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A Step-by-step Approach
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Table of Contents
Chapter 1 Algebraic expressions
5
1 Collecting
...................................................................................................................................
like terms
5
Chapter 2 Fractions
11
1 Adding
...................................................................................................................................
and subtracting fractions
11
2 Reducing
...................................................................................................................................
fractions
15
Chapter 3 Complex numbers
22
1 Introduction
................................................................................................................................... 22
2 Operating
................................................................................................................................... 23
Chapter 4 Rational expressions
28
1 Adding
...................................................................................................................................
and subtracting
28
Chapter 5 Radicals
33
1 Adding
...................................................................................................................................
and subtracting
33
2 Multiplying
...................................................................................................................................
radicals
39
3 Rationalizing
...................................................................................................................................
denominators
40
4 Combining
...................................................................................................................................
like radical terms
43
5 Combining
...................................................................................................................................
operations
44
Chapter 6 Equations
47
1 Linear
...................................................................................................................................
equations
47
Graphing linear
..........................................................................................................................................................
equations
51
2 Absolute
...................................................................................................................................
value equations
54
Chapter 7 Systems of equations
57
1 Solving
...................................................................................................................................
systems by elimination
57
2 Solving
...................................................................................................................................
systems by substitution
58
Chapter 8 Inequalities
62
1 Linear
...................................................................................................................................
inequalities
62
Graphing linear
..........................................................................................................................................................
inequalities
64
2 Simultaneous
...................................................................................................................................
inequalities
69
Chapter 9 Polynomials
73
1 Adding
...................................................................................................................................
and subtracting
73
2 Factoring
................................................................................................................................... 76
Contents
3
3 Evaluating
................................................................................................................................... 80
4 Higher
...................................................................................................................................
degree polynomials
81
5 Completing
...................................................................................................................................
the square
82
Chapter 10 Functions
86
1 Adding
...................................................................................................................................
and subtracting
86
2 Linear
...................................................................................................................................
functions
88
Graphing linear
..........................................................................................................................................................
functions
89
3 Quadratic
...................................................................................................................................
functions
91
Graphing quadratic
..........................................................................................................................................................
functions
92
4 Composition
................................................................................................................................... 95
5 Exponential
...................................................................................................................................
functions
96
Chapter 11 Trigonometry
100
1 Angles
...................................................................................................................................
and degree measures
100
2 Trigonometric
...................................................................................................................................
functions
102
3
Chapter
1
Algebraic expressions
Algebraic expressions
1
Algebraic expressions
1.1
Collecting like terms
5
Like terms in an algebraic expression are terms with identical symbolic or variable parts. Thus
‘3x’ and ‘7x‘ are like terms because both contain the symbolic part ‘x’
‘5x 2 yz’ and ‘13x 2 yz’ are like terms because both contain the symbolic part ‘x 2 yz’
‘4x 2 ’and ‘7x’ are not like terms because even though the symbol present in both is ‘x’, the
symbolic part ‘x 2 ’ is not identical to the symbolic part ‘x’.
Algebraic expressions that contain like terms can be simplified by combining each group of like
terms into a single term. The reason why this is possible and valid is quite easy to see. For instance,
consider the expression
3x + 7x
which is the sum of two like terms, representing the accumulation of three x’s and another seven x’s.
Clearly, the end result is a total of ten x’s. In notation
3x + 7x = (3 + 7)x = 10x
This process of combining (or collecting ) like terms can be performed for each group of like terms
that appear in an expression. The net effect will be that the original expression can now be written
with fewer terms, yet which are entirely equivalent to the terms in the original expression.
Example:
Simplify: 5x 2 + 9 – 3x + 4x 2 + 8x + 7.
solution:
This expression has six terms altogether. However, we notice that
· two of the terms have the literal part ‘x 2 ’ and so are like terms – we can replace
5x 2 + 4x 2 by (5 + 4)x 2 = 9x 2
two of the terms have the same literal part ‘x’ and so are also like terms. We can replace
-3x + 8x by (-3 + 8)x = 5x
two of the terms are just constants, and so obviously can be combined arithmetically:
9 + 7 = 16.
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Algebra e-Book
So
5x 2 + 9 – 3x + 4x 2 + 8x + 7
= 5x 2 + 4x 2 + (-3x) + 8x + 9 + 7
= (5 + 4)x 2 + (-3 + 8)x + 16
= 9x 2 + 5x + 16.
Thus, in simplest form, the original six term expression can be rewritten as
9x 2 + 5x + 16
consisting of just three terms.
Algebrator can easily solve these type of problems:
Algebraic expressions
Clicking on the "Solve Once" button will show the first step of the solution process:
7
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An explanation can be obtained by clicking on the "Explain" button:
Algebraic expressions
9
Chapter
2
Fractions
Fractions
2
Fractions
2.1
Adding and subtracting fractions
11
Addition and subtraction of algebraic fractions works basically the same way as does addition and
subtraction of simple numerical fractions:
(i.) rewrite all of the fractions as mathematically equivalent fractions, but having the same
denominator (ideally, the simplest such common denominator).
(ii.) combine the numerators of these fractions as indicated (adding and subtracting), retaining the
common denominator.
When we work with algebraic fractions, there is then often a third very important step:
(iii.) simplify the resulting fraction as much as possible. It is always necessary to make sure that once
the addition/subtraction step is completed, you make a reasonable effort to simplify the result.
Usually this simplification step will be made very much easier if you take care in step (i.) to make
sure that you have used the simplest possible common denominator. We will illustrate the process
for determining that simplest possible denominator when the fractions to be added/subtracted are
algebraic fractions. You will see that the strategy is essentially the same as the strategy used to
determine the least common denominator when adding or subtracting numerical fractions, except that
now symbolic factors may occur as well as numerical factors.
Example 1: Combine and simplify:
solution:
Here, the two denominators share some factors in common and so the simplest common
denominator for these fractions is not simply the product of the two denominators. Noting that
15x = (3)(5)(x)
and
12x 2 = (2 2)(3)(x 2)
we conclude that the simplest common denominator here is (2 2)(3)(5)(x 2) = 60x 2. So
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The numerator is a difference of two terms, and so we try multiplying to remove the brackets
followed by collection of like terms in hope of being able to achieve some simplification.
When the like terms were combined, we noticed that the two terms remaining shared a common
factor of 2x, which has been factored out in the last line. Thus, our original difference of two
fractions becomes
as the final answer. Since what is left in the numerator cannot be factored further, this is as simple as
we can get for the final answer.
Example 2:
Simplify
solution:
As mentioned before, the instruction to “simplify” here means combine the two fractions into a single
fraction, and reduce that resulting single fraction to as simple a form as possible. We’ve now done
several examples of this type, so here we’ll more or less just show the work without including a lot
of discussion. This would be a good example for you to try as a practice exercise before you look at
our solution given below.
Fractions
13
The two denominators factor as
6xy 2 = (2)(3)(x)(y 2)
and
15x 2 y = (3)(5)(x 2)(y)
Thus, the simplest common denominator is
(2)(3)(5)(x 2)(y 2) = 30x 2 y 2
So, going back to the original problem
There is no obvious way to factor this numerator, so it does not appear that any further simplification
of this fraction is possible. So, the last form above must be the requested final answer.
Example 3:
Simplify
solution:
Some people experience an overwhelming temptation here to simply cancel the x’s:
confusing this addition problem with the much simpler multiplication problem
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where the cancellation to get the simple result ‘1’ is justified. However, no such quick cancellations
can be done in the addition problem we have here because the two x’s aren’t even part of the same
fraction! In fact, the first term here is not even a fraction.
Perhaps the clearest way to proceed is to turn the first term into a fraction with denominator equal to
1, and then apply the methods of the previous examples for adding two fractions. We get
This is the final answer, since the numerator cannot be factored further and so there is no additional
simplification possible.
Example 4:
Combine and simplify:
solution:
Again, we start by turning this problem into a problem with two fractions, and then we just apply the
methods already illustrated in the previous examples.
Fractions
15
The simple form of this last fraction is obtained from the previous step by collecting like terms in the
numerator. The terms -5y + 5y add up to zero y’s, and the two numbers sum to -28. The numerator
of the final fraction shown above, y 2 – 28, cannot be factored further to give a form that leads to
cancellations with the denominator, so this must be the required final answer.
2.2
Reducing fractions
To reduce a numerical fraction to simplest form means to rewrite it as an equivalent fraction which
has the smallest possible denominator. People refer to this operation as reducing a fraction to lowest
terms.
Recall that starting with a fraction, we can get an equivalent fraction by either:
(i) multiplying the numerator and denominator by the same nonzero value
or
(ii) dividing the numerator and denominator by the same nonzero value.
Obviously if the goal of simplification is to find an equivalent fraction with a smaller denominator, we
will have to use the second principle: dividing the numerator and denominator by the same nonzero
value.
The strategy for finding the values to divide into the numerator and denominator is quite systematic:
Step 1: Factor both the numerator and denominator into a product of prime factors using the
method described in the previous note in this series.
Step 2: If the numerator and denominator both have a prime factor which is the same, then divide
the current numerator and denominator by that value. The result will be the numerator and
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denominator of an equivalent fraction, but with a smaller denominator.
Repeat step 2 as often as possible. When the numerator and denominator have no further prime
factors in common, they form the desired equivalent fraction which has the smallest denominator.
We have then obtained the simplest form of the original fraction.
Example:
Reduce the fraction
to simplest form.
solution:
For this example, we’ll go through the process step-by-step in some detail. Then we’ll illustrate the
familiar “shortcuts” in a couple of examples.
The prime factorizations of the numerator and denominator here are easily obtained:
42 = 2 × 21 = 2 × 3 × 7
70 = 2 × 35 = 2 × 5 × 7
So, the numerator and denominator of the fraction we are given both contain the prime factor 2.
Thus
The new fraction, 21 / 35 , is simpler than the old fraction, 42 / 70 , because its denominator, 35, is
smaller than the original denominator of 70. Still, 21 / 35 is equivalent to 42 / 70 because it was
obtained by dividing both the numerator and denominator of 42 / 70 by the same value, 2.
But 21 / 35 is still not in simplest form because the factorization of its numerator and denominator
(shown in brackets above) indicates that they both still contain a common prime factor of 7. So
Thus, we have
These two fractions, the original 42 / 70and the final 3 / 5 , are equivalent because we got 3 / 5 by
dividing the numerator and denominator of 42 / 70 by the same values (2 and 7 in turn, or
Fractions
17
effectively, 14, if you think of doing it in one step). Furthermore, the numerator and denominator of 3
/ 5clearly do not share any further common prime factors, and so this simplification process cannot
be carried further. Therefore,
3 / 5 is the simplest form of 42 / 70 .
Example:
Reduce the fraction
to simplest form.
solution:
Yes! This is the same problem as the first one. What we want to do here is show the “shortcut” form
of the strategy for simplifying fractions. We begin as before by rewriting both the numerator and the
denominator as a product of prime factors:
Now, if you study the steps of the previous example, you will see that dividing the numerator and
denominator by 2 results in those two factors disappearing from each. We indicate this by drawing
“slashes” through them:
Dividing the numerator and denominator in the resulting factored equation by 7 again just results in
those two factors of 7 disappearing from each. So again,
Now there are no common factors left in the numerator and denominator, so the process ends, and
we conclude that 3 / 5 is the simplest form of 42 / 70 .
In practice, this whole process is typically done in a single step, crossing out pairs of factors without
rewriting intermediate forms:
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Algebra e-Book
Algebrator can easily solve these type of problems:
Fractions
19
Clicking on the "solve step" button once shows the first step of the solution process. An explanation
can be obtained by clicking on the "Explain" button.
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Algebrator shows an explanation for any step you want.
Chapter
3
Complex numbers
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Algebra e-Book
3
Complex numbers
3.1
Introduction
In this section we define a set of numbers that has the real numbers as a subset.
Definition
The equation 2x = 1 has no solution in the set of integers, but in the set of rational numbers, 2x = 1
has a solution. The situation is similar for the equation x2 = -4. It has no solution in the set of real
numbers because the square of every real number is nonnegative. However, in the set of complex
numbers x2 = -4 has two solutions. The complex numbers were developed so that equations such as
x2 ?4 would have solutions.
The complex numbers are based on the symbol
. In the real number system this symbol has no
meaning. In the set of complex numbers this symbol is given meaning. We call it i. We make the
definition that
Complex Numbers
The set of complex numbers is the set of all numbers of the form a + bi, where a and b are real
numbers,
, and i2 = -1.
In the complex number a + bi, a is called the real part and b is called the imaginary part. If b ? 0,
the number a + bi is called an imaginary number.
In dealing with complex numbers, we treat a + bi as if it were a binomial, with i being a variable.
Thus we would write 2 + (-3)i as 2 - 3i. We agree that 2 + i3, 3i + 2, and i3 + 2 are just different
ways of writing 2 + 3i (the standard form). Some examples of complex numbers are
For simplicity we write only 7i for 0 + 7i. The complex number 9 + 0i is the real number 9, and 0 +
0i is the real number 0. Any complex number with b = 0 is a real number. For any real number a,
a + 0i a. The set of real numbers is a subset of the set of complex numbers. Take a look at the figure
below.
Complex numbers
3.2
23
Operating
A complex number is an expression of the form x + iy where x and y are real numbers.
Either x or y may be equal to zero: real numbers and imaginary numbers are special cases of
complex numbers.
Given a complex number z = x + iy the real number x is known as the real part of z and the real
number y is known as the imaginary part of z.
The notation Re is used for real part and Im is used for imaginary part.
Note that the imaginary part of z is a real number y (NOT iy).
Adding and Subtracting Complex Numbers
To add or subtract complex numbers, we add or subtract their real and imaginary parts separately:
Examples:
1) (2 + 3i) + (4 + i) = 6 + 4i
2) (3 - 5i) - 7i = 3 - 12i
3) (5 - 4i) + (3 + 2i) - (8 + i) = -3i
4) 12 - 4i + (3 + 4i) = 15
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Multiplying complex numbers
To multiply complex numbers, we use the usual rules and the identity i2 = -1:
Examples:
1) (2 + 3i)(4 +
5i)
= (2 × 4) + (3i × 4) + (2 × 5i) + (3i × 5i)
= 8 12i + 10i - 15
= -7 + 22i
2) (2 +
3i)3
= (2 + 3i)(2 + 3i)(2 + 3i)
= (-5 + 12i)(2 + 3i)
= -46 + 9i
The complex conjugate
Given a complex number z = x + iy, the complex conjugate is given by z* = x - iy (pronounced “z
star”):
To find the complex conjugate of any expression, replace i by –i.
Examples:
1) (3 + 4i)* = 3 - 4i
2) (5 - 2i)* = 5 + 2i
3) (6i)* = -6i
4) [(2 + 4i)*]* = [2 - 4i]* = 2 + 4i
Complex numbers
25
The last expression demonstrates that
A complex number and its complex conjugate have the property that:
Examples:
1) (3 + 4i) + (3 + 4i)* = 3 + 4i + 3 - 4i = 6
2) (3 + 4i) - (3 + 4i)* = 3 + 4i - 3 + 4i = 8i
3) 5i + (5i)* = 5i - 5i = 0
4) 5i - (5i)* = 10i
Complex conjugate of a product
The modulus of a complex number
A complex number z = x + iy multiplied by its complex conjugate is a real number:
The modulus of a complex number z = x + iy is written | z | and is equal to the positive square root of
the sum of the squares of its real and imaginary parts:
Examples:
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The modulus is a real number
The quotient of complex numbers
In order to simplify an expression with a complex number in the denominator, we multiply both the
numerator and denominator by the complex conjugate:
In this way, we get a real number in the denominator (the modulus squared of a + bi).
Examples:
In general:
Chapter
4
Rational expressions
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4
Rational expressions
4.1
Adding and subtracting
Definition A rational expression is a ratio (fraction) of two polynomials. All rational expressions are
valid for all real numbers except those values that make the denominator zero.
are rational expressions whereas
are
not.
When adding (or subtracting) rational expressions with a common denominator, the process is
straightforward: Simply add (or subtract) the numerators, and put the result in a rational expression
with the same denominator; i.e.,
where a, b and c are polynomials.
Example 1
Example 2
Note that there is no restriction on x because
, as will be shown in another paper.
When performing any operation with rational expressions of a single variable, we must make sure
our final answer is simplified:
a. All polynomials are in descending order; i.e., the largest power of the variable appears on the left
and as we proceed to the right, the powers of the variable decrease.
b. All leading terms (highest degree terms) in the denominator have positive coefficients.
Rational expressions
29
c. All like (similar) terms have been combined.
d. The greatest common factor of the numerator and denominator is 1; i.e., all common factors have
been divided out.
e. Unless told otherwise, the numerator is to multiplied out and simplified, while the denominator may
be left in factored form.
The expressions,
, are considered to be simplified. The
expressions,
are not considered to be simplified. The following
shows how to convert the latter pair of expressions into simplified forms.
[Factor out -1 from denominator]
[Put in descending order]
[Factor out -1 from (-x + 1)]
[Eliminate the leading neg sign in denom.]
[Simplify]
Example 3
Substract:
Solution: The factorizations of the denominators are
Hence the LCD is:
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LCD = (x - 2)(x + 2)(x + 1)
Use these factorizations to convert the rational expressions to rational expressions that have the
LCD for their denominators, perform the subtraction, and simplify the result.
[Acceptable as being simplified]
[Preferrable]
Note that the above is valid for all
into
result.
For example, let x = 3. Then we obtain
and
. What this means is that if we substitute any value of
, and evaluate the expressions, we obtain the same
Rational expressions
31
Chapter
5
Radicals
Radicals
5
Radicals
5.1
Adding and subtracting
33
The procedure for adding and subtracting square roots which may contain algebraic expressions is
very similar to the one we usewhen adding and subtracting numerical square roots:
· simplify the square root in each term of the expression
· combine terms whose square roots are identical
Read the following examples:
Example 1:
Simplify
solution:
We have
Thus
as the final answer.
The method should be fairly obvious by now. Use the remaining three examples as practice
problems. Try to do them yourself before looking at our solutions.
Example 2:
Simplify
solution:
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Algebra e-Book
First, we check to simplify each of the two square roots individually:
Thus
as the final answer.
Algebrator can easily solve these type of problems:
Radicals
Clicking on the "solve step" button once shows the first step of the solution process.
35
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An explanation can be obtained by clicking on the "Explain" button.
Radicals
37
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Example 3:
Simplify
solution:
So
as the final answer.
Example 4:
Simplify
solution:
Each square root must first be simplified:
and
Therefore
Radicals
39
This is as simple as we can get the expression.
5.2
Multiplying radicals
The product rule for radicals,
index, such as
allows multiplication of radicals with the same
Caution
The product rule does not allow multiplication of radicals that have different indices. We cannot use
the product rule to multiply
Example 1
Multiplying radicals with the same index
Multiply and simplify the following expressions. Assume the variables represent positive numbers.
Solution
Product rule for
radicals
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Product rule for
radicals
Simplify.
Simplify.
Product rule for radicals
Simplify
Rationalize the denominator.
We find a product such as
by using the distributive property as we do when
multiplying a monomial and a binomial. A product such as
found by using FOIL as we do for the product of two binomials.
5.3
Rationalizing denominators
Examples with Solutions
Example 1:
Simplify
solution:
can be
Radicals
41
There is no factorization of the expression in the square root possible here, so the only simplification
necessary is to rationalize the denominator of the fraction. Since the square root in the denominator
is already as simple as possible, we need to just multiply the numerator and denominator by that
square root to achieve the required result.
Although there is a t in both the numerator and the denominator, it is not a factor in the denominator,
and therefore we cannot cancel the t’s between the numerator and denominator of this last form. In
fact, there is obviously no common factors to cancel between the numerator and denominator of this
last form, so this last expression is as simple as possible and so it must be the required final answer.
NOTE: Sometimes as they focus on simplifying fractions such as the final result of Example 1
above, people note that the numerator and denominator contain a power of the same expression – in
this case of (s + t). Without thinking they do the following:
or
Superficially, it may appear that one or both of these lines represents a useful simplification, but in
fact, both lines really just amount to reversing the simplification previously done. The second line
obviously just gives back the original problem, and the first line does as well, though in a somewhat
less recognizable form. So, while it is very useful to break the simplification of complicated algebraic
expressions down into a series of relatively simple independent steps, it is also important to review
your entire solution to make sure that you haven’t lost track of what the problem actually requires
you to accomplish, and make sure that that is indeed what your final answer has accomplished.
Example 2:
Rationalize the denominator in
and simplify the result as much as possible.
solution:
If we simply follow the steps illustrated in the first and second examples above, we get
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Algebra e-Book
which has a rational denominator, but clearly can be simplified further. First
8x 3 = 2 2 · 2 · x 2 · x
so
Thus,
as the final simplified result.
Alternatively, we could have noted that the square root in the denominator of the original fraction is
not in simplest form, and so we could have started by simplifying both the square root and the
fraction itself first:
Now the denominator of this fraction can be rationalized by multiplying the numerator and
denominator by
:
giving the same final simplified result as we obtained earlier.
Example 3:
Radicals
43
Simplify
solution:
No common factors can be found to cancel between the numerator and denominator here. Also, the
square roots which appear here obviously cannot be simplified further. So, all that’s left to do to
simplify this expression is to rationalize the denominator. This is accomplished by multiplying the
numerator and denominator by
, giving:
as the final result (since there are clearly no common factors that can be cancelled between the
numerator and denominator of this final expression). Notice that we used brackets explicitly to
ensure that the entire numerator was multiplied by
5.4
in the first step here.
Combining like radical terms
Before we can combine like radical terms, we sometimes need to simplify the individual radicals to
obtain like terms.
Example 1
Find:
First simplify
The radicand, 24, has a perfect
square factor, 4.
Replace
with
Combine the like radical terms,
Simplify.
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So,
Example 2
Find:
Solution
First simplify
The radicand,
32, has a perfect cube factor, 8.
Replace
with
Multiply
Combine the like radical terms,
Also, combine the like terms -5 and
-2.
Simplify.
So,
5.5
Combining operations
Now we will simplify some radical expressions that involve more than one operation.
Example 1
Simplify:
Solution
Factor each radicand.
Write each radical as the product of
radicals.
Radicals
Simplify square roots of perfect
squares.
For each term, multiply the radicals.
Combine like radical terms.
Therefore,
Example 2
Simplify:
Solution
To remove the parentheses, distribute
Multiply
radicand.
by
Then factor the
Simplify the second term:
Combine like radical terms.
Thus,
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Chapter
6
Equations
Equations
6
Equations
6.1
Linear equations
47
Equations that can be written in the form ax + b = 0 where a and b are real numbers, with a 0, are
linear equations. Examples oflinear equations include 5y + 9 = 16 and 8x = 4
The following properties are used to solve linear equations.
PROPERTIES OF EQUALITY
For all real numbers a, b, and c:
1. If a = b then a + c = b + c Addition property of equality (The same number may be addedto both
sides of an equation.)
2. If a = b then ac = bc Multiplication property of equality (Both sides of an equation may be
multiplied by the same number.)
Solving Linear Equations
EXAMPLE
(a) If x -2 = 3 then x = 2 + 3 = 5 Addition property of equality
(b) If x/2=3 then x = 2·3 = 6 Multiplication property of equality
The following example shows how these properties are used to solve lineare quations. Of course,
the solutions should always be checked by substitution inthe original equation.
EXAMPLE
Solve 2x - 5 + 8 = 3x + 2(2-3x)
Solution
2x - 5 + 8 = 3x + 4-6x Distributive property
2x + 3 = -3x + 4 Combine like terms
5x + 3 = 4 Add 3x to both sides
5x = 1 Add -3 to both sides
Multiply both sides by .
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Check by substituting in the original equation. The left side becomes 2(1/5)-5+8 and the right side
becomes 3(1/5)+2(2-3(1/5)). Verify that both of these expressions simplify to 17/5.
Algebrator can easily solve these type of problems:
Clicking on the "solve step" button once shows the first step of the solution process.
Equations
An explanation can be obtained by clicking on the "Explain" button.
49
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Equations
6.1.1
Graphing linear equations
Using the Slope-Intercept Method
Objectives
1. Determine the slope of a line from the graph.
2. Determine the slope of a line from two points on the line.
3. Determine the slope of a line from its linear equation in 2 variables.
4. Graph a linear equation in 2 variables using the slope-intercept method.
The slope of a line is the ratio of the amount of rise to the amount of run,
Slope is positive if the line rises.
Slope is negative if the line falls.
Slope is zero if the line is horizontal.
Slope is undefined if the line is vertical.
Slope of a line through 2 given points:
Given 2 ordered pairs, (x1, y1) and (x2, y2)
Slope intercept form of a linear equation: y = mx + b
m is the slope
b is the y-intercept
Note that the coefficient of y must be 1!!
To Graph when you know a point and the slope:
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1. Plot the known point.
2. Count the rise and run from the known point. For a positive rise, count upward. For a negative
rise, count downward. For a positive run, count to the right. For a negative run, count to the left.
( hint: if the slope is a whole number, make it a fraction over one)
3. Place the second point where the next ordered pair is located based on your counting above.
4. Draw a line (not a segment!) connecting the 2 points.
Examples: Graph a line that contains
a) the point (1, -3) with
b) the point (2, 5) with m = -4
To graph a line using the slope-intercept method:
1. Put the equation in slope-intercept form (solve for y )
2. Determine the slope and y -intercept from the equation.
Equations
3. Plot the y -intercept.
4. Locate the next point in the line by counting the slope.
5. Draw a line (not a segment) connecting the 2 points.
Examples:
c) y = -2 x + 4
d) 6 x + 3 y = -9
e) 4y = 14
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Absolute value equations
Solving an Equation of the Form | z| = |w|
Here are several examples of solving equations of the form |z| = |w|.
Example 1
Solve: |4x - 34| = |6x + 14|
Solution
|4x - 34| = |6x + 14|
Replace 4x - 34 with z and 6x + 14 with w:
|z| = |w|
So:
z = w or z = -w
Substitute 4x - 34 for z and 6x + 14 for w.
Now, solve for x.
4x - 34 = 6x + 14
or
4x = 6x + 48
or
-2x = 48
or
or
So:
x = -24
or
4x - 34
=
Check x = -24
|4x - 34| = |6x + 14|
Check x = -3
|4x - 34| = |6x + 14|
-6x - 14
4x - 34
= -6x + 20
4x = 20
10x = 2
x=
Let's check the solutions:
-(6x + 14)
Equations
Is |4(-24) - 34| = |6(-24) + 14| ?
Is |4(2) - 34| = |6(2) + 14|
Is |-130| = |-130| ?
Is |-26| = |26| ?
Is 130 = 130 ? Yes
55
Is 26 = 26 ? Yes
So, the solutions are x = -24 and x = 2.
Example 2
Solve: |3x - 4| = |3x + 16|
Solution
|3x - 4| = |3x + 16|
Replace 3x - 4 with z and 3x + 16 with w:
|z| = |w|
So:
z = w or z = -w
Substitute 3x - 4 for z and 3x + 16 for w.
Now, solve for x.
3x + 4 = 3x + 16
or
3x - 4 = -(3x + 16)
3x = 3x + 20
or
3x - 4 = -3x - 16
or
6x = -12
0 = 20
x = -2
Since 0 = 20 is a contradiction, the left equation does not lead to a solution.
Check x = -2
|3x - 4| = |3x + 16|
Is |3(-2) - 4| = |3(-2) + 16| ?
Is |-10| = |10| ?
Is 10 = 10 ? Yes
Thus, -2 is the only solution.
Chapter
7
Systems of equations
Systems of equations
7
Systems of equations
7.1
Solving systems by elimination
Example
Use elimination to find the solution of this system.
x - 3y = -17 First equation
-x + 8y = 52 Second equation
Solution
Add the two equations.
Simplify. The x-terms have been eliminated.
To solve for y, divide both sides by 5.
3
x - y = -x + 8 =
y
5
2
0x + 5 =
y
3
5
5y = 35
y =7
To find the value of x, substitute 7 for y in either
of the original equations. Then solve for x.
We will use the first equation.
Substitute 7 for y.
Multiply.
Add 21 to both sides.
The solution of the system is (4, 7).
To check the solution, substitute 4 for x and 7
for y into each original equation. Then simplify.
In each case, the result will be a true statement.
The details of the check are left to you.
1
7
x - 3y = -17
x - 3(7) = -17
x - 21 = -17
x =4
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In the two original equations in the previous
example, the coefficients of x were opposites.
Thus, when the equations were added, the x-terms
were eliminated.
3
1x - y = -
1
7
- 1x + 8 =
y
5
2
5 =
y
3
5
When the coefficients of neither variable are opposites, we choose a variable. Then we multiply both
sides of one (or both) equations by an appropriate number (or numbers) to make the coefficients of
that variable opposites.
Note:
The Multiplication Principle of Equality enables us to multiply both sides of an equation by the same
nonzero number without changing the solutions of the equation.
7.2
Solving systems by substitution
Graphing is not always the best way to find the solution of a system of equations. It may be difficult
to read the coordinates of the point of intersection. This is especially true when the coordinates are
not integers.
Instead we can use algebraic methods to solve the system. One algebraic method for finding the
solution of a linear system is the substitution method.
Procedure — To Solve a Linear System By Substitution
Step 1 Solve one equation for one of the variables in terms of the other variable.
Step 2 Substitute the expression found in Step 1 into the other equation. Then, solve for the
variable.
Step 3 Substitute the value obtained in Step 2 into one of the equations containing both variables.
Then, solve for the remaining variable.
Step 4 To check the solution, substitute it into each original equation. Then simplify.
Example
Use substitution to find the solution of this system.
Systems of equations
59
2x + y = 4 First equation
3x + y = 7 Second equation
Solution
Step 1 Solve one equation for one of the variables in terms of the other variable.
Either equation may be solved for either variable.
For instance, let’s solve the first equation for y.
2x + y = 4
Subtract 2x from both sides.
y = -2x + 4
The equation y = -2x + 4 means that y and the
expression -2x + 4 are equivalent.
Step 2 Substitute the expression found in Step 1 into the other equation. Then, solve for the
variable.
3x + y = 7
Substitute -2x + 4 for y in the second equation.
Combine like terms.
Subtract 4 from both sides.
Now we know x = 3.
3x + (-2x + = 7
4)
=7
x+ 4
=3
x
Next, we will find y.
Step 3 Substitute the value obtained in Step 2 into one of the equations containing both variables.
Then, solve for the remaining variable.
We will use the equation from Step 1.
y = -2x + 4
Substitute 3 for x.
y = -2(3) +
4
y
= -2
Simplify.
The solution of the system is x = 3 and y = -2.
The solution may also be written as (3, -2).
Step 4 To check the solution, substitute it into each original equation. Then simplify.
Substitute x = 3 and y = -2 into both of the original equations:
First equation
Second equation
2x +
y =4
3x +
y =7
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Is
2(3) + (-2) = 4 ?
Is
6 -
Is 3(3) +
2 =4?
Is
4 = 4 ? Yes
Is
9 -
(- = 7 ?
2)
=7?
2
= 7 ? Yes
7
Since (3, -2) satisfies both equations, it is the solution of the system.
Note:
If we graphed the system, the lines would intersect at the point (3, -2).
Chapter
8
Inequalities
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8
Inequalities
8.1
Linear inequalities
To write that one number is greater than or less than another number, we use the following symbols.
INEQUALITY SYMBOLS
< means is less than
means is less than or equal to
> means is greater than
means is greater than or equal to
An equation states that two expressions are equal; an inequality states that they are unequal. A linear
inequality is an inequality that can be simplified to the form ax < b (Properties introduced in this
section are given only for <, but they are equally valid for >, or .) Linear inequalities are solved
with the following properties.
PROPERTIES OF INEQUALITY
For all real numbers a, b, and c:
1. If a < b then a + c < b + c
2. If a < b and if c > 0 then ac < bc
3. If a < b and if c < 0 then ac > bc
Pay careful attention to property 3; it says that if both sides of an inequality are multiplied by a
negative number, the direction of the inequality symbol must be reversed.
Solving Linear Inequalities
EXAMPLE
Solve 4 -3y
7 + 2y.
Solution
Use the properties of inequality.
4 -3y + (-4)
-3y
7 + 2y + (-4) Add -4 to both sides.
3 + 2y
Remember that adding the same number to both sides never changes the direction of the inequality
symbol.
Inequalities
-3y + (-2y)
-5y
63
3 + 2y + (-2y) Add -2y to both sides.
3
Multiply both sides by -1/5. Since -1/5 is negative, change the direction of the inequality symbol.
CAUTION
It is a common error to forget to reverse the direction of the inequality sign when multiplying or
dividing by a negative number. For example, to solve -4x 12 we must multiply by -1/4 on both
sides and reverse the inequality symbol to get x -3.
The solution y -3/5 in the previous example represents an interval on the number line. Interval
notation often is used for writing intervals. With interval notation, y -3/5 is written as
.
This is an example of a half-open interval, since one endpoint, -3/5, is included. The open interval (2,
5) corresponds to 2 < x < 5, with neither endpoint included. The closed interval [2, 5] includes both
endpoints and corresponds to 2 x 5.
The graph of an interval shows all points on a number line that correspond to the numbers in the
interval. To graph the interval
, for example, use a solid circle at -3/5 since -3/5 is part of
the solution. To show that the solution includes all real numbers greater than or equal to -3/5 draw a
heavy arrowpointing to the right (the positive direction). See Figure 1.
Graphing Linear Inequalities
EXAMPLE
Solve -2 < 5 + 3m < 20. Graph the solution.
Solution
The inequality -2 < 5 + 3m < 20 says that 5 + 3m is between -2 and 20. Solve this inequality with
an extension of the properties given above.Work as follows, first adding -5 to each part.
-2 + (-5) < 5 + 3m + (-5) < 20+ (-5)
-7 < 3m < 15
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Now multiply each part by 1/3.
A graph of the solution is given in Figure 2; here open circles are used to show that -7/3 and 5 are
not part of the graph.
(Some textbooks use brackets in place of solid circles for the graph of a closed interval, and
parentheses in place of open circles for the graph of an open interval.)
8.1.1
Graphing linear inequalities
You have probably already studied linear equations. We now turn our attention to linear inequalities.
Definition
A linear inequality is a linear equation with the equal sign replaced by an inequality symbol.
Linear Inequality
If A, B, and C are real numbers with A and B not both zero, then Ax + By = C is called a linear
inequality. In place of =, we can also use =, <, or >.
Graphing Linear Inequalities
Consider the inequality -x + y > 1. If we solve the inequality for y, we get y > x + 1.
Which points in the xy-plane satisfy this inequality? We want the points where the y-coordinate is
larger than the x-coordinate plus 1. If we locate a point on the line y = x + 1, say (2, 3), then the ycoordinate is equal to the x-coordinate plus 1. If we move upward from that point, to say (2, 4), the
y-coordinate is larger than the x-coordinate plus 1. Because this argument can be made at every
point on the line, all points above the line satisfy y > x + 1. Likewise, points below the line satisfy
y < x + 1. The solution sets, or graphs, for the inequality y > x + 1 and the inequality y < x + 1 are
the shaded regions shown in the figures below. In each case the line y = x + 1 is dashed to indicate
that points on the line do not satisfy the inequality and so are not in the solution set. If the inequality
symbol is = or =, then points on the boundary line also satisfy the inequality, and the line is drawn
solid.
Inequalities
65
Every nonvertical line divides the xy-plane into two regions. One region is above the line, and the
other is below the line. A vertical line also divides the plane into two regions, but one is on the left
side of the line and the other is on the right side of the line. An inequality involving only x has a
vertical boundary line, and its graph is one of those regions.
Graphing a Linear Inequality
1. Solve the inequality for y, then graph y = mx + b.
y > mx + b is satisfied above the line.
y = mx + b is satisfied on the line itself.
y < mx + b is satisfied below the line.
2. If the inequality involves x and not y, then graph the vertical line x = k.
x > k is satisfied to the right of the line.
x = k is satisfied on the line itself.
x < k is satisfied to the left of the line.
Example 1
Graphing linear inequalities
Graph each inequality.
a)
b) y
- 2x + 1
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c) 3x - 2y < 6
Solution
a) The set of points satisfying this inequality is the region below the line
. To show this
region, we first graph the boundary line
. The slope of the line is and the y-intercept is
(0, -1). Start at (0, -1) on the y-axis, then rise 1 and run 2 to get a second point of the line. We
draw the line dashed because points on the line do not satisfy this inequality. The solution set to the
inequality is the shaded region shown in the figure below.
b) Because the inequality symbol is , every point on or above the line satisfies this inequality. To
show that the line y = - 2x + 1 is included, we make it a solid line. See the figure below.
c) First solve for y:
3x <6
2y
-2y < -3x + 6
Inequalities
y
67
Divide by -2 and reverse the
inequality.
To graph this inequality, use a dashed line for the boundary
the line. See the figure below for the graph.
and shade the region above
Caution In Example 1(c) we solved the inequality for y before graphing the line. We did that
because < corresponds to the region below the line and > corresponds to the region above the line
only when the inequality is solved for y.
Algebrator can easily solve these type of problems:
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Clicking on the "Graph All" button shows the graphical representation.
Inequalities
69
Graphing multiple inequalities together is one of the many features of Algebrator.
8.2
Simultaneous inequalities
If we join to simple inequalities with the connective "and" or the connective "or", we get a compound
inequality. A compound inequality using the connective "and" is true if and only if both simple
inequalities are true.
Example 1
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Compound inequalities using the connective "and"
Determine whether each compound inequality is true.
a) 3 > 2 and 3 < 5
b) 6 > 2 and 6 < 5
Solution
a) The compound inequality is true because 3 > 2 is true and 3 < 5 is true.
b) The compound inequality is false because 6 < 5 is false.
A compound inequality using the connective "or" is true if one or the other or both of the simple
inequalities are true. It is false only if both simple inequalities are false.
Example 2
Compound inequalities using the connective "or"
Determine whether each compound inequality is true.
a) 2 < 3 or 2 > 7
b) 4 < 3 or 4 = 7
Solution
a) The compound inequality is true because 2 < 3 is true.
b) The compound inequality is false because both 4 < 3 and 4 = 7 are false.
If a compound inequality involves a variable, then we are interested in the solution set to the
inequality. The solution set to an "and" inequality consists of all numbers that satisfy both simple
inequalities, whereas the solution set to an "or" inequality consists of all numbers that satisfy at least
one of the simple inequalities.
Example 3
Solutions of compound inequalities
Inequalities
Determine whether 5 satisfies each compound inequality.
a) x < 6 and x < 9
b) 2x - 9 = 5 or -4x = -12
Solution
a) Because 5 < 6 and 5 < 9 are both true, 5 satisfies the compound inequality.
b) Because 2 · 5 - 9 = 5 is true, it does not matter that -4 · 5 = -12 is false. So 5 satisfies the
compound inequality.
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Chapter
9
Polynomials
Polynomials
9
Polynomials
9.1
Adding and subtracting
73
Objective Learn how to add and subtract polynomials.
This lesson is not difficult conceptually, but it is important that you get sufficient practice in addition
and subtraction of polynomials.
Adding Polynomials
Let's begin with an example in order to see how to add polynomials.
Example 1
Add x 3 + x + 1 and 3x 3 + x 2 + 2x .
Solution
First, write the polynomials side by side.
( x 3 + x + 1 ) + ( 3x 3 + x 2 + 2x )
The monomials x 3 and x each occur twice, so group the terms together and combine them.
x 3 + x + 1 + 3x 3 + x 2 = ( x 3 + 3x 3 ) + x 2 + ( x + 2x )
+ 2x
+1
= 4x 3 + x 2 + 3x + 1
This polynomial is simpler than the original two polynomials written side by side.
Like Terms
To be explicit about the steps taken to add polynomials, let's talk a little bit about like terms.
Definition of Like Terms
When we are given two polynomials, we say the monomials in the polynomials are like terms if they
contain exactly the same number of occurrences of each variable.
Example 2
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Name the like terms in x 4 + 4x 3 + 6x 2 + 4 x + 1 and x 3 + x 2 + 1.
Solution
There are several pairs of like terms, shown in the diagram below.
The pairs of like terms are connected by arrows. So, the like terms are 4x 3 and x 3 , 6x 2 and x 2 , 1
and 1.
Example 3
Name the like terms in ab + 2a + 2b and ab - a + 1.
Solution
In this case, there are two pairs of like terms, ab and ab , 2a and -a , shown in the diagram below.
Whenever there are like terms, collect them and add them together to get a single term. In Example
2, add 4x 3 and x 3 together to get 5x 3 . In the same way, add 6x 2 and x 2 together to get 7x 2 , and
finally, 1 + 1 = 2.
When we do this, we get a simpler polynomial.
( x 4 + 4x 3 + 6x 2 + 4 x + 1 ) + ( x 3 + x 2 + 1 ) = x 4 + 5x 3 + 7x 2 + 4x + 2
In Example 3, collect the ab and a terms to get
( ab + 2a + 2b ) + ( ab - a + 1) = 2ab + a + 2b + 1.
Key Idea
Sums of polynomials can be simplified by adding together like terms. In the same way, differences of
polynomials can be simplified by collecting together all pairs of like terms.
Polynomials
75
Subtracting Polynomials
Example 4
Simplify ( x 3 - 3x 2 + 3 x - 1) - ( x 2 + 2x + 1).
Solution
First write this as an addition expression by adding the additive inverse.
( x 3 - 3x 2 + 3 x - 1) - ( x 2 + 2x + 1) = ( x 3 - 3x 2 + 3 x - 1) + ( -x 2 - 2x - 1)
To simplify the difference, collect all pairs of like terms. These are shown in the diagram below.
Collect the terms to get the following polynomial.
x 3 + ( - 3 - 1) x 2 + (3 - 2) x + ( - 1 - 1) = x 3 - 4x 2 + x - 2
There is another method of adding and subtracting polynomials that is similar to adding and
subtracting numbers. Write the polynomials one over the other, with like terms lined up in columns.
To find the difference, either subtract the coefficients in each of the columns, or add the additive
inverse.
Example 5
Subtract 4x 3 + 5x 2 + 6x + 7 and 2x 3 + 3x + 4.
Solution
Write the polynomials with the like terms in columns.
Notice that a zero coefficient is added for each missing term. Now add the coefficients in each
column.
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9.2
Algebra e-Book
Factoring
After studying this lesson, you will be able to:
· Factor various types of problems.
Steps of Factoring:
1. Factor out the GCF
2. Look at the number of terms:
· 2 Terms: Look for the Difference of 2 Squares
· 3 Terms: Factor the Trinomial
· 4 Terms: Factor by Grouping
3. Factor Completely
4. Check by Multiplying
This section is a review of the types of factoring we've covered so far. Follow the steps listed above
to factor the problems.
Example 1
Factor 3x 2 - 27
1 st : Look for a GCF....the GCF is 3 so we factor out 3: 3( x 2 - 9)
2 nd : Look at the number of terms in the parenthesis. There are 2 terms and it is the difference of 2
squares. We factor the difference of 2 squares (keeping the 3). 3(x + 3) ( x - 3)
3 rd : Now, make sure the problem is factored completely. It is.
4 th : Check by multiplying.
Example 2
Factor 9y 2 - 42y + 49
Polynomials
77
1 st : Look for a GCF....the GCF is 1 so we don't have to worry about that.
2 nd : Look at the number of terms. There are 3 terms so we factor the trinomial.
-make 2 parentheses
-using the sign rules, we know the signs will be the same because the constant term is positive
- we also know they will be negative because the inside/outside combination must equal -58y
-find the factors of the 1 st term: 1y, 9y and 3y, 3y . Let's try 3y, 3y
-find the factors of the constant term: 1, 49 and 7, 7. Let's try 7, 7 (3y - 7) (3y - 7)
-check the inside/outside combination: inside we have -21y and outside we have -21y which adds
up to -42y
3 rd : Now, make sure the problem is factored completely. It is.
4 th : Check by multiplying.
Algebrator can easily solve these type of problems:
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Clicking on the "solve step" button once shows the first step of the solution process.
An explanation can be obtained by clicking on the "Explain" button.
Polynomials
Clicking on the "Solve All" button will show all steps in the solution process.
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Example 3
Factor x 3 - 5x 2 - 9x + 45
1 st : Look for a GCF....the GCF is 1 so we don't have to worry about that.
2 nd : Look at the number of terms. There are 4 terms so we factor by grouping.
Group the terms (x 3 - 5x 2 ) + (- 9x + 45 )
Take the GCF of the each group: x 2 (x - 5 )(- 9(x - 5 ))
Take the GCF of the entire problem: (x - 5 )(x 2 -9)
3 rd : Now, make sure the problem is factored completely. It isn't. We can factor the second
parenthesis.
(x - 5 )(x + 3)(x - 3)
4 th : Check by multiplying.
9.3
Evaluating
A polynomial is an algebraic expression. Like other algebraic expressions involving variables, a
polynomial has no specific value unless the variables are replaced by numbers. A polynomial can be
evaluated without using funcion notation.
Value of a Polynomial
a) Find the value of -3x4 - x3 + 20x + 3 when x = 1.
b) Find the value of -3x4 - x3 + 20x + 3 when x = -2.
Solution
a) Replace x by 1 in the polynomial:
-3x4 - x3 +
20x + 3
= -3(1)4 - (1)3 + 20(1) + 3
= -3 - 1 + 20 + 3
Polynomials
81
= 19
So the value of the polynomial is 19 when x = 1.
b) Replace x by -2 in the polynomial:
-3x4 - x3 + 20x +
3
= -3(-2)4 - (-2)3 + 20(-2) +
3
= -3(16) - (-8) - 40 + 3
= -48 + 8 - 40 + 3
= -77
So the value of the polynomial is -77 when x = -2
9.4
Higher degree polynomials
It is not necessary always to use substitution to factor polynomials with higher degrees or variable
exponents. In the next example we use trial and error to factor two polynomials of higher degree and
one with variable exponents. Remember that if there is a common factor to all terms, factor it out
first.
Example 1
Higher-degree and variable exponent trinomials
Factor each polynomial completely. Variables used as exponents represent positive integers.
a) x8 - 2x4 - 15
b) -18y7 + 21y4 + 15y
c) 2u2m - 5um - 3
Solution
a) To factor by trial and error, notice that x8 = x4 · x4. Now 15 is 3 · 5 or 1 · 15. Using 1 and 15
will not give the required -2 for the coefficient of the middle term. So choose 3 and -5 to get the -2
in the middle term:
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x8 - 2x4 - 15 = (x4 - 5)(x4 + 3)
b) -18y7 + 21y4 +
15y
= 3y(6y6 - 7y3 - 5)
Factor out the common factor -3y first.
= 3y(2y3 + 1)(3y3 - 5)
Factor the trinomial by trial and error.
c) Notice that 2u2m = 2um · um and 3 = 3 · 1. Using trial and error, we get
2u2m - 5um - 3 = (2um + 1)(um - 3).
9.5
Completing the square
The essential part of completing the square is to recognize a perfect square trinomial when given its
first two terms. For example, if we are given x2 + 6x, how do we recognize that these are the first
two terms of the perfect square trinomial x2 + 6x + 9? To answer this question, recall that x2 + 6x +
9 is a perfect square trinomial because it is the square of the binomial x + 3:
(x + 3)2 = x2 + 2 · 3x + 32 = x2 + 6x + 9
Notice that the 6 comes frommultiplying 3 by 2 and the 9 comes from squaring the 3. So to find the
missing 9 in x2 + 6x, divide 6 by 2 to get 3, then square 3 to get 9. This procedure can be used to
find the last term in any perfect square trinomial in which the coefficient of x2 is 1.
Rule for Finding the Last Term
The last term of a perfect square trinomial is the square of one-half of the coefficient of the middle
term. In symbols, the perfect square trinomial whose first two terms are
.
Helpful hint
Review the rule for squaring a binomial: square the first term, find twice the product of the two
terms, then square the last term. If you are still using FOIL to find the square of a binomial, it is time
to learn the proper rule.
Example 1
Polynomials
83
Finding the last term
Find the perfect square trinomial whose first two terms are given.
Solution
a) One-half of 8 is 4, and 4 squared is 16. So the perfect square trinomial is x2 + 8x + 16.
b) One-half of -5 is
c) One-half of
d) One-half of
is
, and
, and
is
squared is
squared is
, and
. So the perfect square trinomial is
.
. So the perfect square trinomial is
. So the perfect square trinomial is
Another essential step in completing the square is to write the perfect square trinomial as the square
of a binomial. Recall that
a2 + 2ab + b2 = (a + b)2 and a2 - 2ab + b2 = (a - b)2.
Example 2
Factoring perfect square trinomials
Factor each trinomial.
Solution
a) The trinomial x2 + 12x + 36 is of the form a2 + 2ab + b2 with a = x and b = 6. So x2 + 12x + 36
= (x + 6)2. Check by squaring x + 6.
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is of the form a2 - 2ab + b2 with a = y and
b) The trinomial
Check by squaring
c) The trinomial
. So
.
is of the form a2 - 2ab + b2 with a = z and
. So
Chapter
10
Functions
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10
Functions
10.1
Adding and subtracting
Recall that a function is a rule that assigns exactly one output number to each input number.
We can combine functions in several ways in order to make new functions.
Definion — Sum and Difference of Two Functions
Given two functions, f(x) and g(x):
The sum of f and g, written (f + g)(x), is defined as (f + g)(x) = f(x) + g(x).
The difference of f and g, written (f - g)(x), is defined as (f - g)(x) = f(x) - g(x).
The domains of (f + g)(x) and (f - g)(x) consist of all real numbers that are in the domain of both f(x)
and g(x).
Example 1
Given f(x) = 0.5x + 4 and g(x) = 0.25x + 1, find the sum (f + g)(x).
Solution
Use the definition for the sum of
functions.
(f + g)(x) = f(x) + g(x)
Substitute for f(x) and g(x).
= (0.5x + 4) + (0.25x +
1)
Combine like terms.
= 0.75x + 5
So, (f + g)(x) = 0.75x + 5.
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Example 2
Given f(x) = 5x2 + 6x - 12 and g(x) = 8x - 15, find the difference (f - g)(x).
Solution
Use the definition for the difference of
functions.
(f - g)(x) = f(x) - g(x)
Substitute for f(x) and g(x).
= (5x2 + 6x - 12) - (8x 15)
Remove parentheses.
= 5x2 + 6x - 12 - 8x + 15
Combine like terms.
= 5x2 - 2x + 3
So, (f - g)(x) = 5x2 - 2x + 3.
You have already learned to evaluate a function for a specific value of the input.
For example, if f(x) = 2x + 1, then we can
find f(3) by substituting 3 for x in the
f(x) = 2x + 1
function rule.
f(3) = 2(3) + 1
=7
Likewise, we can evaluate the sum or difference of two functions for a given number. There are two
methods that are typically used.
Procedure — To Evaluate the Sum or Difference of Functions
Step 1 Find (f + g)(x) or (f - g)(x).
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Step 2 Use x = a to find (f + g)(a) or (f - g)(a).
10.2
Linear functions
You have learned that a function is a rule that assigns to each input number, x, exactly one output
number, y.
A linear function is a rule that can be written in the form:
y = Ax + B or f(x) = Ax + B
For example, the following are linear functions:
y= x
or
f(x) = x
y = -2x + 6
or
f(x) = -2x + 6
The following functions are not linear functions:
y = x2 - 7 or f(x) = x2 - 7 (Not linear since x is squared.)
(Not linear since x is in the denominator.)
For any linear function y = Ax + B:
• The domain is all real numbers, which is the interval (-8 , +8 ).
This is because we can multiply any input x, by any real number A, and then add any real number B.
• If A ? 0, then the range is all real numbers, (-8 , +8 ).
• If A = 0, then the graph is a horizontal line. In that case, the range is the single number B.
In applications of linear functions, however, the domain and range may be restricted to only a
portion of the real numbers.
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10.2.1 Graphing linear functions
We can graph a linear function by calculating two ordered pairs, plotting the corresponding points on
a Cartesian coordinate system, and then drawing a line through the points. We typically calculate and
plot a third point as a check.
To find ordered pairs, we select values for x and then calculate the corresponding values for y. Thus,
the output value y depends on our choice of the input value x. For this reason, the variable y is
frequently called the dependent variable and the variable x is called the independent variable.
Example 1
Make a table of at least three ordered pairs that satisfy the function f(x) = 3x - 1. Then, use your
table to graph the function.
Solution
To make a table, select 3 values for x. We’ll let x = -2, 0, and 2.
Substitute the values of x into the function and simplify.
x
f(x) = 3x - 1
-2
f(-2) = 3(-2) - 1 = -6 - 1 =
-7
0
2
f(0) = 3(0) - 1 = 0 - 1 = -1
f(2) = 3(2) - 1 = 6 - 1 = 5
(x, y)
(-2, -7)
(0, -1)
(2, 5)
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Now, plot the points (-2, -7), (0, -1), and (2, 5) and then draw a line through them.
Two important characteristics of the graph of a linear function are its y-intercept and its slope.
• The y-intercept is the point where the line crosses the y-axis.
• The slope measures the steepness or tilt of the line. Slope is defined as the ratio of the rise to the
run of the line. When moving from one point to another on the line, the rise is the vertical change and
the run is the horizontal change.
The linear function, f(x) = Ax + B, is another way of writing the familiar slope-intercept form for the
equation of a line, y = mx + b. In f(x) = Ax + B the slope of the line is given by A and the y-intercept
is given by B.
f(x) = Ax + B is equivalent to y = mx + b
This means that the graphs of all linear functions are straight lines. That is why such functions are
called linear.
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Example 2
Graph the function:
Solution
This has the form of a linear
function.
Thus, the slope is
and the y-intercept is b = -3.
To graph the line, first plot the y-intercept; that is, plot the point (0, -3).
From this point rise in the y-direction 5 units (the numerator of the slope) and run in the x-direction 4
units (the denominator of the slope). The new location (4, 2) is a second point on the line.
Finally, connect the plotted points.
10.3
Quadratic functions
A Quadratic (or Second-Degree) Function is a function of the form f(x) = ax2 + bx + c, where a,
b and c are real numbers and a ? 0. This form is called the General Form of a Quadratic
Function.
The graph of a quadratic function is a parabola with a "turning point" called the vertex.
The domain of a quadratic function is all real numbers.
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f(x) = x2 is concave up while f(x) = -x2 + 4x + 21 concave down.
So, when looking at the general form for a quadratic function, f(x) = ax2 + bx + c, what does a tell
us about the graph?
The sign of a tells us whether the parabola is concave up or concave down.
There is another form of a quadratic function, called the Standard Form of a Quadratic Function
y = a(x - h)2 + k is a quadratic function with vertex (h, k) and axis of symmetry x = h. It opens up if
a > 0 and down if a < 0. It is wider than f(x) = x2 if -1 < a < 1, and narrower if a > 1 or if a < -1.
Now if the function is in General Form, f(x) = ax2 + bx + c, the vertex is not as obvious. There is a
helpful formula, however:
Given a quadratic function f(x) = ax2 + bx + c, the vertex is the point:
When you graph a Quadratic Function (parabola), you should show:
· Vertex
· Axis of Symmetry
· x-intercepts (or ) if it has any zeros
· y-intercept
10.3.1 Graphing quadratic functions
The Graph of f(x) = Ax2 + Bx + C
To graph a quadratic function, first calculate several ordered pairs. Then, plot the corresponding
points on a Cartesian coordinate system. Finally, connect the points with a smooth line.
Example 1
Make a table of five ordered pairs that satisfy the function f(x) = x2. Then, use the table to graph the
function.
Solution
To make a table, select 5 values for x. We will let x = -2, -1, 0, 1, and 2. Substitute those values of
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x into the function and simplify.
x
f(x) = x2
(x ,y)
-2
f(-2) = (-2)2 = 4
(-2, 4)
-1
f(-1) = (-1)2 = 1
(-1, 1)
0
f(0) = (0)2 = 0
(0, 0)
1
f(1) = (1)2 = 1
(1, 1)
2
f(2) = (2)2 = 4
(2, 4)
Now, plot the points and connect them with a smooth curve.
The graph of a quadratic function, f(x) = Ax2 + Bx + C, has a distinctive shape called a parabola.
The sign of the coefficient of x2, A, determines whether the graph opens up or down.
• When A is positive, the parabola opens up.
• When A is negative, the parabola opens down.
Graph the functions and state the domain and range of each.
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a. f(x) = x2 + 3
b. f(x) = x2 - 5
Solution
a. Graph The graph of f(x) = x2 + 3 is related to the graph of y = x2:
For each input value x, the output of f(x) = x2 + 3 is 3 more than the output of f(x) = x2.
Thus, the graphs have the same shape, but the graph of f(x) = x2 + 3 is shifted up 3 units.
Domain Since we can square any real number, the domain is all real numbers or (-8 , +8 ).
Range We can see from the graph that the smallest value of y is 3. Therefore, the range is y = 3 or
[3, +8 ).
b. Graph To graph f(x) = x2 - 5, we note that for each input value x, the output of f(x) = x2 - 5 is 5
less than the output of f(x) = x2. Thus, the graph is shifted down 5 units.
Domain Since we can square any real number, the domain is all real numbers or (-8 , +8 ).
Range We can see from the graph that the smallest value of y is -5. Therefore, the range is y = -5
or [-5, +8 ).
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95
Composition
Another way to combine two functions is called the composition of functions. This is where the
output of one function becomes the input of a second function.
Definition— Composition of Functions
Let f(x) and g(x) represent two functions. The composition of f and g, written (f ? g)(x), is defined as
(f ? g)(x) = f[g(x)]
Here, g(x) must be in the domain of f(x). If it is not, then f[g(x)] will be undefined.
Note:
In the composition (f ? g)(x), the output from g(x) is the input for f(x).
Example 1
Given f(x) = 3x - 5 and g(x) = x2 - 7, find:
a. (f ? g)(x)
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b. (g ? f)(x)
Solution
a. Use g(x) as the input for f(x).
(f ? g)(x) = f[g(x)]
Replace g(x) with x2 - 7.
= f[x2 - 7]
In f(x), replace x with x2 - 7.
= 3(x2 - 7) - 5
Remove parentheses.
= 3x2 - 21 - 5
Subtract.
= 3x2 - 26
So, (f ? g) = 3x2 - 26.
b Use f(x) as the input for g(x).
.
Replace f(x) with 3x - 5.
(g ? f)(x) = g[f(x)]
= g[3x - 5]
In g(x), replace x with 3x - 5.
= (3x - 5)2 - 7
Square the binomial.
= 9x2 - 15x - 15x + 25 - 7
Combine like terms.
= 9x2 - 30x + 18
So, (g ? f)(x) = 9x2 - 30x + 18.
Notes:
In (f ? g)(x) = f[g(x)], we substitute the expression for g(x) into f(x).
In (g ? f)(x) = g[f(x)], we substitute the expression for f(x) into g(x).
In this example:
(f ? g)(x) = 3x2 - 19
(g ? f)(x) = 9x2 - 30x + 18.
In most cases, (f ? g)(x) and (g ? f)(x) do NOT yield the same result.
10.5
Exponential functions
Definition of an Exponential Function
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Previously, you have studied functions that have terms where the base is a variable and the exponent
is a constant. For example,
In this lesson, you will study exponential functions. Exponential functions have terms where the
base is a constant and the exponent contains a variable.
Definition — Exponential Function
An exponential function is a function that has the form:
f(x) = bx
where b and x are real numbers, b > 0, and b ? 1.
The domain is all real numbers.
The range is all positive real numbers.
Note:
We restrict the values of the base, b, so that the function f(x) = bx is a one to one function. ? 1.
A multiple of an exponential function is also an exponential function. This includes the following
forms:
General form
Example
f(x) = A · bx
f(x) = 5 · 2x
f(x) = C + A · bx
f(x) = -3 + 7 · 5x
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Note the similarities and differences between the graphs of linear, quadratic, and exponential
functions:
Linear: f(x) = 2x (straight
line)
Quadratic: f(x) = x2
(parabola)
Exponential: f(x) = 2x (a
curve)
Chapter
11
Trigonometry
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11
Trigonometry
11.1
Angles and degree measures
An angle has three parts: an initial ray, a terminal ray, and a vertex (the point of intersection of
the two rays), as shown in the figure below.
An angle is in standard position if its initial ray coincides with the positive x-axis and its vertex is at
the origin. We assume that you are familiar with the degree measure of an angle. It is common
practice to use θ (the Greek lowercase theta) to represent both an angle and its measure. Angles
between 0º and 90º are acute, and angles between 90º and 180º are obtuse.
Positive angles are measured counterclockwise, and negative angles are measured clockwise. For
instance, the figure below shows an angle whose measure is -45º. You cannot assign a measure to
an angle by simply knowing where its initial and terminal rays are located. To measure an angle, you
must also know how the terminal ray was revolved. For example, this figure shows that the angle
measuring -45º has the same terminal ray as the angle measuring 315º. Such angles are coterminal.
In general, if θ is any angle, then
θ + n(360 ), n is a nonzero integer
is coterminal with θ.
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An angle that is larger than 360º is one whose terminal ray has been revolved more than one full
revolution counterclockwise, as shown in the following figure.
NOTE
It is common to use the simbol θ to refer to both an angle and its measure. For instance, in the
previous figure, you can write the measure of the smaller angle as θ = 45 .
Radian Measure
To assign a radian measure to an angle θ, consider θ to be a central angle of a circle of radius 1, as
shown in the figure below.
The radian measure of θ is then defined to be the length of the arc of the sector. Because the
circumference of a circle is 2πr, the circumference of a unit circle (of radius 1) is 2π. This implies
that the radian measure of an angle measuring is 360 . In other words, 360 = 2π radians.
Using radian measure for θ, the length s of a circular arc of radius r is s = rθ, as shown in the figure
below.
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You should know the conversions of the common angles shown in the next figure. For other angles,
use the fact that is equal to radians.
11.2
Trigonometric functions
There are two common approaches to the study of trigonometry. In one, the trigonometric functions
are defined as ratios of two sides of a right triangle. In the other, these functions are defined in terms
of a point on the terminal side of an angle in standard position. We define the six trigonometric
functions, sine, cosine, tangent, cotangent, secant, and cosecant (abbreviated as sin, cos, etc.),
from both viewpoints.
Definition of the Six Trigonometric Functions
Right triangle definitions, where
(see the figure below)
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Circular function defiinitions, where θ is any angle (see the figure below).
The following trigonometric identities are direct consequences of the definitions ( is the Greek letter
phi).
Trigonometric Identities (Note that sin2θ is used to represent (sin θ)2.)
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