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Elementary Algebra
Final Exam Review
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Section 2.1: The Additive Property of Equality
Definition: Linear Equation in One Variable
A linear equation in one variable can be written in the form
Ax + B = C
For real numbers A, B, and C, and C  0.
The addition property of equality: The same number may be added to (or subtracted from) each side of an
equation without changing the equation. Algebraic statement:
If
then
A = B,
A + C = B + C.
Section 2.2: The Multiplication Property of Equality
The multiplication property of equality: You can multiply (or divide) each side of an equation by the same
NONZERO number without changing the solution. Algebraic statement:
If
A = B,
then AC = BC
are equivalent equations as long as C  0.
Section 2.3: More on Solving Linear Equations
Definition of a Linear Equation: A linear equation (after simplification) has only numbers and the variable to
the first power. Algebraic description of a linear equation: Ax + B = C, and A  0.
Steps for solving a linear equation:
1. Simplify each side separately.
a. Use the distributive property to remove parentheses.
b. Combine like terms.
2. Isolate the variable term.
a. Use the addition property of equality to get all variable terms on one side.
b. Use the addition property of equality to get all numbers on the other side.
3. Isolate the variable.
a. Use the multiplication property of equality to get: x = a number.
4. CHECK YOUR WORK!!
Special situations in equation solving:
1. Sometimes an equation has NO SOLUTIONS. In this case, the variables disappear after you try to
get them all on one side, and you are left with a false number statement. That means that the equation
can’t be solved for any value of the variable.
Example: 2t + 1 = 2t
2. Sometimes an equation has INFINITELY MANY SOLUTIONS. In this case, the variables
disappear after you try to get them all on one side, but you are left with a true number statement. That
means that the equation is solved by ANY value of the variable.
Elementary Algebra
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Section 2.4: An Introduction to Applications of Linear Equations
Six Steps for Doing Word Problems (Applied Problems):
1. Read the problem carefully, and draw a picture if you can. Label the picture with all given information.
2. Assign a variable, and write any other unknown quantities in terms of the variable.
3. Write an equation, starting with a verbal equation if it helps.
4. Solve the equation.
5. State the answer, and verify that it makes sense.
6. Check the answer in the words of the original problem.
Section 2.5: Formulas and Applications from Geometry
Perimeter: Perimeter is the distance around the outside of a two-dimensional shape.
The perimeter of a shape with straight sides is the sum of the lengths of its sides.
The perimeter of a circle is called its circumference.
The formula for the circumference of a circle is: C = 2r
Pi is an irrational numeric constant, and it is approximately given by:   3.14
Perimeter and circumference are measured in units of length, like feet or inches or yards or meters.
Area: Area is the measure of how many squares of a given dimension it takes to cover a geometric shape.
Area is measured in units of length2, like feet2 (i.e., square feet), or inches2 (i.e., square inches), or
yards2 (i.e., square yards), or meters2 (i.e., square meters).
Some geometric formulas for areas:
Rectangle
Arect  lw
Circle
Acirc  r 2
Triangle
Trapezoid
Atri 
Atrap 
1
bh
2
1
 B  b h
2
Elementary Algebra
Final Exam Review
Sphere
Asphere  4 r 2
Box
Abox  2lw  2lh  2wh
Cylinder
Acyl  2 rh  2 r 2
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Volume: Volume is the measure of how many cubes of a given dimension it takes to fill a geometric shape.
Volume is measured in units of length3, like feet3 (i.e., cubic feet), or inches3 (i.e., cubic inches), or yards3 (i.e.,
cubic yards), or meters3 (i.e., cubic meters).
Some geometric formulas for volumes:
Box
Vbox  lwh
Cylinder
Vcyl  r 2 h
Sphere
4
Vsphere  r 3
3
Elementary Algebra
Final Exam Review
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1
Bh
3
B is the base area.
The base can be any
shape (not just a
square or rectangle)!
V pyr 
Pyramid
Cone
Vcone 
1
1
Bh  r 2 h
3
3
Angle facts:
The sum of the interior angles of a triangle is 180.
The sum of the interior angles of a rectangle is 360.
Angles that are on opposite sides of intersecting lines are equal.
Two angles that make a straight line are called supplementary, and add up to 180.
Two angles that make a right angle are called complementary, and add up to 90.
Section 2.6: Ratios and Proportions
Ratio: A fancy word for a fraction. This comes from the fact that when you divide one quantity by another
quantity, you get a “rate”.
Proportion: An equation where two ratios are equal.
Fast trick for solving proportions:
a c
If  , then ab = cd.
b d
Section 2.7: More About Problem Solving
part
whole
Interest Problems: Interest = Principal  Interest Rate  I = pr.
Percent Problems: A percent is just a fraction: % 
Money Problems: number  value of one item = total value.
Speed Problems: distance = rate  time  d = rt. Draw a picture with all distances, rates (speeds), and times
labeled.
Elementary Algebra
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Section 2.8: Solving Linear Inequalities
Ax + B < C
Properties of Inequalities: An inequality can be transformed into an equivalent inequality by:

adding or subtracting any quantity to both sides (the addition property of inequalities), or

multiplying or dividing by any positive quantity (the multiplication property of inequalities).

If both sides are multiplied or divided by a negative quantity, then the inequality symbol gets reversed.
Steps for Solving a Linear Inequality:
1. Simplify each side separately.
2. Isolate the variable terms on one side using the addition property of inequalities.
3. Isolate the variable using the multiplication property of inequalities.
Interval Notation:

Put the lower bound of the interval first

Then a comma

Then the upper bound of the interval.

Use a parenthesis next to the bound if it is not included in the interval.

Use a square bracket next to the bound if it is included in the interval.
Section 3.1: Reading Graphs: Linear Equations in Two Variables
Graph: c vs. n
 cost goes on vertical axis (dependent variable)
 number goes on horizontal axis (independent variable)
Section 3.2: Graphing Linear Equations in Two Variables
Technique #1 for graphing a line: BRUTE FORCE
Solve the linear equation for the dependent variable (i.e., y)
Pick some random values of the independent variable (i.e., x), and calculate y; make a table of the values.
Graph the points and connect them with a straight line.
Technique #2 for graphing a line: USE THE INTERCEPTS (uses the fact that you only need 2 points to
draw a line)
1. Find the x-intercept (which is where the line crosses the x-axis) by setting y = 0 and solving for x.
2. Find the y-intercept (which is where the line crosses the y-axis) by setting x = 0 and solving for y.
3. Graph the points and connect them with a straight line.
Technique #3 for graphing a line: RECOGNIZE SPECIAL CASES
1. Horizontal line: The graph of the equation y = k is a horizontal line with y-intercept (0, y) and no
x-intercept.
2. Vertical line: The graph of the equation x = k is a vertical line with x-intercept (x, 0) and no
y-intercept.
3. Line through the Origin: The graph of the equation Ax + By = 0 is a slanted line through the origin
(0, 0). To graph the line, you’ll need a second point, so pick a random value for x, calculate y, then
graph.
Elementary Algebra
Final Exam Review
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Section 3.3: The Slope of a Line
Definition: Slope of a Line
The slope of a line is defined: slope 
vertical change in y (rise)
horizontal change in x (run)
Slope Formula:
The slope of the line through the points (x1, y1) and (x2, y2) is m 
y2  y1
x2  x1
Finding the Slope of a Line from its Equation:
 Solve the linear equation for the dependent variable (i.e., y)
 The coefficient of the x term is the slope.
Technique #4 for graphing a line: SLOPE INTERCEPT METHOD
Solve the linear equation for the dependent variable (i.e., y)
The constant term is the y-intercept. Plot it.
The coefficient of the x term is the slope. Use it to count up and over (or down and backward) to plot more
points on the line.
Facts about slope:
1. A steep line has a large numerical slope.
2. A not so steep line has a small numerical slope.
3. The “dividing line” is the 45 line, which has a slope of m = 1.
4. Lines that go “uphill” (or are increasing as you read the graph from left to right) have a positive slope.
5. Lines that go “downhill” (or are decreasing as you read the graph from left to right) have a negative
slope.
6. Horizontal lines have a slope of m = 0.
7. Vertical lines have an undefined (or infinite) slope.
8. Parallel lines have the same slope.
9. Perpendicular lines have slopes whose product is -1.
Section 3.4: Equations of a Line
Technique #4 for graphing a line: SLOPE INTERCEPT METHOD
Solve the linear equation for the dependent variable (i.e., y) and simplify. Your equation will be in “slopeintercept” form: y = mx + b.
The constant term (b) is the “y-intercept”. Plot it (i.e., (0, b)).
The coefficient of the x term (m) is the slope. Use it to count up and over to plot more points on the line.
Elementary Algebra
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Technique #5 for graphing a line: POINT SLOPE METHOD This method is used when a slope and a point
OR two points are given.
1. Graph the line using given data.
2. Find the equation for the line using the “point-slope form” for a line: y – y1 = m(x – x1). Note: if you are
only given two points, then you’ll have to calculate the slope yourself first.
Section 3.5: Graphing Linear Inequalities in Two Variables
Examples of Linear Inequalities:
To graph a linear inequality:
1. Graph the boundary line
a. If the inequality uses  or , then draw a solid line to show that the line itself satisfies the
inequality.
b. If the inequality uses just < or >, then draw a dashed line to show that the line does not satisfies
the inequality.
2. Shade the appropriate side.
a. Pick any point not on the line for a test point, then plug its x and y values into the inequality.
b. If the inequality is true for the test point, then shade in that side of the line.
c. If the inequality is false for the test point, then shade in the opposite side of the line.
Section 3.6: Introduction to Functions
Big Idea: There are 4 main ways to specify a relation between numbers in mathematics: listing related pairs
of numbers as ordered pairs, drawing a graph of the relationship between the numbers, giving a verbal
description between related quantities, or specifying a calculation that relates numbers.
1. Listing related numbers as ordered pairs:
a. After 1 hour, the car has traveled 65 miles  (1 hr, 65 mi)
b. After 2 hours, the car has traveled 130 miles  (2 hr, 130 mi)
c. After 3 hours, the car has traveled 195 miles  (3hr, 195 mi)
2. Drawing a graph of the relationship:
Elementary Algebra
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3. Give a verbal description:
a. The car travels 65 miles for every hour it has traveled.
4. Specify a calculation, but give it a name.
a. Let t represent the number of hours the car has been traveling.
b. Let d represent the distance traveled.
c. The fact that distance is a function of time can be written in function notation as d(t) = 65t.
The set of all first components in all the ordered pairs of any relation (i.e., all the x’s) is called the domain
of the relation.
The set of all second components in all the ordered pairs of any relation (i.e., all the y’s) is called the range
of the relation.
A function is a special type relationship where each element in the domain corresponds to exactly one
element in the range.
If a vertical line intersects the graph of a function at more than one point, then the graph is not the graph of a
function.
Section 4.1: Solving Systems of Linear Equations by Graphing
A system of linear equations is a grouping of two or more linear equations, each of which contains the same
variables.
Examples:  y  36 x  200
2 x  3 y  4
 y  36 x  200



 y  40 x
 y  40 x
3 x  y  5
The solution to a system of linear equations consists of values for both of the variables x and y that make both
equations in the system be true.
Example:
The system of equations  y  36 x  200 has solution (x, y) = (50, 2000) because:

 y  40 x
Elementary Algebra
2000 = 36(50) + 200
Final Exam Review
AND
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2000 = 40(50)
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.
There are three different cases for the solution to a system of two equations in two variables. We describe these
cases using the words:

Consistent, which means that there is at least one solution (no solutions  inconsistent)

Dependent, which means that the graphs of the lines are the same (different lines  independent)
Three Possible Cases for Solutions of a System of Two Linear Equations in Two Variables:
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 and the equations are called independent.
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.
We can identify which of these three cases a system of equations will fall into (without graphing) by putting
both equations into slope-intercept form, and comparing the slopes and y-intercepts.
Section 4.2: Solving Systems of Linear Equations by Substitution
 Yesterday, we saw that the lines  y  36 x  200 cross at the point (50, 2000), which means that if you play

 y  40 x
50 games at either club, the cost will be $2,000. Here is how to solve this system using substitution.
Since the first equation already tells us that y = 36x + 200, we can replace the y in the second equation with
36x + 200:
y  40 x
36 x  200  40 x
36 x  200  36 x  40 x  36 x
200  4 x
200 4 x

4
4
50  x
Now that we know x = 50, we can replace the x in y = 36x + 200 to get:
Elementary Algebra
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y  36 x  200
 36  50   200
 1800  200
y  2000
So, we just derived that the solution to this system is (50, 2000).
Solving a System of Two Linear Equations Using the Substitution Method
Solve one of the equations for one of the variables in terms of the other variable.
Replace that variable in the other equation with the expression you just derived.
Solve your new equation (that should have only one variable in it).
Substitute that answer into the first equation and solve it to find the value of the original variable.
Check to see if those coordinates are the solution.
Section 4.4: Applications of Linear Systems
Solving an Applied Problem with Two Variables
1. Read the problem carefully.
2. Draw a picture of the situation, if you can.
3. Assign variables to represent the unknown quantities.
4. Write two equations using both variables.
5. Solve the system of equations.
6. State the answer.
7. Check the answer in the words of the original problem.
Section 4.5: Solving Systems of Linear Inequalities
Solving a System of Linear Inequalities:
Graph each inequality separately (using the method from section 3.5)
Graph the line with either a solid or dashed line
Shade the appropriate side of the line
Shade the region that is the intersection of all the inequalities.
Elementary Algebra
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Section 5.1: The Product and Power Rules for Exponents
Rules for Exponents:
For positive integers m and n:
Rule Name
Product Rule
Rule
a m  a n  a mn
a 
 a mn
3 
 ab 
 a mb m
2 p
m n
Power Rules
Example
6 2  6 4  6 2  4  66
m
2 4
m
5
 32(4)  38
 25 p 5
2
am
a

 
bm
b
2
5 5

 
2
 3 3
Section 5.2: Integer Exponents and the Quotient Rule
Rules for Exponents:
For positive integers m and n and any nonzero real numbers a and b:
New Rules
Rule Name
Zero Exponent Definition
Negative Exponent Definition
Negative to Positive Rules
Quotient Rule
Rule
a0  1
Example
170 = 1; (-4)0 = 1
1 1
32  2 
3
9
5
2
3
4
 5
2
4
3
1
an
bn
 m
a
an 
am
bn
m
a
b
   
b
a
m
a
 a mn
an
 ab 
m
m
m
 a mb m
am
a
   m
b
b
5
5
 2
 3
   
 3
 2
8
2
 283  25
3
2
5
 2 p   25 p 5
2
2
5 5
   2
 3 3
Elementary Algebra
Final Exam Review
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Section 5.3: An Application of Exponents: Scientific Notation
Examples:
4,280 = 4.28  103
2.2  102 = 220
93,000,000 = 9.3  107
4.56  108 = 456,000,000
753,658,554  7.54  109
128,000,000,000,000 = 1.28  1014
0.032 = 3.2  10-2
0.000 259 828  2.60  10-4
0.000 000 089 = 8.9  10-8
0.000 000 000 000 001 = 1  10-15
7.2  10-3 = 0.007 2
8.975  10-6 = 0.000 008 975
Section 5.4: Adding and Subtracting Polynomials; Graphing Simple Polynomials
Polynomial vocabulary:
A polynomial in x is a sum of finite terms of the form axn.
Example: 4x3 + 6x2 – 5x – 8
Notice that the polynomial is written in descending powers of the variable.
The coefficient of any term is the number part
The degree of a term is the exponent on the variable.
The degree of a polynomial is the greatest exponent.
A polynomial with one term is called a monomial.
A polynomial with two terms is called a binomial.
A polynomial with three terms is called a trinomial.
Polynomials with more terms aren’t usually given specific names.
Section 5.5: Multiplying Polynomials
Multiplying two monomials: Use the rules of exponents to combine common bases.
Multiplying a monomial and a polynomial: Use the distributive property.
Multiplying two binomials: Use the FOIL acronym: multiply FIRST terms, then OUTSIDE terms, then
INSIDE terms, then LAST terms. This method ONLY WORKS FOR BINOMIALS!
Elementary Algebra
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Multiplying two polynomials: Multiply every term of the first polynomial with every term of the second
polynomial, then add up the products. There are three techniques for doing this:
1. Technique #1 for multiplying two polynomials: Write out every combination of products, then
combine like terms.
2. Technique #2 for multiplying two polynomials: Write out the multiplication vertically, then carry
it out like a long-hand numerical multiplication problem.
3. Technique #3 for multiplying two polynomials: Write out the polynomials on a grid, multiply out
each pair of monomials, then add like terms.
Section 5.6: Special Products (of Polynomials)
Recognizing Special Binomial Products
When multiplying a pair of “conjugate” binomials, or when multiplying a binomial with itself, you can use the
following formulas:
Product of conjugate binomials
The pattern: the middle terms always add to zero, and you are left with the square of the first term minus
the square of the second term. Verbally, we can say that the product of conjugate binomials is the
square of the first term minus the square of the second term. Mathematically, we describe this pattern
as:
The Product of Conjugate Binomials, or, a Difference of Two Squares
(A + B)(A – B) = A2 – B2
Square of a binomial
The pattern:

The answer is a trinomial.

The first term of the trinomial is the square of the first term of the binomial.

The second term of the trinomial is twice the product of the first and last terms of the binomial.

The last term of the trinomial is the square of the last term of the binomial.
This pattern is described mathematically as:
The Square of a Binomial, or, a Perfect Square Trinomial
(A + B)2 = A2 + 2AB + B2
(A – B)2 = A2 – 2AB + B2
Elementary Algebra
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Section 5.7: Dividing Polynomials
Dividing a polynomial by a monomial: Divide the monomial into each term of the polynomial, and cancel
ab a b
 
when possible. This is allowable because of the arithmetic fraction rule that
c
c c
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.
Section 6.1: The Greatest Common Factor; Factoring by Grouping
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 
Elementary Algebra
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Section 6.2: Factoring Trinomials
Steps for Factoring Quadratic Trinomials of the Form x2 + bx + c:

List all factors of c.

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 sums of these factors are:
o (+1) + (-12) = -11
o (-1) + (+12) = +11
o (+2) + (-6) = –4
o …The rest don’t matter because:

The factors that add up to –4 are: (+2) and (-6)

 x2 – 4x – 12 = (x+ 2)(x – 6)
DON’T FORGET TO FACTOR OUT THE GREATEST COMMON FACTOR BEFORE YOU START
THESE STEPS!
Section 6.3: More on Factoring Trinomials
Steps for Factoring Quadratic Trinomials of the Form ax2 + bx + c:

Compute ac.

List all factors of ac

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

Note: THE ORDER IN WHICH YOU WRITE THE PAIR DOES NOT MATTER!

Take your re-written polynomial and factor it by grouping.

Check your work by multiplying out your factors.
DON’T FORGET TO FACTOR OUT THE GREATEST COMMON FACTOR BEFORE YOU START
THESE STEPS!
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)
Elementary Algebra
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Section 6.4: Special Factoring Rules
DON’T FORGET TO FACTOR OUT THE GREATEST COMMON FACTOR BEFORE YOU START!
1. Difference of Squares: x2 – y2 = (x + y)(x – y)
a. Pattern of the polynomial:
i. the first term is positive, and a perfect square
ii. the second term is negative, and a perfect square
iii. the answer is the sum of the squared quantities times the difference of the squared
quantities.
2. Perfect Squares: x2 + 2xy + y2 = (x + y)2
x2 – 2xy + y2 = (x – y)2
a. Pattern of the polynomial:
i. the first term is positive, and a perfect square
ii. the last term is positive, and a perfect square
iii. the middle term can be positive or negative, but it is twice the product of the two terms
that get squared
iv. the answer is the square of either the sum or difference of the squared quantities
3. Difference or Sum of Cubes: x3 – y3 = (x – y)(x2 + xy + y2)
x3 + y3 = (x + y)(x2 – xy + y2)
a. Pattern of the polynomial:
i. the first term is positive, and a perfect cube
ii. the second term can be positive or negative, but it is a perfect cube
iii. the answer is:
Section 6.5: Solving Quadratic Equations by Factoring
Big Idea for section 6.5: Your skills in factoring can be applied to solving a new kind of nonlinear equation:
the quadratic equation, and this will require two new ways of thinking about how to solve equations:
1. We don’t try to isolate the variable right away.
2. You can get two different answers that are both solutions to the equation.
A quadratic equation in standard form is an equation where a polynomial of degree 2 is equal to zero.
Algebraic way of saying this: ax2 + bx + c = 0.
Note: the right hand side MUST be zero.
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.
A consequence of the zero-product property is that some equations have more than one solution.
Steps for Solving a Quadratic Equation:
1. Get the equation in standard form by expanding the polynomial equation (if needed), collecting all terms
on one side, and combining like terms.
2. Factor the quadratic polynomial.
Elementary Algebra
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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.
Example: Solve the quadratic equation 2x2 – 5x + 7 = x2 + 1
1. Get the equation in standard form.
2 x2  5x  7  x2  1
2 x2  5x  7  x2  1  x2  1  x2  1
x2  5x  6  0
2. Factor the quadratic polynomial.
x2  5x  6  0
 x  2  x  3  0
3. Set each factor equal to zero.
 x  2  x  3  0
x  2  0 OR x  3  0
4. Solve each first degree equation.
x  2  0 OR x  3  0
x  2 OR x  3
5. Check answers in original equation.
2 x2  5x  7  x2  1
2  22  5  2  7  22  1
8  10  7  4  1
55
2 x2  5x  7  x2  1
2  32  5  3  7  32  1
18  15  7  9  1
10  10
Section 6.6: Applications of Quadratic Equations
Big Idea for section 6.6: There are many word problems that require you to solve a quadratic equation to find
the answer.
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Section 7.1: The Fundamental Property of Rational Expressions
To simplify rational expressions:
4. Factor the numerator.
5. Factor the denominator.
6. Cancel any common factors.
Example:
 x  1 x  1
x2 1

2
x  2 x  1  x  1 x  1

 x  1
 x  1
Section 7.2: Multiplying and Dividing Rational Expressions
P R PR
To multiply rational expressions:  
Q S QS
1. The numerator of the answer is the product of the numerators.
2. The denominator of the answer is the product of the denominators.
3. Hint: it helps to write everything in factored form, because then it is easy to see what cancels in your
final answer.
Example:
 x  1 x  6   6  x  3
x2  7 x  6
6 x  18
 2

3x  6
x  2 x  15
3 x  2
 x  5  x  3


6  x  1 x  6  x  3
3  x  2  x  5  x  3
6  x  1 x  6 
3  x  2  x  5 
Elementary Algebra
Final Exam Review
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P R P S PS
   
Q S Q R QR
1. Multiply the first rational expression times the reciprocal of the second rational expression.
To divide rational expressions:
Example:
a b  a 
ab  a 2
a b
a b
 2


2
a  1 a  2a  1  a  1 a  1  a  1 a  1

a  a  b   a  1 a  1

 a  1 a  1  a  b 

a  a  b  a  1 a  1
 a  1 a  1 a  b 

a  a  1
a 1
Section 7.3: Least Common Denominators
To write rational expressions with least common denominators:
1. Factor each denominator into prime factors.
2. List each different denominator factor the greatest number of times it appears in any one of the
denominators.
3. Multiply numerator and denominator of each given rational expression by factors from the list that do
not occur in the denominator of that given rational expression.
Example:
1
1
z 1
z 1



2
z  4 z z  z  4  z  1 z  z  4  z  1
4
4
z
4z

 
z  3z  4  z  4  z  1 z z  z  4  z  1
2
Elementary Algebra
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Section 7.4: Adding and Subtracting Rational Expressions
Adding Rational Expressions with the Same Denominator:
P R PR
P
R
If
and
are rational expressions, then  
. That is to say, to get the answer simply put the sum
Q Q
Q
Q
Q
of the numerators over the common denominator.
Adding (or Subtracting) Rational Expressions with Different Denominators:
Find the least common denominator (LCD).
Rewrite each rational expression as an equivalent expression with the LCD as the denominator.
Put the sum (or difference) of the numerators over the LCD.
Simplify the answer using the fundamental property of rational expressions.
Example:
2k
3
2k
3
 2


2
k  5k  4 k  1  k  4  k  1  k  1 k  1

2k
k 1
3
k 4



 k  4  k  1 k  1  k  1 k  1 k  4

2k  k  1  3  k  4 
 k  4  k  1 k  1

2k 2  2k  3k  12
 k  4  k  1 k  1

2k 2  5k  12
 k  4  k  1 k  1

2k 2  2k  3k  12
 k  4  k  1 k  1

 2k  3 k  4 
 k  4  k  1 k  1
Elementary Algebra
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Section 7.5: Complex Fractions
Method #1 To simplify a complex fraction:
1. Simplify the numerators and denominators into single fractions.
2. Change the complex fraction into a division problem.
3. Invert the second fraction and multiply.
Examples:
1
2
1
1
2
2 3 
2
4 3 1
4
4 1

 2 2
12 1

4 4
5
 2
13
4
5 13
 
2 4
5 4
 
2 13
10

13
1
x 1

x  x x
1 x2 1
x

x
x x
x 1
 2x
x 1
x
x 1 x2 1


x
x
x 1 x


x x2 1
 x  1  x

x   x  1 x  1
1

1
x 1
Elementary Algebra
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Section 7.6: Solving Equations with Rational Expressions
To solve an equation with rational expressions:
1. Multiply each side of the equation by the LCD.
2. Solve the resulting equation.
3. Check your answers by plugging them back in to the original equation. Remember that zero
denominator answers are not allowed.
Example:
6
1
4

 2
5a  10 a  5 a  3a  10
6
1
4


5  a  2  a  5  a  5  a  2 
6
1
4
 5  a  5  a  2  
 5  a  5  a  2  
 5  a  5  a  2 
5  a  2
a 5
 a  5 a  2 
6  a  5  5  a  2   4  5
6a  30  5a  10  20
a  40  20
a  60
Section 7.7: Applications of Rational Expressions
To solve word problems:
1. Draw a picture.
2. Label the picture with the unknowns and the given information.
3. Write an equation that relates all of the information.
4. Solve the equation.
5. Check your answer.
Special equations:
1. Distance, Rate, and Time Relationship: d = rt.
2. Rate of Work: If a job can be done in t units of time, then the rate of work is:
1
job per unit of time.
t
Elementary Algebra
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Section 7.8: Variation
Direct Variation:
1. y varies directly as x means the equation: y = kx.
2. Or, we also say y is proportional to x.
3. k is called the constant of variation (or proportionality constant).
Direct Variation as a Power:
1. y varies directly as the nth power of x means the equation: y = kxn.
Inverse Variation:
1. y varies inversely as x means the equation: y 
k
x
2. y varies inversely as the nth power of x means the equation: y 
k
xn
Elementary Algebra
Final Exam Review
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Section 8.1: Evaluating Roots
ROOTS “UNDO” POWERS!
Square Roots:
If a is a positive real number, then
a is the principal square root of a, and
- a is the negative square root of a.
For nonnegative a,
a a =a
and (- a )  (- a ) = a.
Application of Square Roots: the Pythagorean Theorem
Pythagorean Theorem:
a 2  b2  c2
Higher order roots:
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.
Elementary Algebra
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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 8.2: Multiplying, Dividing, and Simplifying Radicals
Product Rule for Square Roots: For nonnegative real numbers a and b,
a  b  ab (the product of two square roots is the square root of the products)
and ab  a  b (vice-versa; the square root of a product equals the product of the square roots).
This second equation is very useful for simplifying radicals, if you can think of the number under the square
root as a product of a perfect square and another number.
Criteria for a Simplified Radical Expression:

There are as few radicals in the expression as possible.

The radicands are as small as possible.
Quotient Rule for Square Roots: For nonnegative real numbers a and b, and b  0,
a
a

(the square root of a quotient is the quotient of the square roots)

b
b

And
a
a
(the quotient of two square roots is the square root of the quotient)

b
b
Elementary Algebra
Final Exam Review
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The Square Root of a Square: For any number a,


a 2  a (the square root of a square is the absolute value of squared value)
Notice:
 4
2
 16  4  4
Product Rule for Radicals: For all real numbers for which the indicated roots exist,
 n a  n b  n ab

n
n
a na

b
b
Section 8.3: Adding and Subtracting Radicals
Big Idea: Radicals can only be added or subtracted when they are like radicals. Like 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
Criteria for a Simplified Radical Expression:

There are as few radicals in the expression as possible.

The radicands are as small as possible.
Section 8.4: Rationalizing the Denominator
Criteria for a Simplified Radical Expression:

There are as few radicals in the expression as possible.

The radicands are as small as possible.

The radicand has no fractions.

No denominator contains a radical.
Section 8.5: More Simplifying and Operations with Radicals
Criteria for a Simplified Radical Expression:

There are as few radicals in the expression as possible.

The radicands are as small as possible.

The radicand has no fractions.

No denominator contains a radical.
Elementary Algebra
Final Exam Review
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To multiply radical expressions with terms, use the distributive property.
Examples:


3
5 42 5  4 5 2 5 5


6  5 2 6  3 5  3 6  2 6  3 6 3 5  5  2 6  5 3 5
 4 5  2 10
 6 6 6 9 6 5 2 5 6 3 5 5
 20  4 5
 6  6  9 30  2 30  3  5
 36  7 30  15
 21  7 30
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 8.6: Solving Equations with Radicals
Big Idea: To solve an equation with a radical, isolate the radical, then raise both sides to an appropriate power
to get rid of the radical.
Section 8.7: Using Rational Numbers as Exponents
We have seen that a square root “undoes” or “cancels” a power of 2:
25  52  5
But notice what raising a power of 2 to the power of ½ does:
25 2   52   52 2  51  5
1
1
1
2
So, a square root produces the same result as a power of ½.
The same comparison for cube roots and a power of
3
1
3
:
8  3 23  2
8 3   23   23 3  21  2
1
1
3
1
Thus, taking the nth root of a number is the same thing as raising it to a power of
n
a a
1
n
Other results based on this idea and the rules of exponents:
n
a m  a m n  a n  a
1
m
1
n m
   a
 a
1
m
n
n
m
1
n
:
Elementary Algebra
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Section 9.1: Solving Quadratic Equations by the Square Root Property
The Square Root Property
If k is a positive number and x 2  k then x  k or x   k . The solution set is  k , k , which can be




written as  k .
To Solve a Quadratic Equation Using the Square Root Property:

Isolate the square.

Use the square root property.
Section 9.2: Solving Quadratic Equations by Completing the Square
To solve quadratic equations that can’t be factored, 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, to solve x 2  2 x  1  0 , the first two terms are identical to the first two terms of the perfect square trinomial
2
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:
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.
1 
i.e., Find the number b in x  bx  c and compute  b 
2 
Add that number to both sides of the equation.
2
2
1 
1 
i.e., x 2  bx   b   c   b 
2 
2 
2
2
Elementary Algebra
Final Exam Review
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Write the resulting perfect square trinomial as the square of the binomial .
2
2
1 

1 
i.e.,  x  b   c   b 
2 

2 
Use the square root property to solve the equation.
Section 9.3: Solving Quadratic Equations by the Quadratic Formula
Quadratic Formula
The solutions of the quadratic formula ax 2  bx  c  0 (a  0) are:
b  b2  4ac
b  b2  4ac
and
x
2a
2a
2
b  b  4ac
or, in compact form, x 
.
2a
x
Section 9.4: Complex Numbers
Definition: The b
The imaginary unit, denoted by i, is the number whose square root is -1. That is:
i2 = -1
OR
i  1
Thus, for any positive real number b, b  i b .
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.
If a = 0 and b  0 the number bi is called a pure imaginary 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.
Note: under this definition, the real numbers are a subset of the complex numbers.
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
Elementary Algebra
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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” (like using FOIL):
Definition: conjugate of a complex number
The conjugate of a complex number a + bi is a – bi.
Product of conjugates:
 a  bi  a  bi   a2  b2
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

Section 9.5: More on Graphing Quadratic Equations; Quadratic Functions
Big Idea: Defining a new set of numbers that includes the square root of negative numbers lets us write down
answers to many types of equations that we couldn’t otherwise.
Big Skill: You should be able to accurately graph quadratic functions.