedures for rewriting fractions
Fraction cancellation:
Fraction building:
A
B
AC
BC
=
A
B,
used to reduce fractions.
A
C
=B
, used to add or subtract fractions.
C
Stanley Ocken
M19500 Precalculus Chapter 1.4: Rational Expressions
Quiz Review
Welcome
Algebra with numerical fractions
Algebra with polynomial fractions
Exercises
Quiz Review
AC
A
The fraction building rule B
= BC
says that you can rewrite a fraction by multiplying
both numerator and denominator by the same expression C. When we do this, we say
we are building the fraction with building factor C.
Warning: The ONLY thing you can do to both numerator and denominator is to
multiply both by the same expression C. If you do anything else, you will change the
fraction: that’s something you never want to do.
A+C
B+C
A
A2
fraction: B
6= B
2
If you add C to top and bottom, you change the fraction:
If you square both the top and bottom you change the
In other words
A
B
=
A+C
B+C
and
A
B
=
A2
B2
A
B
6=
are INCORRECT identities.
DON’T INVENT IDENTITIES.
Stanley Ocken
M19500 Precalculus Chapter 1.4: Rational Expressions
Welcome
Algebra with numerical fractions
Algebra with polynomial fractions
Exercises
Procedure
To multiply fractions,
• factor the numerators
• factor the numerators
• multiply the fractions to get a single fraction
• reduce the resulting fraction completely.
2
2x +7x−15
Example 12: Find 4xx+3
2 −9 · x2 +7x+12
Solution:
(2x−3)(x+5)
x+3
(2x+3)(2x−3) · (x+3)(x+4)
=
=
=
(x+3)(2x−3)(x+5)
(2x+3)(2x−3)(x+3)(x+4)
(2x−3)(x+5)
(x+3)
(2x+3)
(2x−3)
(x+3)(x+4)
(x+5)
(2x+3)(x+4)
Stanley Ocken
M19500 Precalculus Chapter 1.4: Rational Expressions
Quiz Review
Welcome
Algebra with numerical fractions
Algebra with polynomial fractions
Exercises
Procedure
To divide by a fraction, multiply by its reciprocal: X ÷
A
B
=X·
B
A
2
x +7x+12
Example 13: Find 4xx+3
2 −9 ÷ 2x2 +7x−15
Solution: Multiply the first fraction by the reciprocal of the second:
2 +7x−15
x+3
· 2x
4x2 −9
x2 +7x+12
=
=
=
=
(2x−3)(x+5)
x+3
(2x+3)(2x−3) · (x+3)(x+4)
(x+3)(2x−3)(x+5)
(2x+3)(2x−3)(x+3)(x+4)
(x+3)
(2x−3)(x+5)
(2x+3)
(2x−3)
(x+3)(x+4)
(x+5)
(2x+3)(x+4)
Stanley Ocken
M19500 Precalculus Chapter 1.4: Rational Expressions
Quiz Review
Welcome
Algebra with numerical fractions
Algebra with polynomial fractions
Exercises
Quiz Review
Adding polynomial fractions
The following very basic rule follows from the Distributive Law:
Procedure
Adding fractions with the same denominator:
A
C
+
B
C
=
A+B
C
Fraction building allows us to add fractions with different denominators by rewriting
them as fractions with the same denominator:
AD
C
BC
A
C
AD
BC
AD+BC
A
B = BD and D = BD ⇒ B + D = BD + BD =
BD . This gives us a
Procedure for adding fractions
A
C
AD + BC
+
=
B D
BD
As with the case of numerical fractions, this formula is correct but inefficient.
Stanley Ocken
M19500 Precalculus Chapter 1.4: Rational Expressions
Welcome
Algebra with numerical fractions
Algebra with polynomial fractions
Exercises
Quiz Review
Building polynomial fractions
Procedure for polynomial fraction building
A
Suppose you want to rewrite B
as a fraction with denominator D.
• Find C so that BC = D . Then
A
AC
• B
= BC
= AC
D
Example 14: Rewrite
3
x2
as a fraction with denominator x5 .
Solution: Since x2 · x3 = x5 , then
Example 15: Rewrite
ab
x2 y 3
3
x2
=
3·x3
x2 ·x3
=
3x3
x5
.
as a fraction with denominator x5 y 8 .
3 y5
abx3 y 5
3
= x3·x
.
2 ·x3 y 5 =
x2 y 3
x5 y 8
a b
c
x , y , and z with denominator
cxy
cxy
c
z = zxy = xyz
Solution: Since x2 · x3 = x5 and y 3 · y 5 = y 8 then
Example 16: Rewrite each of the three fractions
Solution:
a
x
=
ayz
xyz
=
ayz
xyz
b
y
=
bxz
yxz
=
Stanley Ocken
bxz
xyz
M19500 Precalculus Chapter 1.4: Rational Expressions
xyz.
Welcome
Algebra with numerical fractions
Algebra with polynomial fractions
Exercises
3
as a fraction with denominator 50c8 d11 .
5c2 d3
Solution: We want to build the denominator 5c2 d3 up to 50c8 d11 .
In other words, solve 5c2 d3 · X = 50c8 d11 for X.
Find X by building its three factors separately:
Example 17: Rewrite
5 · ? = 50 , c2 · ? = c8 , and d3 · ? = d11
5 · 10 = 50 , c2 · c6 = c8 , and d3 · d8 = d11 .
The product of the blue building factors is 10 · c6 · d8 . Therefore:
3
3
10 · c6 · d8
30c6 d8
=
·
=
5c2 d3
5c2 d3 10 · c6 · d8
50c8 d11
Stanley Ocken
M19500 Precalculus Chapter 1.4: Rational Expressions
Quiz Review
Welcome
Algebra with numerical fractions
Algebra with polynomial fractions
Exercises
Quiz Review
Adding polynomial fractions: LCM and LCD once more
When we found the LCM of whole numbers, we used prime power factorizations. To
imitate that procedure, we need to understand what is meant by a prime polynomial.
Definitions
Polynomial R is a factor of polynomial P if there is a polynomial Q with
P = Q · R.
A polynomial P is prime provided its only factors are the obvious ones:
P, −P, 1, and −1.
Examples of prime polynomials are x, y, x + 4, x2 + 1, 3x + 10, and x2 + 3x − 12.
It should be obvious that the first five do not factor. The following result tells us
(without trial and error) that the last polynomial does not factor either.
Procedure: the quadratic polynomial factoring criterion
2 + bx + c with integer coefficients factors as a product of
A quadratic polynomial ax√
polynomials if and only if b2 − 4ac is a whole number.
Stanley Ocken
M19500 Precalculus Chapter 1.4: Rational Expressions
Welcome
Algebra with numerical fractions
Algebra with polynomial fractions
Exercises
Quiz Review
Example 18: Use the quadratic factoring criterion to decide if each of the following
polynomials factors: • x2 + 19x + 34
• x2 + 19x + 35
• x2 + 10x + 30.
Solution:
2
2
x
√ + 19x + 34√= ax + bx + c. Here
√ c = 34 and
√ a = 1; b = 19,
2
2
b − 4ac = 19 − 4 · 1 · 34 = 361 − 136 = 225 = 15 is a whole number.
Therefore x2 + 19x + 34 factors.
2
2
x
√ + 19x + 35√= ax + bx + c. Here
√ c = 35 and
√ a = 1; b = 19,
2
2
b − 4ac = 19 − 4 · 1 · 35 = 361 − 140 = 221
√
To decide if 221 is a whole number, examine nearby squares 142 = 196 and
2 < 221 < 152 . Taking square roots gives
152 =√225. Therefore 14√
14 < 221 < 15. Thus 221 is not a whole number and so
x2 + 19x + 35 does not factor.
2
2
x
√ + 10x + 30√= ax + bx + c. Here
√ a = 1; b = 10,
√ c = 30 and
b2 − 4ac = 102 − 4 · 1 · 30 = 100 − 120 = −20, the square root of a
negative number, is undefined and is certainly not a whole number. Therefore
x2 + 10x + 30 does not factor.
Stanley Ocken
M19500 Precalculus Chapter 1.4: Rational Expressions
Welcome
Algebra with numerical fractions
Algebra with polynomial fractions
Exercises
Quiz Review
Now that we know what it means for a polynomial to be prime, we can translate the
LCM method for integers (use prime power factorizations) to find the LCM of
polynomials. However, it’s not always easy to factor polynomials into prime
polynomials. Nevertheless, we have the following:
Procedure for finding the Least Common Multiple (LCM) of polynomials
Write the prime power factorization of each polynomial.
For each prime polynomial that appears, choose the highest power of that prime.
The LCM is the product of those highest powers.
If the factorizations involve powers of non-prime polynomials, the above procedure will
produce a common multiple, albeit possible not the least common multiple.
Definition
The Least Common Denominator (LCD) of polynomial fractions is the Least Common
Multiple (LCM) of the denominators of those fractions.
In order to add polynomial fractions, we will need to find their LCD.
Stanley Ocken
M19500 Precalculus Chapter 1.4: Rational Expressions
Welcome
Algebra with numerical fractions
Algebra with polynomial fractions
Exercises
Example 19: Find the LCM of x3 y 4 and x2 y 3 z 5
Each of these is already written as a prime power factorization.
The highest power of x in the above two factors is x3 (in x3 y 4 )
The highest power of y in the above two factors is y 4 (in x3 y 4 )
The highest power of z in the above two factors is z 5 (in x2 y 3 z 5 )
Answer: The LCM is the product of highest powers: x3 y 4 z 5
Example 20: Find the LCM of x2 + 3x + 2 and x2 + 4x + 4
This example requires factoring the polynomials.
x2 + 3x + 2= (x + 2)(x + 1) and x2 + 4x + 4=(x + 2)2 .
Highest power of (x + 1) is (x + 1) = (x + 1)1 ,appearing in (x + 2)(x + 1).
Highest power of (x + 2) is (x + 2)2 appearing in (x + 2)2 .
Answer: The LCM is the product of highest powers:
(x + 1)(x + 2)2 .
Check that this answer is a multiple of each denominator:
LCM = (x + 1)(x + 2)2 = (x2 + 3x + 2)(x + 2)
LCM = (x + 1)(x + 2)2 = (x2 + 4x + 4)(x + 1)
Stanley Ocken
M19500 Precalculus Chapter 1.4: Rational Expressions
Quiz Review
Welcome
Algebra with numerical fractions
Algebra with polynomial fractions
Exercises
Procedure for adding polynomial fractions
The LCD of the fractions is the LCM of their denominators.
Find building factors: what multiple of each denominator gives the LCD?
Multiply numerator and denominator of each fraction by its building factor.
Add the built-up fractions by adding their numerators.
Reduce that answer to lowest terms.
3z
4
Example 21: Find the sum 3 4 + 2 3 5
x y
x y z
The fractions’ LCD is the LCM of the denominators = x3 y 4 z 5 (from above)
Find building factors: x3 y 4 z 5 = x3 y 4 z 5 and x2 y 3 z 5 xy = x3 y 4 z 5
Multiply top and bottom of each fraction by that building factor:
3z
3zz 5
3z 6
4
4xy
4xy
=
=
and 2 3 5 = 2 3 5
= 3 4 5
3
4
3
4
5
3
4
5
x y
x y z
x y z
x y z
x y z xy
x y z
4xy
3z 6 + 4xy
3z 6
+
=
x3 y 4 z 5
x3 y 4 z 5
x3 y 4 z 5
The answer is a reduced fraction since numerator does not factor.
Add the fractions:
Stanley Ocken
M19500 Precalculus Chapter 1.4: Rational Expressions
Quiz Review
Welcome
Algebra with numerical fractions
Algebra with polynomial fractions
Example 22: Use the LCM from Example 16 to simplify
Solution:
x2
Exercises
Quiz Review
x
x+3
+ 2
+ 3x + 2 x + 4x + 4
LCD= the LCM found above = (x + 1)(x + 2)2
x(x + 2)
x
x
x2 + 2x
=
=
=
x2 + 3x + 2
(x + 2)(x + 1)
(x + 2)(x + 1)(x + 2)
(x + 2)2 (x + 1)
2
(x + 3)(x + 1)
x+3
x + 4x + 3
x+3
=
=
=
x2 + 4x + 4
(x + 2)2
(x + 2)2 (x + 1)
(x + 2)2 (x + 1)
Add the rewritten fractions by adding their numerators:
x2 + 2x
x2 + 4x + 3
2x2 + 6x + 3
+
=
2
2
(x + 2) (x + 1) (x + 2) (x + 1)
(x + 2)2 (x + 1)
Try to reduce the fraction by factoring the numerator. To see if that is possible, apply the
Quadratic
Factoring
Criterion: √
2x2 + 6x + 3√= ax2 +√bx + c with a = 1, b = 6, and c = 3. Then
√
√
b2 − 4ac = 62 − 4 · 2 · 3 = 36 − 24 = 12 = 2 3. Since this is not a whole number, the
numerator does not factor and therefore the fraction cannot be reduced.
Answer:
x
x+3
2x2 + 6x + 3
+
=
x2 + 3x + 2 x2 + 4x + 4
(x + 2)2 (x + 1)
Stanley Ocken
M19500 Precalculus Chapter 1.4: Rational Expressions
Welcome
Algebra with numerical fractions
Algebra with polynomial fractions
Exercises
Quiz Review
Special issues with minus signs
Minus signs can be annoying, especially when we deal with fractions. The nicest way
7
7
to write −0.7 as a fraction is − 10
. Also OK is −7
10 . However, −10 is inelegant: convert
it to one of othe other forms as soon as possible.
When you reduce a polynomial fraction, or find the LCM of polynomials, make sure
that all polynomial factors have a positive leading coefficient. The basic identity you
need to know is that a − x = −(x − a) = −1(x − a).
Example 23: Reduce each of the following fractions: a)
Solutions:
−1
(x−4)
1·
(x−4)
a)
4−x
x−4
b)
x2 −4x+4
4−x2
=
=
4−x
x−4
b)
x2 −4x+4
4−x2
= −1
(x−2)2
−(x2 −4)
(x−2)(x−2)
= −1
=
(x−2)(x+2)
(x−2)
−1(x+2)
Stanley Ocken
= − x−2
x+2 or
2−x
x+2
M19500 Precalculus Chapter 1.4: Rational Expressions
Welcome
Algebra with numerical fractions
Algebra with polynomial fractions
Exercises
Quiz Review
Complex Fractions
Definition
E
in which E and/or F contains a
F
fraction (which might be written as a negative power).
These contained fractions are called nested fractions.
x
x
1+
z+
x+y
y
Examples of complex fractions include
and
1
b
a
xy +
+
4
x x+y
A complex fraction is an expression of the form
Procedure
To simplify a complex fraction: multiply both numerator and denominator by the
LCD of all the nested fractions.
Stanley Ocken
M19500 Precalculus Chapter 1.4: Rational Expressions
Welcome
Algebra with numerical fractions
Algebra with polynomial fractions
Exercises
Quiz Review
y + x−2
without negative exponents.
z
y + x12
1
by x2 :
Solution Method 1. Rewrite x−2 as 2 , multiply top and bottom of
x
z
1
y + 2 x2
y + x12
x2 y + 1
x
=
=
z
zx2
x2 z
Example 24: Rewrite
Solution Method 2: Multiply top and bottom of original fraction
y + x−2
by x2 :
z
y + x−2 x2
y + x−2
yx2 + x−2 x2
yx2 + 1
=
=
=
z
zx2
zx2
x2 z
Stanley Ocken
M19500 Precalculus Chapter 1.4: Rational Expressions
Welcome
Algebra with numerical fractions
Example 25: Rewrite
x−1 +y −1
x−2 −y −2
Algebra with polynomial fractions
Exercises
Quiz Review
as a reduced fraction.
Reminder: to reduce a fraction: factor top and bottom, then remove common factors.
Solution Method 1: Rewrite by converting powers with negative exponents to
fractions. Then multiplying top and bottom by x2 y 2 will cancel all denominators.
1
1
2 2
1
1
+
−1
−1
x
y x y
xy
x +y
xy 2 + yx2
xy
(y
+
x)
x + y
=
=
=
=
=
1
1
−2
−2
2
2
1
(x
+ y)(x − y)
x −y
x −y
x−y
+ y2
+ 1 x2 y 2
x2
x2
y2
Solution Method 2: Multiplying x−2 or x−1 by x2 will remove negative exponents.
The same idea works for y. Therefore we multiply top and bottom by x2 y 2 to get
x−1 + y −1 x2 y 2
x−1 x2 y 2 + y −1 x2 y 2
x−1 + y −1
xy 2 + yx2
=
=
=
x−2 − y −2
(x−2 − y −2 ) x2 y 2
(x−2 x2 y 2 − y −2 x2 y 2 )
x2 − y 2
This gets rid of the negative powers. Now reduce the answer, just as before:
xy
(y
+
x)
xy 2 + yx2
=
=
x2 − y 2
(x
+y)(x − y)
Stanley Ocken
xy
x−y
M19500 Precalculus Chapter 1.4: Rational Expressions
Welcome
Algebra with numerical fractions
Algebra with polynomial fractions
Exercises
1
a+h
Quiz Review
− a1
as a reduced fraction.
h
1
and a1 are nested in the numerator of the given fraction.
Solution: The fractions a+h
Multiply top and bottom by their LCD, which is a(a + h).
Example 26: Rewrite
1
1
−
a+h
a
h a(a + h)
The original problem:
Multiply top and bottom by LCD:
=
Apply the distributive law:
=
Rewrite the products:
=
Cancel common factors:
=
Distribute minus sign, cancel:
=
1
1
−
a+h
a
a(a + h)h
1
1
− a(a + h)
a(a + h)
a+h
a
a(a + h)h
a(a + h)
a(a + h)
−
a+h
a
a(a + h)h
a − (a + h)
a(a + h)h
−h
−1
=
a(a + h)h
a(a + h)
Stanley Ocken
M19500 Precalculus Chapter 1.4: Rational Expressions
Welcome
Algebra with numerical fractions
Algebra with polynomial fractions
Exercises
Quiz Review
1 )=a
(a
+
h)(
You should learn to condense this by multiplying and cancelling a
a+
h
1
and a(a
+ h)( a ) = a + h in your head:
1
1
−
a+h a
h a(a + h)
=
1
1
−
a+h a
a(a + h)h
=
a − (a + h)
a(a + h)h
=
−h
a(a + h)h
=
−1
a(a + h)
Stanley Ocken
M19500 Precalculus Chapter 1.4: Rational Expressions
Welcome
Algebra with numerical fractions
Algebra with polynomial fractions
Exercises
Exercises for Chapter 1.4: Rational expressions
to find an answer checker.
Click on
Wolfram Calculator
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to see how to check various types of algebra problems.
1. Find each sum by first finding the LCD of the fractions to be added.
a)
3
100
c)
1
x2 +2x+1
e)
1
2x2 +4x+2
g)
1
2−x
i)
+
5
200
+
x
x2 −x−6
+
+
−
1
x2 −2x+1
+
2
x−2
7
800
+
1
x2 −2x+1
+
1
x+2
x
x2 −4
−
2
x−3
1
x2 −1
+
1
x2 −1
b)
1
xy
d)
x
x−4
f)
x
x2 −4
h)
x
x2 +x−2
j)
+
3
2x−4
Stanley Ocken
1
yz
3
x+6
−
+
+
1
xz
+
1
x−2
−
2
x2 −5x+4
1
8−4x
+
16
4−2x
M19500 Precalculus Chapter 1.4: Rational Expressions
Quiz Review
Welcome
Algebra with numerical fractions
Algebra with polynomial fractions
Exercises
Quiz Review
2. For each of the following polynomials,
• Try to factor the polynomial by using the methods you learned in other math
courses.
• Use the Quadratic Polynomial Factoring Criterion to decide if the polynomial can
be factored. If it can, find the factorization.
a) 6x2 + 17x + 12
d) 9x2 + 28x + 20
g) x2 + 30x + 200
b) 6x2 + 18x + 12
e) 8x2 + 18x + 9
h) x2 − 32x − 420
c) 9x2 + 27x + 20
f) 8x2 + 18x + 11
i) 8x2 − 19x + 12
3. Multiply or divide as indicated.
a)
x2 −x−6
x2 +2x
·
x3 +x2
x2 −2x−3
4. Find the sum
2x2
b)
x2 −1
x2 −4
·
2x3 −4x2 +2x
x2 −4x+4
c)
x2 +2x
x2 −x−6
÷
x3 +x2
x2 −2x−3
3
5
3
+
+
2
4
+ 8x + 8
12x + 48x3 + 48x2
3x2 (2x + 4)
Stanley Ocken
M19500 Precalculus Chapter 1.4: Rational Expressions
Welcome
Algebra with numerical fractions
Algebra with polynomial fractions
Exercises
Quiz Review
If your answer is a fraction, it must be reduced completely.
A
C
Find 45 + 67 by substitution in the formula B
+D
= AD+BC
BD
7
11
A
C
Find 2000
+ 3000
by substitution in the formula B
+D
= AD+BC
BD .
3
Rewrite 5 as a fraction with denominator 50.
3
Rewrite 120
as a fraction with denominator 18000.
3
Rewrite 5·22 ·73 as a fraction with denominator 50 · 28 · 711 .
Example
Example
Example
Example
Example
1:
2:
3:
4:
5:
Example
Example
Example
Example
Example
6: Find the LCM of 2000 and 3000 by using the above method.
7: Find the LCM of 35 and 77.
7
11
8: Find 2000
+ 3000
by using the above procedure for finding the LCD.
2
5
9: Find 35 + 77 .
7
7
3
10: Find 20
+ 16
+ 24
Example 11: Rewrite the complex fraction
Example 12: Find
x+3
4x2 −9
·
2x2 +7x−15
x2 +7x+12
Stanley Ocken
5+7−2
3
without negative exponents.
Example 13: Find
x+3
4x2 −9
÷
x2 +7x+12
2x2 +7x−15
M19500 Precalculus Chapter 1.4: Rational Expressions
Quiz Review
Welcome
Algebra with numerical fractions
Example 14: Rewrite
Example 15: Rewrite
Algebra with polynomial fractions
3
as a
x2
ab
as
x2 y 3
Exercises
Quiz Review
fraction with denominator x5 .
a fraction with denominator x5 y 8 .
Example 16: Rewrite each of the three fractions xa , yb , and zc with denominator xyz.
Example 17: Rewrite 5c23d3 as a fraction with denominator 50c8 d11 .
Example 18: Use the quadratic factoring criterion to decide if each of the following
polynomials factors: x2 + 19x + 34 and x2 + 19x + 35.
Example 19: Find the LCM of x3 y 4 and x2 y 3 z 5
Example 20: Find the LCM of x2 + 3x + 2 and x2 + 4x + 4
Example 21: Find the sum
3z
x3 y 4
+
4
x2 y 3 z 5
Example 22: Use the LCM from Example 16 to simplify
Example 23: Reduce the fractions: a)
Example 24: Rewrite
Example 25: Rewrite
Example 26: Rewrite
4−x
x−4
b)
x2 −4x+4
4−x2
x
x2 +3x+2
+
x+3
x2 +4x+4
y+x−2
without negative exponents.
z
x−1 +y −1
as a reduced fraction.
x−2 −y −2
1
− a1
a+h
h
as a reduced fraction.
Stanley Ocken
M19500 Precalculus Chapter 1.4: Rational Expressions