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Section 2.1: Rectangular Coordinates and Graphs of Equations
Definition: Equation in Two Variables
An equation in two variables (like x and y) is a statement in which the algebraic expressions involving x and y
are equal. The expressions are called the sides of the equation.
Definition: Graph of an Equation in Two Variables
The graph of an equation in two variables (like x and y) is the set of all ordered pairs (x, y) in the xy-plane
that satisfy the equation.
An x-intercept of a graph is the x-coordinate of a point on the graph that crosses or touches the x-axis.
A y-intercept of a graph is the y-coordinate of a point on the graph that crosses or touches the y-axis.
y




      







x


y = x^4 - 5x^2 + 4

This graph has x-intercepts at x = -2, x = -1, x = +1, and x = +2, and a y-intercept at y = 4.
You should also know how to read points off of a graph
Intermediate Algebra
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Section 2.2: Relations
A relation is a “link” from elements of one set to elements of another set.
If x and y are two elements in these set, and if a relation exists between x and y, then we say:
x corresponds to y,
or y depends on x,
and we write x  y.
We may also write a relation where y depends on x as an ordered pair (x, y).
We can define relations by:

Maps

Sets of ordered pairs

Graphs

Equations
The domain of a relation is the set of all inputs to the relation (the domain is all the x values on a graph).
The range of a relation is the set of all outputs of the relation (the range is all the y values on a graph).
Section 2.3: An Introduction to Functions
Definition: Function
A function is a relation in which each element of the domain (the inputs, or x-values) of the relation corresponds
to exactly one element in the range (the outputs, or y-values) of the function.
Vertical Line Test
A set of points in the xy-plane is the graph of a function if and only if every vertical line intersects the graph in
at most one point.
When we want to represent a function with an equation in two variables x and y, it is customary to solve the
equation for y, so that y = an algebraic expression involving x only. The algebraic expression then gets a name
(usually f), and the variable x is listed after the function name to indicate that it is the domain variable.
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Section 2.4: Functions and Their Graphs
To determine the domain of a function f(x):
1. Assume all real numbers is the domain, then start removing numbers from that set:
2. Exclude any values of x that are not valid to plug into the function, like values of x that would make the
denominator be zero.
3. Exclude any values of x that do not make sense in terms of real-world applications of the function
Obtaining Information from the Graph of a Function f(x):
1. The domain can be obtained from the extent of the graph in the x direction.
2. The range can be obtained from the extent of the graph in the y direction.
3. The intercepts can have meaningful real-world interpretations.
4. Coordinates can be used to relate meaningful real-world events.
Relating a Function f(x) to its Graph:
1. If you are given an x value, then the corresponding point (x, y) on the graph is obtained by computing
y = f(x). Thus, (x, y) = (x, f(x)) are points on the graph.
2. If you are given a y value, then the corresponding point (x, y) on the graph is obtained by solving the
equation y = f(x) for an x value.
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Section 3.1: Linear Equations and Linear Functions
Definition: Linear Equation
A linear equation (in two variables) is an equation of the form
Ax + By = C
Where A, B, and C are real numbers and A and B both cannot be zero. This is called the standard form for the
equation.
The graph of a linear equation is a line (hence the name).
Graphing a linear equation by plotting points:
Solve for the y variable, make a table of values, then plot the values (you only need 2 points… make them
spread apart for a more accurate graph)
Graphing a linear equation using intercepts:
1. Let y = 0 and solve for x (for the x-intercept)
2. Let x = 0 and solve for y (for the y-intercept)
Equation of a Vertical Line
A vertical line is given by an equation of the form
x=a
Where a is the x-intercept.
Equation of a Horizontal Line
A horizontal line is given by an equation of the form
y=b
Where b is the y-intercept.
Definition: Linear Function
A linear function is a function of the form
f(x) = mx + b
Where m, and b are real numbers.
The graph of a linear function is a line.
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Section 3.2: Slope and Equations of Lines
Definition: Slope
The slope m between two points with coordinates (x1, y1) and (x2, y2) is defined by the formula
y y2  y1
m

.
x x2  x1
If x1 = x2, then the line between the two points is a vertical line, and the slope m is undefined.
Properties of Slope:

m > 0  the line slants upward from left to right

m < 0  the line slants downward from left to right

m = 0  the line is horizontal

m = undefined  the line is vertical

m is a large number  the line is steep

m is a small number  the line is almost horizontal
Slope as an average rate of change:
The slope of the line between two points on a graph represents the average rate of change over that interval
Quick Graphing of a line given a point and a slope:
If you are told a point on the line and the slope of the line through that point, then you can draw the line quickly
by locating a second point by going up by y and over by x from the given point, then drawing the line
connecting them.
Definition: Point-Slope Form of a Line
An equation for a nonvertical line with slope m and that contains the point (x1, y1) is:
y  y1  m  x  x1  (Point-Slope Form of a Line)
Quick Graphing of a line given the y-intercept and the slope:
If you are told the y-intercept of a line and the slope of the line, then you can draw the line quickly by plotting
the y-intercept, then locating a second point by going up by y and over by x from the y-intercept, and finally
drawing the line connecting them.
Definition: Slope-Intercept Form of an Equation of a Line
An equation for a nonvertical line with slope m and y-intercept b is:
y  f  x   mx  b (Slope-Intercept Form of a Line)
How to Find the Equation of a Line When Give Two Points on the Line:
y y
1) Compute the slope m  2 1 .
x2  x1
2) Plug the slope and the coordinates of either given point into the point-slope form.
Section 3.3: Parallel and Perpendicular Lines
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Definition: Relationship between the slopes of parallel lines
Two nonvertical lines are parallel if and only if their slopes are equal and they have different y-intercepts.
Vertical lines are parallel if they have different x-intercepts.
Definition: Relationship between the slopes of perpendicular lines
Two nonvertical lines are perpendicular if and only if the product of their slopes is -1. Alternatively, their
slopes are negative reciprocals of one another. Vertical lines are perpendicular to horizontal lines.
Section 3.4: Linear Inequalities in Two Variables
Support Skill: Determining if an ordered pair satisfies a linear inequality
 Plug in the numbers and see if you get a true inequality.
Main Skill: Graphing Linear Inequalities in Two Variables
 Step 1: Write the inequality as an equality, then graph the equation using a dashed line if it is a strict
inequality (< or >), or using a solid line if it is not a strict inequality ( or ).
 Step 2: Pick a test point and see if the test point satisfies the inequality.

If the test point satisfies the inequality, then shade the half of the plane containing the test point.

If the test point does not satisfy the inequality, then shade the other half.
Section 3.5: Building Linear Models
Using linear equations to model real-world data.
1. First, graph the ordered pairs that make up the data. The resulting graph is called a scatter plot.
The scatter plot should must like a “fuzzy line” if you want to model it with a linear equation.
2. To model the data with a linear equation, you need to find two points that you can use to get the
equation of a line. You can get the two points using one of two techniques:
a. pick two “representative” points to serve as two points
b. draw a line through the data so that about half the points are above the line and half are below
(i.e., the “line of best fit”), then pick two points on the line that cross near the intersection of grid
lines on the graph.
3. Once you have the two points, use our standard technique to find the equation of the line containing
them. You can then use this equation to make predictions.
Section 4.1: Systems of Linear Equations in Two Variables
A system of linear equations is a grouping of two or more linear equations, each of which contains one or
more variables.
A solution of a system of linear equations consists of values for the variables that are solutions to ALL of the
equations in the system.
Geometric/Visual Interpretation of a System of Two Linear Equations in Two Variables:
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INTERSECT: The lines intersect at one point, and thus the system has exactly one solution. This type of
system is called consistent and the equations are called independent.
PARALLEL: The lines never intersect (i.e., they are parallel to one another), and thus the system has no
solutions. This type of system is called inconsistent.
COINCIDENT: The lines lie on top of each other, and thus the system has infinitely many solutions. This type
of system is called consistent and the equations are called dependent.
Solving a System of Two Linear Equations by Graphing

Graph both the lines.

Read the coordinates of the intersection point off the graph.

Check to see if those coordinates are the solution.
Solving a System of Two Linear Equations Using Substitution

Solve one of the equations for one of the unknowns.

Substitute the expression for that unknown into the other equation.

Solve the resulting equation in one unknown.

Substitute that solution into the first equation to solve for the remaining variable.

Check your answer.
Solving a System of Two Linear Equations Using Elimination

Multiply one or both equations by nonzero constants so that the coefficients of one of the variables are
additive inverses.

Add the two equations to obtain a new equation in one unknown.

Solve the resulting equation in one unknown.

Substitute that solution into either of the original equations and solve for the remaining variable.

Check your answer.
Intermediate Algebra
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Section 4.2: Problem Solving: Systems of Two Linear Equations Containing Two Unknowns
Big Skill: You should be able to write out a model for a real-world situation involving two equations in two
unknowns, then solve that system to get an answer.
Section 4.3: Systems of Linear Equations in Three Variables
Example of a system of three linear equations in three unknowns:
y  z  2
 x 

 x  2 y  3z  12
2 x  2 y  z  9

A linear equation with three variables describes a two-dimensional plane embedded in three dimensions:
When two planes intersect, the intersection is usually a line and when three planes intersect, the intersection is
usually a point:
y
x
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Geometric/Visual Interpretation of a System of Three Linear Equations in Three Variables:
Exactly one solution: A consistent system with independent equations where the planes intersect at a single
point.
No solution: An inconsistent system where the planes are either all parallel, or intersect along parallel lines.
Inconsistent systems yield a false equation (like 0 = 3) after trying to solve them.
Infinitely many solutions: A consistent system with dependent equations where the planes all intersect along
the same line, or are all coincident. Dependent systems yield one or more equations of 0 = 0 after applying
Gaussian elimination.
Example of a system of three linear equations in three unknowns that is in triangular form:
x  2 y  z  1

y  2z  5


z  3

Notice: the name triangular form comes from the “blank” triangular space in the lower left corner due to no x
or y variables. The goal of solving a system of linear equations using elimination is to get the system into
triangular form, because a triangular form system is really easy to solve using back-substitution.
Key fact behind the technique of elimination:

Multiplying an equation in a system by a constant, or adding two equations in a system together results
in a new system of equations called a transformed system, and the solution to the transformed system is the
same as the solution to the original system.
Steps for Solving a System of Three Linear Equations in Three Unknowns Using Elimination

Eliminate the x variable from the second and third equations using elimination.

Eliminate the y variable from the third equation using elimination.

Solve for z in the third equation.

Substitute z into the second equation to find the solution for y, then substitute y and z into the first
equation to find the solution for x.
Rules for showing your work:

Draw an arrow from one transformed system to the next, and write on the arrow what you did to
transform the system.

Any equation that is unchanged gets copied from one system to the next.
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Example:
y  z  2
 x 

 x  2 y  3 z  12
2 x  2 y  z  9

 (1)(eqn. #1) is placed in row#1
 2 x  2 y  2 z  4

y  4 z  14


 4y 
z  5

 (4)(eqn. #2) is placed in row#2
y  z 
2
 x 

 x  2 y  3 z  12
 2 x  2 y  z  9

 (eqn. #1)  (eqn. #2) is placed in row#2
 2 x  2 y  2 z  4

4 y  16 z  56


 4y 
z  5

 (eqn. #2)  (eqn. #3) is placed in row#3
y 
z 
2
 x 

y  4 z  14

 2x  2 y 
z  9

 2 x  2 y  2 z  4

4 y  16 z  56


 17 z  51

 (eqn. #3)  ( 17) is placed in row#3
 (2)(eqn. #1) is placed in row#1
 2 x  2 y  2 z  4

y  4 z  14

 2x  2 y 
z  9

 (eqn. #1)  (eqn. #3) is placed in row#3
 2 x  2 y  2 z  4

y  4 z  14


 4y 
z  5

 2 x  2 y  2 z  4

4 y  16 z  56


z  3

Substitute into equation #2:
4 y  16  3  56  y  2
Substitute into equation #1:
2 x  2  2   2  3  4  x  1
Intermediate Algebra
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Section 4.4: Using Matrices to Solve Systems
Example of a system of three equations in three unknowns (that happens to be in triangular form) and
that system’s augmented matrix:
 1 2 1 1 
x  2 y  z  1



y  2 z  5  0 1 2 5


 0 0 1 3
z  3

Steps for Solving a System of Three Linear Equations in Three Unknowns Using an Augmented Matrix

Eliminate the x coefficients from the second and third rows using elimination.

Eliminate the y coefficient from the third row using elimination.

Make the coefficient of z in the third equation equal to one by dividing by the appropriate number.

Convert the triangular matrix back into a system of equations.

Substitute z into the second equation to find the solution for y, then substitute y and z into the first
equation to find the solution for x.
Rules for showing your work:

Draw an arrow from one transformed matrix to the next, and write on the arrow what you did to
transform the matrix.

Any row that is unchanged gets copied from one system to the next.
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Example:
y  z  2
 x 

 x  2 y  3 z  12
2 x  2 y  z  9

 convert to augmented matrix
2 2 2 4 


 0 1 4 14 
 0 4 1 5
 (4)(row #2) is placed in row#2
 1 1 1 2 


1 2 3 12 
 0 2 1 9 
 (1)(row #1) is placed in row#1
2 2 2
4


 0 4 16 56 
 0 4 1 5
 (row #2)  (row #3) is placed in row#3
1 1 1 2 


 1 2 3 12 
 2 2 1 9 
 (row #1)  (row #2) is placed in row#2
2 2 2
4


 0 4 16 56 
 0 0 17 51
 (row #3)  (17) is placed in row#3
 1 1 1 2 


 0 1 4 14 
 2 2 1 9 
 (2)(row #1) is placed in row#1
 2 2 2
4


 0 4 16 56 
 0 0
1
3
 convert back to a system of equations
2 2 2 4 


 0 1 4 14 
 2 2 1 9 
 (row #1)  (row #3) is placed in row#3
 2 x  2 y  2 z  4

4 y  16 z  56


z  3

2 2 2

 0 1 4
 2 4 1
4

14 
5
Substitute into equation #2:
4 y  16  3  56  y  2
Substitute into equation #1:
2 x  2  2   2  3  4  x  1
Intermediate Algebra
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Section 4.5: Determinants and Cramer's Rule
Definition: Determinant of a 2  2 matrix
a b
a b 
Suppose that a, b, c, and d are real numbers. The determinant of the 2  2 matrix 
, written as
,

c d
c d 
a b
is
 ad  bc .
c d
Cramer’s Rule for solving a system of two linear equations in two unknowns
ax  by  s
The solution to the system of equations 
is given by
 cx  dy  t
s
x
b
a s
Dy
t d
c t
D
 x and y 

a b
a b
D
D
c d
c d
Provided that D 
a b
c d
 ad  bc  0
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Definition: Determinant of a 3  3 matrix
 a1,1 a1,2

The determinant of the 3  3 matrix  a2,1 a2,2
 a3,1 a3,2
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a1,3 
a1,1

a2,3  , written as a2,1
a3,3 
a3,1
a1,2
a1,3
a2,2
a2,3 , is calculated using the
a3,2
a3,3
determinants of 2  2 matrices as follows:
a1,1
a1,2
a1,3
a2,1
a2,2
a2,3  a1,1
a3,1
a3,2
a3,3
a2,2
a2,3
a3,2
a3,3
 a1,2
a2,1
a2,3
a3,1
a3,3
 a1,3
a2,1
a2,2
a3,1
a3,2
.
Notice a pattern: the 2  2 determinants are what remains after the row and column corresponding to the matrix
element in front of the 2  2 determinant is blocked out.
Cramer’s Rule for solving a system of three linear equations in three unknowns
 a1 x  b1 y  c1 z  d1

The solution to the system of equations a2 x  b2 y  c2 z  d 2
a x  b y  c z  d
3
3
3
 3
with
a1 b1 c1
d1 b1 c1
a1 d1 c1
a1 b1 d1
D  a2 b2 c2  0 , Dx  d 2 b2 c2 , Dy  a2 d 2 c2 , and Dz  a2 b2 d 2
a3 b3 c3
d3 b3 c3
a3 d3 c3
a3 b3 d3
is given by
D
D
D
x  x , y  y , and z  z .
D
D
D
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FINAL EXAM Review
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Section 4.6: Systems of Linear Inequalities
Solving a System of Linear Inequalities:

Graph each inequality separately

Choose the region that is the intersection of all the inequalities.
Example:
 2x

The graph of the solution of the system 3x
5 x


4y

8

2y

4 is:

3y

6
Section 5.1: Adding and Subtracting Polynomials
A monomial in one variable is the product of a constant and a variable raised to a nonnegative integer power.

General form of a monomial in one variable: axk.

The constant a is called the coefficient and k is the degree of the monomial.

Examples of monomials in one variable:
2x2
-3y5
m6
4z
-7
A binomial is the sum of two monomials, like 3p2 – 6p
A trinomial is the sum of three monomials, like p2 – 6p + 9
A polynomial is a monomial or the sum of monomials, like 4p3 + 2p2 + 7p – 12

A polynomial is in standard form when it is written with the terms in descending order of degree.

The degree of the polynomial is the highest degree of any of its terms.
A monomial in many variables is the product of a constant and two or more variables raised to nonnegative
integer powers.

General form of a monomial (in two variables): axmyn

The constant a is called the coefficient and m + n is the degree of the monomial.
To add polynomials, combine like terms.
To subtract polynomials, combine like terms.
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A polynomial function is a function whose rule is a polynomial.
 The domain of a polynomial function is all real numbers.
 The degree of a polynomial function is the value of the largest exponent on the variable.
 A polynomial function of degree one or zero is called a linear function, like f  x   0.5x 1 , or
f  x  3 .

A polynomial function of degree two is called a quadratic function, like f  x   x2  2 x  9 .

A polynomial function of degree three is called a cubic function, like f  x   2x3  7 x2  5x  3 .
If f and g are two functions, then

The new function that can be made by adding them together is called f + g: (f + g)(x) = f (x) + g(x).
The new function that can be made by subtracting them is called f – g: (f – g)(x) = f (x) – g(x).
Section 5.2: Multiplying Polynomials
Skill #1: Multiplying Monomials
When multiplying monomials:
 Multiply the coefficients to get a single new coefficient.
 Add exponents of common variables to get simplified variable factors.
Example:
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 2a b  6a b    2  6     a
3
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 

 a2    b  b4 
 aaaaa   bbbbb 
2 4
3
  12   a 3 2  b14 
 12a 5b5
Skill #2: Multiplying a Monomial and a Polynomial
When multiplying a monomial and a polynomial, use the extended form of the distributive property:
Extended form of the Distributive Property:
a  b1  b2  b3   bn   ab1  ab2  ab3   abn
Example:
2 x 2  x 2  3x  5  2 x 2  x 2  2 x 2  3x  2 x 2   5
 2 x 4  6 x3  10 x 2
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Skill #3: Multiplying a Binomial and a Binomial
When multiplying a binomial and a binomial, you can use one of four techniques:
1. Distributive Property Technique
Example:
 3x  4  2 x  9    3x  4   2 x   3x  4    9 
 3 x  2 x  4  2 x  3 x   9   4   9 
 6 x 2  8 x  27 x  36
 6 x 2  19 x  36
2. Vertical Multiplication Technique
Example:
 3x  4  2 x  9  
3x  4
 2x  9
 27 x  36
6x2

8x
6 x 2  19 x  36
3. Table Multiplication Technique (not in book)
Example:
To Calculate 3429:

20
9
30
600
270
4
80
36
= 600 + 80 + 270 + 36
= 986
4. FOIL Technique (ONLY works for binomials)
FOIL  First, Outside, Inside, Last
To Calculate  3x  4 2 x  9  :

2x
-9
= 6x2 + 8x – 27x – 36
= 6x2 – 19x – 36
3x
6x2
-27x
4
8x
-36
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Skill #4: Multiplying a Polynomial and a Polynomial
When multiplying a binomial and a binomial, you can use one of two techniques:
1. Distributive Property Technique
Example:
 2 x  3  x 2  5 x  2    2 x  3   2 x  3   5 x    2 x  3   2 
 2 x  x 2  3  x 2  2 x  5 x  3  5 x  2 x   2   3    2 
 2 x3  3x 2  10 x 2  15 x  4 x  6
 2 x3  13x 2  11x  6
2. Table Multiplication Technique (not in book)
Example:
x2
5x
-2

2x
2x3
10x2
-4x
3
3x2
15x
-6
Notice that like terms line up on the diagonals…
Skill #5: Recognizing Special Binomial Products
When multiplying a pair of “conjugate” binomials, or when multiplying a binomial with itself, you can use the
following formulas:
1. The Product of Conjugate Binomials, or, a Difference of Two Squares
(A + B)(A – B) = A2 – B2
2. The Square of a Binomial, or, a Perfect Square Trinomial
(A + B)2 = A2 + 2AB + B2
(A – B)2 = A2 – 2AB + B2
If f and g are two functions, then

The new function that can be made by multiplying them together is called f  g: (f  g)(x) = f (x)  g(x).
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Section 5.3: Dividing Polynomials; Synthetic Division
Dividing a polynomial by a monomial: Divide the monomial into each term of the polynomial, and cancel
when possible.
Dividing a polynomial by a polynomial using long division:
Long division of polynomials is a lot like long division of numbers:
a. Arrange divisor and dividend around the dividing symbol, and be sure to write them in
descending order of powers with all terms explicitly stated (even the terms with zero
coefficients).
b. Divide leading terms, then multiply and subtract.
c. Repeat until a remainder of order less than the divisor is obtained.
Dividend
Remainder
 Quotient 
Divisor
Divisor
Example:
Compute (5x2 + 7x +9) ÷ (x + 6)
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Dividing a polynomial by a binomial using synthetic division: THIS IS A SHORTCUT THAT ONLY
WORKS WHEN THE DIVISOR IS A LINEAR BINOMIAL (I.E., THE DIVISOR IS x – c) !!!
Synthetic division is a shorthand way to divide a polynomial by the linear factor x – c:
a. Write c outside the division bar and the coefficients of the dividend inside the bar.
b. Bring the leading coefficient of the dividend straight down.
c. Compute c times the number in the bottom row, and write the answer in the middle row to the
right.
d. Add and repeat until all coefficients are used up.
Example:
Compute (2x3 – 3x2 – 4x + 11) ÷ (x – 2) using synthetic division.
Definition: Quotient of Functions
If f and g are two functions, then the new function that can be made by taking their quotient is called
f  x
 f 
defined as:    x  
, provided g(x)  0.
g  x
g
f
, and is
g
The Remainder Theorem
If the polynomial P(x) is divided by x – c, then the remainder is the value P(c). This is because when we divide
a polynomial by x – c, the remainder must be of degree less than x – c, which means the remainder has degree
of zero, which is just a numeric constant R. So:
The Factor Theorem
If P(x) is a polynomial function, then x – c is a factor of P(x) if and only if R = P(c) = 0.
In other words, if P(c) = 0, then P(x) can be written as P(x) = (x – c)(Quotient(x)).
(This can be used to see if a divisor divides evenly into a dividend quickly.)
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Section 5.4: Greatest Common Factor; Factoring by Grouping
In algebra, when you are asked to factor a polynomial, you are supposed to find other polynomials whose
product is the polynomial you are factoring.
Examples:

Factor x 2  2 x  1 :
x2  2 x  1   x  1 x  1

Factor x3  3x 2  3x  1 :

Factor x 3  1 :
x3  3x2  3x  1   x  1 x  1 x  1
x3  1   x  1  x 2  x  1

The above examples are prime factorizations because each of the factors are prime polynomials. We say
that the polynomials above have been factored completely.
In algebra, a polynomial with integer coefficients is prime when it can not be written as the product of two
other polynomials with integer coefficients (other than itself and 1).
Examples:

7 is prime

 x  1 is prime

x
2
 x  1 is also prime
Factoring the greatest common factor:
1. Identify the greatest common factor (GCF) in each term.
2. Rewrite each term as the product of the GCF and the remaining factor.
3. Use the Distributive Property to factor out the GCF.
4. Use the Distributive Property to verify that the factorization is correct.
Example:
10 x 2 y 2  15 xy 3  25 x3 y 4  5 xy 2  2 x  5 xy  3 y  5 xy  5 x 2 y 2
 5 xy   2 x  3 y  5 x 2 y 2 
Factoring by grouping:
1. Group terms that have a common factor. You may need to rearrange the terms.
2. In each grouping, factor out the common factor.
3. Factor out the common factor that remains.
4. Check your work.   .
Example:
5 y  5 z  ay  az   5 y  5 z    ay  az 
 5 y  z   a  y  z 
  5  a  y  z 
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Section 5.5: Factoring Trinomials
Steps for Factoring Quadratic Trinomials of the Form x2 + bx + c:

Find factors of c that sum to b…
o List all factors of b
o Add up each pair of factors; choose the pair of factors that add up to b

Supposing that the factors are m and n (i.e., mn = c), write the trinomial in factored form as:
x2 + bx + c = (x + m)(x + n)

Check your work by multiplying out your factors.
Example:

Factor x2 – 4x – 12

All the factors of -12 are: (+1)(-12), (-1)(+12), (+2)(-6), (-2)(+6), (+3)(-4), (-3)(+4)

The factors that add up to –4 are: (+2) and (-6)

 x2 – 4x – 12 = (x+ 2)(x – 6)
Steps for Factoring Quadratic Trinomials of the Form ax2 + bx + c:

Find factors of ac that sum to b…
o List all factors of ac
o Add up each pair of factors; choose the pair of factors that add up to b

Supposing that the factors are m and n (i.e., mn = ac and m + n = b), re-write the trinomial with the linear
term broken up into a sum using m and n:
ax2 + bx + c = ax2 + mx + nx+ c

Take your re-written polynomial and factor it by grouping.

Check your work by multiplying out your factors.
Example:

Factor 2x2 – 9x – 18

Multiply leading term coefficient and constant term: (2)(-18) = -36

All the factors of -36 are: (+1)(-36), (-1)(+36), (+2)(-18), (-2)(+18), (+3)(-12), (-3)(+12), (+4)(-9),
(-4)(+9), (+6)(-6)

The factors that add up to -9 are: (+3) and (-12)

2x2 – 9x – 18 = 2x2 + 3x – 12x – 18
= x(2x + 3) – 6(2x + 3)
= (x – 6) (2x + 3)
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Section 5.6: Factoring Special Products
Product of conjugate binomials, or a Difference of Squares
A2 – B2 = (A + B)(A – B)
Square of a binomial, or a Perfect Square Trinomial
A2 + 2AB + B2 = (A + B)2
A2 – 2AB + B2 = (A – B)2
Factoring a Sum of Two Cubes
A3  B 3   A  B   A2  AB  B 2 
Factoring a Difference of Two Cubes
A3  B 3   A  B   A2  AB  B 2 
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Section 5.7: Factoring: A General Strategy
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Section 5.8: Polynomial Equations
The Zero-Product Property
If the product of two numbers is zero, then at least one of the numbers is zero. That is,
if ab = 0, then a = 0 or b = 0, or both a and b are 0.
Steps for solving any polynomial equation (degree 2 or higher):
1. Expand the polynomial equation (if needed), and collect all terms on one side; combine like terms.
2. Factor the polynomial on the one side.
3. Set each factor from step 2 equal to zero (this is justified by the zero-product property rule).
4. Solve each first degree equation for the variable.
5. Check answers in original equation.
Section 6.1: Multiplying and Dividing Rational Expressions
Examples of rational expressions:
x5
1
x 2  7 x  18
2
2x 1
x 3
x 4
The domain of a rational expression is the set of all numbers which can be plugged into the rational
expression. To find the domain of a rational expression, find all values of the variable that cause the
denominator to be zero, and exclude them from the domain.
A rational expression is simplified when all common factors in the numerator and denominator have been
cancelled.
Examples:
12 2  2  3 2  2  3 3



20 2  2  5 2  2  5 5
 x  5   x  3 x  5
x 2  2 x  15  x  5 x  3



2
2 x  3x  9  2 x  3 x  3  2 x  3  x  3 2 x  3
a c ac
 
b d bd
a c a d ad
To divide rational expressions, use the arithmetic rule    
(i.e., invert and multiply)
b d b c bc
A rational function is a function of the form
p  x
R  x 
q  x
Where p and q are polynomial functions and q is not the zero of the polynomial. The domain consists of all real
numbers except those for which the denominator q is zero.
To multiply rational expressions, use the arithmetic rule
Section 6.2: Adding and Subtracting Rational Expressions
When rational expressions have common denominators, they can be added or subtracted by simply adding
or subtracting the numerators (like with arithmetic fractions).
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To find the least common denominator of a rational expression, begin by factoring each denominator into
its prime factorization. The LCD is the product of every factor from each expression raised to the highest
power found in each expression.
To add or subtract rational expressions, find the LCD, then write each rational expression with the common
denominator, then add or subtract numerators.
Section 6.3: Complex Rational Expressions
Examples of complex rational expressions:
1
3
x
1
1
x
x 1
3

x2 x2
2x  3
1
x2
Method #1: Simplify Numerator/Denominator, then Divide

Convert the numerator of the complex rational expression to a single rational expression.

Convert the denominator of the complex rational expression to a single rational expression.

Divide the two rational expressions.
Example:
1 3x 1
3

x  x x
1
x 1
1

x
x x
3x  1
 x
x 1
x
3x  1 x


x
x 1

3x  1
x 1
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Method #2: Multiply the Numerator and Denominator by the LCD

Find the LCD of all the denominators of all the rational expressions in the complex rational expression.

Multiply both the numerator and the denominator by the LCD.

Simplify the resulting rational expression.
Example:
x 1
3
x 1
3

  x  2  x  2  
  x  2  x  2 
x

2
x

2



x2 x2 
x2
 x2
2x  3
2x  3
 1  x  2  x  2 
  x  2  x  2   1   x  2  x  2 
x2
x2
 x  1 x  2   3  x  2 

 2 x  3 x  2    x  2  x  2 
x 2  3x  3  3x  6
2 x2  7 x  6  x2  4
x2  8
 2
3x  7 x  2


x2  8
 x  2  3x  1
Section 6.4: Rational Equations
Steps for Solving a Rational Equation
1. Determine the domain of the rational equation.
2. Determine the LCD of all the denominators.
3. Multiply both sides of the equation by the LCD and simplify each side of the equation into polynomials.
4. Solve the resulting polynomial equation (using factoring and the zero-product property if needed).
5. Verify your solutions using the original equation. Make sure they are in the domain of the equation (i.e.,
that they are not extraneous solutions).
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Section 6.5: Rational Inequalities
Steps for Solving a Rational Inequality
1. Write the inequality so it is a single rational expression on one side of the inequality and zero on the
other side. Completely factor the numerator and denominator of the rational expression.
2. Determine all numbers that make the factors of the rational expression zero.
3. Use those zeros to separate the real number line into intervals. We do this because the factor will be
positive on one side of the zero and negative on the other side of the zero.
4. Choose a test point in each interval, and determine the sign of the rational expression at that test point.
If the sign for a test value matches the inequality, then the entire interval containing the test point is a
solution.
Section 6.6: Models Involving Rational Expressions
Vocabulary:
a

A ratio of two numbers a and b is:
b

A proportion is an equation where two ratios are equal.

We say that y varies directly with x, or y is directly proportional to x if there is a nonzero number k
such that y  kx . The number k is called the proportionality constant.

We say that y varies inversely with x, or y is inversely proportional to x if there is a nonzero number k
k
such that y  . The number k is called the proportionality constant.
x

We say that y varies jointly with x and z if there is a nonzero number k such that y  kxz . The number k
is called the proportionality constant.

A combined variation is a variation with both direct and inverse variation.
Section 1.1: Linear Equations
The goal in manipulating equations when we solve them is to isolate the variable on one side of the
equation.
When solving equations, proceed opposite the order of operations:
a. Simplify both sides of the equation as much as possible. This may include “clearing fractions”
by multiplying both sides of the equation by the LCD of all fractions in the equation.
b. Look for something to add or subtract to both sides.
c. Look for something to multiply or divide both sides.
d. Look for an exponent to raise both sides to.
Section 1.2:An Introduction to Problem Solving
Translating Phrases and Sentences into Algebraic Form
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English
Math
is, was, are, yields, equals, gives,
results in, is equal to, is equivalent
to
=
product of, of (with a fraction or %) 
sum of
+
difference
-
x more than y
x+y
x less than y
y-x
increase
+
decrease
-
twice x or 2 times x
2x
n times x
nx
Two consecutive integers
n,
n+1
Area of a rectangle: A = Width  Length
Perimeter = sum of the lengths of the sides of a shape
Rate  Time = Amount
Distance = rate  time
Mixture Problems: Portion from Item A + Portion from Item B = Total
part
Percentage problems: percentage 
whole
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Problem Solving with Mathematical Models
1. Identify what you are looking for.
2. Give names to the unknowns.
3. Translate the problem into the language of mathematics. Use pictures to help you when possible.
Solve the equation.
4. Check the reasonableness of your answer.
5. Answer the original question.
Section 1.3: Using Formulas to Solve Problems
Shape
Square
Rectangle
Triangle
Trapezoid
Parallelogram
Circle
Cube
Rectangular Solid
Sphere
Right Circular Cylinder
Cone
Some Geometric Formulas
Formula
Area A  s 2
Perimeter P  4s
Area A  lw
Perimeter P  2l  2w
1
Area A  bh
2
Perimeter P  a  b  c
1
Area A  h  B  b 
2
Perimeter P  a  b  c  B
Area A  ah
Perimeter P  2a  2b
Area A   r 2
Circumference C  2 r   d
Volume V  s 3
Surface Area S  6 s 2
Volume V  lwh
Surface Area S  2lw  2lh  2wh
4
Volume V   r 3
3
Surface Area S  4 r 2
Volume V   r 2 h
Surface Area S  2 r 2  2 rh
1
Volume V   r 2 h
3
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Section 1.4: Linear Inequalities
Properties of Inequalities: An inequality can be transformed into an equivalent inequality by:

adding or subtracting any quantity to both sides, or

multiplying or dividing by any positive quantity.

If both sides are multiplied or divided by a negative quantity, then the inequality symbol gets reversed.
Section 1.5: Compound Inequalities
Definitions:
The intersection of two sets A and B, denoted A  B, is the set of all elements that belong to both set A and set
B.
The union of two sets A and B, denoted A  B, is the set of all elements that belong to either set A or set B, or
belong to both sets.
The word and implies intersection, the word or implies union, because and is more restrictive, while or is more
inclusive.
A compound inequality is formed by joining two inequalities with the word “and” or “or”, and they are used
when a quantity must satisfy two conditions. To solve a compound inequality means to find all values of the
variable that satisfy both inequalities.
Steps for Solving a Compound Inequality with “and”:
1. Solve each inequality separately.
2. Take the intersection of the two solution sets for the solution set of the compound inequality.
Writing Inequalities with “and” Compactly:
If a < b, then we can write the compound inequality
a < x and x < b
more compactly as:
a<x<b
Steps for Solving a Compound Inequality with “or”:
1. Solve each inequality separately.
2. Take the union of the two solution sets for the solution set of the compound inequality.
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Section 1.6: Absolute Value Equations and Inequalities
To solve absolute value equations, use the facts that:
 u  a is equivalent to u = a or u = -a.
 u  v is equivalent to u = v or u = -v.
Steps to solve an absolute value equation:
1. Isolate the absolute value.
2. Write two equations, one using the positive value of the constant, and the other using the negative:
u a
ua
u  a
3. Solve each equation.
To solve absolute value inequalities, use the facts that:
 u  a is equivalent to –a < u < a
 u  a is equivalent to –a  u  a
 u  a is equivalent to u < -a or u > a
 u  a is equivalent to u  -a or u  a
Steps to solve an absolute value inequality:
1. Isolate the absolute value.
2. Write a pair of compound inequalities, depending on the direction of the comparison:
u a
u a
3.
u  a
ua
4. Solve each inequality.
u  a
ua
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Section 7.1: nth Roots and Rational Exponents
Definition: The principal square root of a number
If a is a nonnegative real number, and b is a nonnegative real number such that b2 = a, then b is called the
principal square root of a and is denoted by b  a
Properties of Square Roots

Every positive real number has two square roots, one positive, and one negative.

The square root of zero is zero.

We use the symbol
, called a radical, to denote the nonnegative square root of a real number. The
nonnegative square root is called the principal square root.

The number under the radical is called the radicand.

For any nonnegative real number c,

For any real number c,
 c
2
 c (i.e., squaring “undoes” a square root…)
c 2  c . (i.e., a square root “undoes” squaring …)





Evaluating a square root means finding the number whose square is the radicand.
Sometimes we can “see” how to evaluate a square root based on our knowledge of arithmetic facts.
Usually we have to use a calculator to evaluate a square root.
To evaluate a square root by hand, we can “guess, check, modify the guess, and check again.”
To evaluate the square root c by hand, we can use the “Babylonian” iteration formula of
1
c
xn1   xn   .
2
xn 
More Properties of Square Roots

The square root of a perfect square is a rational number.

The square root of a positive rational number that is not a perfect square is an irrational number.

The square root of a negative number is not a real number. In other words, there is no real number such
that when you multiply it with itself gives a negative answer.
Definition: The principal nth root of a number
n
a , where n is an integer greater than or equal to 2, computes to a number b such that if
n
a  b , then b n  a .
If n is an even number bigger than 2, then a and b must be positive.
If n is an odd number, then a and b can be any real number.
Properties of nth Roots

n
an  a if n is an odd number bigger than 2.

n
a n  a if n is an even number bigger than 2.

Another way to write an nth root is a1/n: a1/ n  n a .

In general, a rational exponent can be thought of in one of two ways: a m / n 
 a
n
m
 n am

A negative exponent “moves” the exponential expression to the other side of the fraction bar, and
changes the exponent from negative to positive:
Intermediate Algebra
am / n 
am / n
1
 m/n
1
a
FINAL EXAM Review
1
am / n

Page 35 of 53
am / n
 am / n
1
n
2
3
4
5
6
7
8
9
10
2^n
4
8
16
32
64
128
256
512
1,024
3^n
4^n
5^n
6^n
7^n
8^n
9^n
10^n
9
16
25
36
49
64
81
100
27
64
125
216
343
512
729
1,000
81
256
625
1,296
2,401
4,096
6,561
10,000
243
1,024
3,125
7,776
16,807
32,768
59,049
100,000
729
4,096
15,625
46,656
117,649
262,144
531,441 1,000,000
2,187
16,384
78,125
279,936
823,543 2,097,152 4,782,969
6,561
65,536
390,625 1,679,616 5,764,801
19,683
262,144 1,953,125
59,049 1,048,576 9,765,625
n
2
3
4
5
6
7
8
9
10
(-2)^n
4
-8
16
-32
64
-128
256
-512
1,024
(-3)^n
(-4)^n
(-5)^n
(-6)^n
(-7)^n
(-8)^n
(-9)^n
(-10)^n
9
16
25
36
49
64
81
100
-27
-64
-125
-216
-343
-512
-729
-1,000
81
256
625
1,296
2,401
4,096
6,561
10,000
-243
-1,024
-3,125
-7,776
-16,807
-32,768
-59,049 -100,000
729
4,096
15,625
46,656
117,649
262,144
531,441 1,000,000
-2,187
-16,384
-78,125 -279,936 -823,543 -2,097,152 -4,782,969
6,561
65,536
390,625 1,679,616 5,764,801
-19,683 -262,144 -1,953,125
59,049 1,048,576 9,765,625
Section 7.2: Simplifying Expressions Using the Laws of Exponents
Formulas for rational exponents:
The following formulas apply when a and b are any real numbers, and m and n are rational numbers.
n
a 0  1 for a  0
a m  a mn
 
1
for a  0
an
a m  a n  a mn
a n 
am
1
 a mn  nm
n
a
a
ab n
 a nb n
n
an
a
   n
b
b
a
 
b
n
b
 
a
n
Intermediate Algebra
FINAL EXAM Review
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A rational expression is considered simplified when:

All the exponents are positive.

Each base occurs only once.

There are no parentheses in the expression (unless there is addition or subtraction of a base involved).

There are no powers written to powers.
Section 7.3: Simplifying Radical Expressions
Big Idea: A radical is in simplest form when:
1. As many powers as possible are pulled out of the radical.
2. The order of the radical is as low as possible.
3. There are no radicals in the denominator.
Vocabulary:
1. Radicand is the expression under the radical.
2. The order (or index) of a radical is the number indicating the root being taken.
Rational exponent Formulas
a
a
n
1
2
1
n
 a
 a
n
a  mn a
1
3
3a
m
n a
an 
m n
a
n
a
 
a n  n am  n a
New formulas
n
a n b  n ab
m
  
n
a
n
b
n
a
b
Steps to simplify a radical expression:
1. Write each factor of the radicand as the product of two factors, one of which is a perfect power of the
radicand.
2. Write the radicand as the product of two radicals (using the product property of radicals).
3. Take the nth root of the perfect power.
Intermediate Algebra
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Section 7.4: Adding, Subtracting, and Multiplying Radical Expressions
Big Idea: Radicals can only be added or subtracted when they are similar. Similar radicals have the same
order and same radicand.
 If the radicals are not similar, then they can not be combined by addition or subtraction.
 Example of similar radicals that can be added:
43 7  53 7  93 7
 Example of radicals that are not similar because of different orders and thus can not be added:
3
757
 Example of radicals that are not similar because of different radicands and thus can not be added:
11  17
 Radicals that do not look similar but can be shown to be similar when they are simplified first:
8  2  42  2  2 2  2  3 2
To multiply radical expressions with more than one term, use the extended form of the distributive property
(i.e., your polynomial multiplying skills)
Sections 7.5: Rationalizing Radical Expressions
The main idea behind all the techniques in this section is to multiply the top and bottom of the algebraic
expression by a quantity that will make the radical in the denominator “disappear”.
Rationalizing a denominator with one term:
Multiply top and bottom by an appropriate power of the denominator so that the denominator becomes a perfect
power of the radical.
Rationalizing a denominator with two (square root) terms:
Multiply top and bottom by the conjugate of the denominator so that the denominator becomes a difference of
squares.
Section 7.6: Functions Involving Radicals
To find the domain of a radical expression:
If the index of the radical is even, then the radicand must be nonnegative (i.e., greater than or equal to zero).
If the index of the radical is odd, then the radicand an be any real numbers.
Intermediate Algebra
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Section 7.7: Radical Equations and Their Application
Solving a radical equation with one radical expression:
1. Isolate the radical expression (i.e., get it on one side of the equation by itself with all other terms on the
other side).
2. Raise both sides of the equation to the power of the index of the radical.
3. Solve the resulting equation.
4. Check your answer. It is common to get answers that are not truly solutions. These answers are called
extraneous solutions.
Example:
3x  2  4

3x  2

2
 42
3 x  2  16
3 x  18
x6
Extraneous solutions arise from raising both sides of an equation to an even power, because that process
eliminates information about the sign of the original equation.
Example:
When trying to solve the equation x  2  x , you will get an extraneous solution because the equations
x  2  x and  x  2  x result in the same equation after squaring:
x  2 x
 x
x  2  x
 x  2 x
 2  x
2
x  2  x
2
2
2
x  4  4x  x2
0  x2  5x  4
0   x  1 x  4 
 x = 1 OR x = 4
 x 
 2  x
2
x  2  x
2
2
Intermediate Algebra
FINAL EXAM Review
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Solving a radical equation with two radical expressions:
1. Isolate ONE of the radical expressions.
2. Raise both sides of the equation to the power of the index of the radical to eliminate the first radical.
3. Solve the resulting radical equation (that now only has one radical).
4. Check your answer. Eliminate any extraneous solutions.
Section 7.8: The Complex Number System
Nonnegativity Property of Real Numbers
For any real number a, a2  0.
Consequence: there is no real number whose square is less than zero. i.e.,
7 is not a real number.
Definition: The imaginary Unit
The imaginary unit, denoted by i, is the number whose square root is -1. That is:
i2 = -1
OR
i  1
Definition: Complex Numbers
A complex number is a number of the form a + bi, where a and b are real numbers and i is the imaginary unit.
The real number a is called the real part of the complex number, and the real number b is called the imaginary
part of the complex number.
The standard form for writing a complex number is the form a + bi.
A pure imaginary number is a complex number of the form bi.
Evaluating Square Roots of Real Numbers
If N is a positive real number, then we define the principal square root of –N, denoted as
 N  Ni ,
where i is the imaginary unit.
 N , as
THE ARITHMETIC OF COMPLEX NUMBERS
Note: Adding, subtracting, multiplying, or dividing complex numbers results in an answer that is also a
complex number.
To add complex numbers, add “like terms”:
 a  bi    c  di    a  c   b  d  i
To subtract complex numbers, subtract “like terms”:
 a  bi    c  di    a  c   b  d  i
To multiply complex numbers, use the distributive property, then combine “like terms”.
Intermediate Algebra
FINAL EXAM Review
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One thing to keep in mind: the product property of radicals does not apply when the radicand is a complex
number, so evaluate the square root of the negative number first, then multiply the radicals.
Correct:
25 4 

25i
Incorrect:
 4i 
25 4 
  5i  2i 
 25 4 
 100
 10i 2

 10
 10

To divide complex numbers, multiply the numerator and denominator by the conjugate of the denominator,
then multiply and simplify. Note: the denominator will always multiply out to a difference of squares, which
will be a real number.
Example:
4  3i 4  3i 1  2i


1  2i 1  2i 1  2i
 4  3i 1  2i 

1  2i 1  2i 

4  8i  3i  6i 2
12   2i 
2
4  11i  6
1  4i 2
2  11i

1 4
2 11i


5
5

To compute powers of i, use an exponent that is the remainder of dividing the original exponent by 4. This
works because the powers of i repeat themselves every four factors of i, as can be seen from making a list of
powers of i.
Intermediate Algebra
FINAL EXAM Review
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Sections 8.1: Solving Quadratic Equations by Completing the Square
The Square Root Property
If x 2  p then x  p or x   p .
To Solve a Quadratic Equation Containing Only a Square and Constant Term:

Isolate the square term.

Use the square root property.
Comparison Example:
Solve x 2  16  0 using factoring.
x 2  16  0
 x  4  x  4   0
x  4  0 OR
x40
x4
OR
x  4
Note: Instead of writing
x  16 OR x   16 ,
mathematicians use the shorthand
“plus-minus” notation of
x   16
.
x  4
Solve x 2  16  0 using the square root property.
x 2  16  0
x 2  16
x  16 OR
x   16
x4
x  4
OR
Intermediate Algebra
FINAL EXAM Review
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Another Example: Note that the square term can be the square of a quantity.
 y  3  100  0
2
 y  3  100
2
y  3  100
OR
y  3   100
y  3  10
OR
y  3  10
y7
OR
y  13
Here is how to write the solution using “plus-minus” notation:
 y  3
2
 100
 y  3
2
  100
y  3  10
y  3  10
y  3  10 OR
y  3  10
y7
y  13
OR
When the quadratic equation is prime (i.e., it can’t be factored), like x 2  2 x  1  0 , we manipulate the
equation so that it becomes the square of a binomial plus a constant. This manipulation involves taking the first
two terms, and finding out what we have to add to them to make a perfect square trinomial, which can be
replaced with the square of a binomial.
So, for x 2  2 x  1  0 , the first two terms are identical to the first two terms of the perfect square
2
trinomial x 2  2 x  1 , which comes from the square of the binomial x + 1:  x  1  x 2  2 x  1 . So, here is
what we do:
Intermediate Algebra
FINAL EXAM Review
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x2  2x 1  0
x2  2x  1
x2  2x  1  1  1
 x  1
2
2
 x  1
2
 2
x 1   2
x  1  2
x  1  2
OR
x  1  2
x  0.414
OR
x  2.414
To Solve a Quadratic Equation by Completing the Square :
(i.e, writing a quadratic trinomial as a perfect square trinomial plus a constant)
 Get the constant term on the right hand side of the equation.
i.e., if x 2  bx  c  0 , then write the equation as x 2  bx  c
 Make sure the coefficient of the square term is 1.
 Identify the coefficient of the linear term; multiply it by ½ and square the result.
2

1 
i.e., Find the number b in x  bx  c and compute  b 
2 
Add that number to both sides of the equation.

1 
1 
i.e., x  bx   b   c   b 
2 
2 
Write the resulting perfect square trinomial as the square of the binomial .

1 

1 
i.e.,  x  b   c   b 
2 

2 
Use the square root property to solve the equation.
2
2
2
2
2
2
Intermediate Algebra
FINAL EXAM Review
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Section 8.2: Solving Quadratic Equations by the Quadratic Formula
The Quadratic Formula
The solution(s) to the quadratic formula ax 2  bx  c  0 (for a  0) are given by the quadratic formula:
x
b  b 2  4ac
2a
b2 – 4ac is called the discriminant. The discriminant is important because it determines the nature of the
solutions (roots) of the quadratic equation.
Examples of the Nature of the Roots of a Quadratic Equation with Rational Coefficients:
1. If b2 – 4ac is positive and a perfect square, then there are two solutions that are real, rational, and
unequal. In this case, you can also solve the quadratic equation by factoring.
2. If b2 – 4ac is positive but not a perfect square, then there are two solutions that are real, irrational, and
unequal.
3. If b2 – 4ac = 0, then there is just one solution (a repeated root) that is real and rational (or we can say
that the two solutions are equal). This case can also be solved by factoring.
4. If b2 – 4ac is negative, then there are two solutions that are complex and unequal.
Intermediate Algebra
FINAL EXAM Review
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Section 8.3: Solving Equations Quadratic in Form
Big Skill: You should be able to solve quadratic-like equations by making an appropriate substitution and then
solving the resulting quadratic equation.
Example:
x4  x2  6  0
Let u = x2.  u2 = (x2)2 = x4.

u2  u  6  0
 u  3 u  2   0
u  3 u  2
u  3 u  2
x 2  3 x 2  2
x 3
x   2i
Section 8.4: Graphing Quadratic Equations Using Transformations
Definition: Quadratic Function
A quadratic function (in general form) is a function of the form f  x   ax2  bx  c where a, b, and c are real
numbers and a  0. The domain consists of all real numbers. The point (0, c) is the y-intercept of the graph of
the quadratic function. The graph of a quadratic function is always a parabola.
A quadratic function (in standard form) is a function of the form f  x   a  x  h   k where a, h, and k are
2
real numbers and a  0. The point (h, k) is the vertex of the graph of the quadratic function.
The graph of every quadratic function can be obtained by transforming the graph of y = x2 with:
1. a vertical shift,
2. a horizontal shift,
3. a reflection about the x-axis,
4. and/or a vertical stretch or compression
Intermediate Algebra
FINAL EXAM Review
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Graph of the Quadratic Function in Standard Form f  x   a  x  h   k :
2
 Locate the vertex of the graph at (h, k).
 If a > 0, sketch the parabola opening up, otherwise
sketch it opening down.
 If |a| > 0, sketch the parabola vertically stretched,
otherwise sketch it vertically compressed.
 The picture to the left shows the steps for graphing
2
y  1.5  x  3  1 :



Locate vertex at (3, -1)
Sketch parabola opening down
Sketch parabola vertically stretched.
Graph of the Quadratic Function in General Form f  x   ax2  bx  c :
 Convert the function to standard form
2
f  x   a  x  h   k by completing the square
 Graph the function f  x   a  x  h   k using
transformations.
2
 To graph y  x 2  2 x  5 :
y   x2  2x   5
y  1   x 2  2 x  1  5
y  1   x  1  5
2
y   x  1  6
2
Intermediate Algebra
FINAL EXAM Review
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Section 8.5: Graphing Quadratic Equations Using Properties
Big Idea: There are formulas that convert between the general form and standard form of a quadratic function.
Big Skill: You should be able to use those formulas to convert between forms so that you can quickly sketch the
graph of a quadratic function
Quadratic function in general form: f  x   ax2  bx  c
Quadratic function standard form: f  x   a  x  h   k
2
h
b
2a
4ac  b 2
k
4a
The Vertex of a Parabola
Any quadratic function in general form f  x   ax2  bx  c (a  0) will have its vertex at the point whose
coordinates are:
 b 4ac  b2 
 ,
 .
4a 
 2a
Two alternative ways to state the vertex coordinates are using the discriminant:
D
 b
D  b 2  4ac    ,  
 2a 4a 
And by plugging the x-coordinate of the vertex into the function (i.e., since y = f(x) ):
 b
 b 
  2a , f   2a  



Intermediate Algebra
FINAL EXAM Review
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The x-Intercepts of the Graph of Parabola
The x-intercepts of a graph are the x values where y = 0:
y0
f  x  0
ax 2  bx  c  0
Thus, the x-intercepts of the graph of a parabola are given by the quadratic formula. We can anticipate the
number of x-intercepts based on the discriminant:
If the discriminant D  b 2  4ac  0 , then the graph of f  x   ax2  bx  c has two different x-intercepts at
b  D
.
2a
If the discriminant D  b 2  4ac  0 , then the graph of f  x   ax2  bx  c has one x-intercept, and the vertex of
x
b
.
2a
If the discriminant D  b 2  4ac  0 , then the graph of f  x   ax2  bx  c has no x-intercepts (the graph does
not cross or touch the x-axis).
the graph will touch the x-axis at x 
To Graph a Quadratic Function Using Its Properties:
b
4ac  b 2
 Use the formulas h  
and k 
to quickly convert the general form of the quadratic equation,
2a
4a
2
f  x   ax2  bx  c , to the standard form f  x   a  x  h   k .
 Graph the standard form using translations.
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Section 9.1 Composite Functions and Inverse Functions
Composite Functions:
Another way to combine functions is to take the answer from a first function and use that answer as the input to
a second function.
f  g  x     f g  x 
g  f  x     g f  x 
To find an inverse function:
Solve the equation f  x   c for x.
The expression you get involving c is your inverse function.
Example:
3
Find the inverse of f  x    x  2   5
f  x  c
 x  2  5  c
3
 x  2  c  5
3
x 2  3 c 5
x  3 c 5  2
f 1  x   3 x  5  2
Recall:
f  f 1  x    x
and
f 1  f  x    x
Section 9.2: Exponential Functions
Definition: Exponential Function
An exponential function is a function of the form
f  x  ax
where a is a positive real number (i.e., a > 0) and a  1. The domain of the exponential function is the set of all
real numbers.
Properties of the Graph of an Exponential Function, f  x   a x , when a > 1





The domain is the set of all real numbers.
The range is the set of all positive real numbers.
There are no x-intercepts.
The y-intercept is 1.
1

The graph of f contains the points  1,  ,  0,1 , and 1, a  .
a

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Properties of the Graph of an Exponential Function, f  x   a x , when a < 1






The domain is the set of all real numbers.
The range is the set of all positive real numbers.
There are no x-intercepts.
The y-intercept is 1.
1

The graph of f contains the points  1,  ,  0,1 , and 1, a  .
a

The graph is a reflection about the y-axis of the graph one obtains when a > 1.
x

  1 1   1   x
The function can be written as f  x   a        
 a    a 


x
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Definition
The number e (Euler’s Number) is approximately, e  2.718 281 827
Property for Solving Exponential Equations
If a u  a v , then u = v.
Applications of Exponential Functions: Radioactive Decay
t
 1  t1/ 2
The amount A of a radioactive isotope left after an amount of time t is given by the formula A  t   A0   ,
 2
where A0 is the initial amount of the isotope, and t1/2 is the half-life of the isotope (i.e., the amount of time for
half of the isotope to decay away).
Applications of Exponential Functions: Compound Interest Formula
A principal amount of money P invested in an account with an interest rate of r and which pays compound
interest n times per year will be worth an amount of money A after t years of compounding according to the
nt
 r
formula A  t   P 1   .
 n
Section 9.3: Logarithmic Functions
Definition: Logarithmic Function
The logarithmic function to the base a, where a > 0 and a  1, is denoted by y  loga  x  (read as “y is the
logarithm to the base of a of x”, or as “y is the exponent on a that gives an answer of x”), and is defined as
x  ay .
y  loga  x  is equivalent to
Exponential Functions and Their Inverses
If 2  128 , then
log 2  2 x   log 2 128 
x
x  log 2 128 
x7
We know this is the correct answer because 27 = 128
The logarithm to the base of 2 found the exponent, that when placed on a base of 2, gives an answer of 128.
The logarithm to the base of 2 is the inverse of the exponential function 2x: If f  x   2x , then
f 1  x   log2  x  .
Notice that we had to write a subscript of 2 in the logarithm to indicate it “undoes” an exponential function with
a base of 2.
Converting Equations Between Exponential and Logarithmic Form
54  b

4  log5  b 
y  log3 81 
3 y  81
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Properties of the Graph of an Logarithmic Function, f  x   loga  x  , when a > 1






The domain is the set of all positive real numbers.
The range is the set of all real numbers.
There are no y-intercepts.
The x-intercept is 1.
1

The graph of f contains the points  , 1 , 1.0  , and  a,1 .
a

The function is always increasing.
Natural and Common Logarithms
Definition: Common Logarithm
The common logarithm y  log  x  is defined and y  log10  x 
Definition: Natural Logarithm
The natural logarithm y  ln  x  is defined and y  loge  x 
Solving Logarithmic Equations
 Re-write the logarithmic equation in exponential form, then use your knowledge of exponents and the fact
that if a u  a v , then u = v.
Section 9.4: Properties of Logarithms
Exponent Rule
b0  1
b1  b
b b
n
n
Corresponding Logarithm Rule
logb 1  0
logb  b   1
Inverse Property of Logarithms:
log b  b n   n
b b   M
Product Rule of Logarithms:
logb  xy   logb  x   logb  y 
log M
bu b v  bu  v
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bu
 bu  v
bv
b 
m n
 b mn
Page 53 of 53
Quotient Rule of Logarithms:
 x
logb    log b  x   log b  y 
 y
Power Rule of Logarithms:
log b  x n   n log b  x 
Change of base Formula:
log a  M  
log b  M 
log b  a 
Section 9.5: Exponential and Logarithmic Equations
One-to-One Property of Logarithms
If loga  M   log a  N  , then M = N.
To solve logarithmic equations, do one of two things:

Simplify both sides to single logarithmic terms with the same base, then use the one-to-one property of
logarithms to solve.

Simplify both sides to a single term, then re-write the equation in exponential form.
To solve exponential equations, do one of two things:
 Simplify both sides to a single exponential term with the same base, then use the one-to-one
property of exponential functions to solve.
 Simplify both sides to a single term, then take the logarithm of both sides.