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NOTES FOR ALGEBRA 2 CP FINAL
Chapter 1: Equations & Inequalities
Section 1.5 ~ Solving Inequalities (pg. 33-39)
Section 1.1 ~ Expressions & Formulas (pg. 6-10)
Examples:

90  5(2r  6)
For evaluating or simplifying expressions, ALWAYS
follow the order of operations!!!
1.
o
o
o
o
o
o
Parentheses
Exponents
Multiplication
Division
Addition
Subtraction
Real Number Properties
For any real numbers a, b, and c
Commutative
Associative
Identity
Inverse
Distributive
 20 w  32  33
 20 w  1
w
1
20
Remember: When multiplying or dividing by a negative
number, change the direction of the inequality symbol.
(Refer to example 2.)

Some of the properties of real numbers are
summarized below.
Property
60  10r
2.
6r
Section 1.2 ~ Properties of Real Numbers (pg. 11-17)

90  10r  30
 4(5w  8)  33
Addition
Multiplication
a+b = b+a
ab=ba
(a+b)+c = a+(b+c)
(a  b)  c = a  (b  c)
a+0 = a = 0+a
a1=a=1a
a+(-a) = 0 = (-a)+a
a  (1/a) = 1 = (1/a)  a
a(b + c) = ab +ac and (b + c)a = ba + ca
When graphing inequalities…
o <, > = open circle
o
 ,  = closed circle
Section 1.6 ~ Solving Compound & Absolute Value Inequalities
(pg. 41-48)

Compound Inequalities
o “and” compound inequalities – connect
o “or” compound inequalities – separate

Absolute Value Inequalities
o If |a| < b, then –b < a < b.
(i.e., “and” compound inequality)
Example: If |2x + 1| < 5, then –5 < 2x + 1 < 5.
o
Section 1.4 ~ Solving Absolute Value Equations (pg. 27-31)
If |a| > b, then a > b or a < -b.
(i.e., “or” compound inequality)
Example: If |2x + 1| > 5, then 2x + 1 > 5 or 2x + 1 < -5.
Examples:
3n  2  4  0
1.
Chapter 2: Linear Relations & Functions
3n  2  4
Section 2.1 ~ Relations & Functions (pg. 58-64)
This equation is NEVER TRUE because the absolute
value of a number is always positive or zero. Therefore,
the equation has no solution.
35  7 4 x  13
2.
5  4 x  13
5  4 x  13
 5  4 x  13
18  4 x
8  4x
9
x
2
2x
Remember: Before making the two equations, make sure
you have the absolute value bars alone on one side of the
equation.
Definitions:

domain – the set of all first coordinates
(x-coordinates) from the ordered pairs

range – the set of all second coordinates
(y-coordinates) from the ordered pairs

function – a special type of relation in which each
element is paired with exactly one element of the
range (i.e., there is exactly one x for every y)
o
One way you can determine if a relation is a
function is by using the vertical line test. If
no vertical line intersects a graph in more
than one point, the graph represents a
function. (pg. 59)
Section 2.2 ~ Linear Equations (pg. 66-70)

Since two points determine a line, one way to graph
a linear equation or function is to find the points at
which the graph interests each axis and connect them
with a line.
o
o
The x-intercept is the value of x when y = 0.
The y-intercept is the value of y when x = 0.
Example: Find the x-intercept and the y-intercept of the
graph of 3x – 4y + 12 = 0. Then graph the equation.
5 x  3 y  15
5 x  3 y  15
5 x  3(0)  15
5(0)  3 y  15
5 x  15
3 y  15
x3
y5
Section 2.4 ~ Writing Linear Equations (pg. 79-84)

There are three ways of writing a linear equation:
o
standard form
y  ax2  bx  c
o
slope-intercept form
y  mx  b


o
m = slope
b = y-intercept (0, b)
point-slope form
y  y1  m( x  x1 )

m = slope

( x1, y1 ) = a point on the line
Section 2.6 ~ Special Functions (pg. 95-101)
When graphing piecewise functions…
The x-intercept is 3, so the graph crosses the x-axis at
(3,0). The y-intercept is 5, so the graph crosses the y-axis
at (0,5).
Section 2.3 ~ Slope (pg. 71-77)

slope =
rise y 2  y1

run x2  x1
Example:
Example: Find the slope of the line that passes through
each (-2, -1) and (2, -3).
slope = 



Step 1: Draw a dotted vertical line for the boundary
Step 2: Draw the line of each equation separately
Step 3: Look at the direction of the inequality symbol and keep
that side of the line in relation to the boundary

<, < = keep everything to the left of the boundary

>, > = keep everything to the right of the boundary
y 2  y1  3  (1)  2
1



x2  x1
2  (2)
4
2
The slope of a line tells the direction in which it rises
or falls.
positive slope
negative slope
zero slope
undefined slope
Parallel lines have the same slope.
Perpendicular lines have opposite reciprocal slopes.
5 if x  4

f ( x )   x if  2  x  4
 3 if x  2

Boundaries
When graphing absolute value functions…
Chapter 3: Systems of Equations & Inequalities
Think of it as the vertex form of a quadratic formula
Section 3.1 ~ Solving Systems of Equations by Graphing
(pg. 116-122)
Vertex form
y  a( x  h)2  k
Absolute value
y  a xh k
o

(h, k) for both formulas represents the vertex
(or tip of the function)
Example: Graph f(x) = |x| - 6 and g(x) = |x – 6| on the same
coordinate plane.
f(x) = |x| - 6
vertex: (h, k) = (0, -6)
g(x) = |x – 6|
vertex: (h, k) = (6, 0)
Once you find the vertex, find several ordered pairs for each
function.
x
–2
–1
0
1
2
|x| – 6
–4
–5
–6
–5
–4
x
4
5
6
7
8
|x – 6|
2
1
0
1
2
Then graph the points and connect them.
One way to solve a system of equations is to graph
the equations on the same coordinate plane. The
point of intersection represents the solution.
Example:
Solve the system of equations by graphing.
x y  2
3x  y  6
Write each equation in slope–intercept
form before graphing.

x+y=2
y = –x + 2

3x – y = 6
y = 3x – 6
The solution of the
system of equations is the
ordered pair that satisfies
both equations.
The lines of each equation intersect at (2,0). Therefore, the
solution of the system is (2,0).
Note: Make sure to substitute the coordinates into each
equation to check that the point is really a solution to both
equations.

Section 2.7 ~ Graphing Inequalities (pg. 102-105)
The relationship between the graph of a system of
equations and the number of its solutions is
summarized below.
Consistent & Independent
- intersecting lines
- one solution
- different slopes
Graphing inequalities is the same as graphing equations
except:


there is a dotted or solid line
o <, > = dotted line
o
 ,  = solid line
one side of the graph is shaded (by testing a point)
o If TRUE, then shade the area that includes
the test point
o If FALSE, then shade the area that does not
include the test point
Consistent & Dependent
- same line
- infinitely many solutions
- same slope & y-intercept
Example: Graph 3x – 2y > 6.
3x  2 y  6
 2 y  3 x  6
3
y  x3
2
y-int: (0, -3)
slope: 3/2
- up 3
- right 2
Inconsistent
- parallel lines
- no solution
- same slope, different y-intercepts
Section 3.2 ~ Solving Systems of Equations Algebraically
(pg. 123-129)

Section 3.3 ~ Solving Systems of Inequalities by Graphing
(pg. 130-135)

One algebraic method is the substitution method.
Using this method, one equation is solved for one
variable in terms of the other. Then, this expression is
substituted for the variable in the other equation.
2x  3y  2
Example: Use substitution to solve
x  2 y  15
The intersecting regions of the graphs of each
inequality represent the solution to a system of
inequalities.
Examples:
.
1.
Step 1: Solve the second equation for y in terms of
x.
yx
2x  y  7
Solution of y > x
Regions 1 & 2
x  2 y  15
x  2 y  15
Solution of 2x + y < 7
Regions 1 & 3
Step 2: Substitute –2y + 15 for x in the first equation
and solve for y.
The region that provides a solution to both inequalities
is the solution of the system. Region 1 is the solution
to the system.
2x  3y  2
2(2 y  15)  3 y  2
2.
 4 y  30  3 y  2
 7 y  30  2
 7 y  28
x y 5
x 4
The inequality |x| < 4
can be written as x < 4
and x > -4.
y4
Step 3: Substitute the value for y in either original
equation and solve for x.
Graph all of the
inequalities on the same coordinate plane and shade
the region or regions that are common to all.
x  2 y  15
x  2( 4)  15
x  8  15
x7
Therefore, the solution of the system is (7, 4).

Another algebraic method is the elimination method.
Using this method, you eliminate one of the variables
by adding the equations.
Example: Use elimination to solve
2x  3y  2
x  2 y  15
.
Step 1: Multiply the second equation by –2.
 2( x  2 y  15)
 2 x  4 y  30
Section 3.5 ~ Solving Systems of Equations in Three Variables
(pg. 145-152)

Solving systems of equations in three variables is similar
to solving systems of equations in two variables. Use
strategies of substitution and elimination.
a  2b  4c  8
Example: Solve the system 2a  b  c  8
.
 a  3b  2c  9
Step 1: Use elimination to make a system of two equations in
two variables.
Multiply the first equation by 2. Then add it to the second
equation to eliminate a.
Step 2: Add the two equations together to eliminate
x by adding down the columns, and solve for
y. Then find x by substituting 4 for y in either
original equation.
2x  3y  2
x  2 y  15
 2 x  4 y  30
x  2( 4)  15
 7 y  28
x  8  15
y4
x7
Therefore, the solution of the system is (7, 4).
a + 2b – 4c = 8
2a – b + c = –8
2a + 4b – 8c = 16
(+) -2a + b – c = 8
5b – 9c = 24
Also, add the first equation to the third equation to
eliminate a again.
a + 2b – 4c = 8
(+) –a – 3b + 2c = –9
–b – 2c = –1

Step 2: Solve the system of two equations.
x
Multiply the second equation by 5. Then add it to the first
equation to eliminate b.
5b – 9c = 24
–b – 2c = –1
5b – 9c = 24
(+) –5b – 10c = –5
– 19c = 19
c = –1
axis of symmetry (& x-coordinate of vertex):
Now, make a table of values that includes the vertex.
x
-4
-5
-6
-7
-8
Use one of the equations with two variables to solve for b.
–b – 2c = –1
–b – 2(–1) = –1
–b + 2 = –1
b=3
b
12

 6
2a
2(1)
Equation with two variables
x2 + 12x + 36
16 – 48 + 36
25 – 60 + 36
36 – 72 + 36
49 –84 + 36
64 – 96 + 36
f(x)
4
1
0
1
4
(x, f(x))
(-4, 4)
(-5, 1)
(-6, 0)
(-7, 1)
(-8, 4)
Replace c with –1.
Then, use the table to plot each point and graph the function.
Multiply.

Simplify.
Step 3: Solve for a using one of the original equations with
three variables.
a + 2b – 4c = 8
a + 2(3) – 4(–1) = 8
a+6+4=8
a = –2
The y-coordinate of the vertex of a quadratic function is
the maximum value or minimum value attained by the
function.
o
Original equation with three variables
o
Replace b with 3 and c with –1.
Multiply.
Simplify.
If a > 0, then the parabola looks like a happy face.
Therefore, the graph opens up and has a minimum
value.
If a < 0, then the parabola looks like a sad face.
Therefore, the graph opens down and has a
maximum value.
Example: Determine whether the function f ( x)  2 x 2  4 x  3
has a maximum or minimum & state that value.
Therefore, the solution is (-2, 3, -1).
Step 1: Look at the value of a.
a = 2 > 0 (positive #)
 happy face  opens up 
minimum
Chapter 5: Quadratic Functions & Inequalities
Step 2: Find the minimum value by finding the coordinates of
the vertex.
Section 5.1 ~ Graphing Quadratic Functions (pg. 236-244)
x

b
4

1
2a
2(2)
The x-coordinate of the vertex is 1.
Consider the graph of y  ax 2  bx  c , where a  0 .
o
The y-intercept is a(0) 2  b(0)  c  c .
o
The equation of the axis of symmetry is x  
o
The x-coordinate of the vertex is 
b .
2a
y  f (1)  2(1)2  4(1)  3  2  4  3  5
b .
2a
The y-coordinate of the vertex is –5, which also represents the
minimum value of the function.
Therefore, the minimum value of the function is –5.
Use this information to graph any quadratic equation.
Section 5.2 ~ Solving Quadratic Equations by Graphing
(pg. 246-251)

When graphing quadratic equations, the solutions to the
equation are where the parabolas cross the x-axis (i.e.,
the x-intercepts of the graph).


Example: Graph the function f ( x)  x 2  12 x  36 .
a = 1, b = 12, c = 36

y-intercept: c = 36
When solving by graphing, a quadratic equation can have:
o One real solution
o Two real solutions
o No real solution
Example: Solve x2 – x – 12 = 0 by graphing.

Graph the related quadratic function f(x) = x2 – x – 12.

The equation of the axis of symmetry is x = –
–1
2(1)
or
1
2
.

Make a table using x–values around
1
. Then, graph each
2
point.
1
x
-1
0
f(x)
-10
-12
2
–12
1
2
-12
-10
1
4
3x 2  9 x
 2 x 2  12 x  16  0
3x 2  9 x  0
 2( x 2  6 x  8)  0
3x( x  3)  0
 2( x  4)( x  2)  0
3x  0
x3  0
x4  0
x2  0
x0
x3
x4
x2
Remember: When solving by factoring, set each factor that
contains a variable equal to zero and then solve for that
variable.
We can see that the zeros of the
function are –3 and 4.
Just in case you forget! Using the Magic X works like this:
Therefore, the quadratic
equation has two real
solutions, and the solutions of
the equation are –3 and 4.
x2  7 x  6
ac
6
a 1
b  7
-6
-1
c6
Section 5.3 ~ Solving Quadratic Equations by Factoring
(pg. 253-258)

-7
b
The following factoring techniques will help factor
polynomials.
Remember: You are finding factor pairs of 6 that add up to –7,
which are –6 and –1!
Factoring Techniques
# of
terms
any #
Technique
Greatest Common
Factor (GCF)
Difference of
Squares
Sum of Cubes
2
Difference of
Cubes
Magic X
(& Grouping when
a  1)
Grouping
3
4 or
more

Formula
a 2  b2  (a  b)(a  b)
Example: Write a quadratic equation that has roots –
a  b  (a  b)(a  ab  b )
a3  b3  (a  b)(a 2  ab  b2 )
3
3
2
Examples:

2. Factor 5 x 2  80 .
* GCF & Diff. of Sq.
3 x 2  12 x  63
5 x 2  80
 3( x 2  4 x  21)
 5( x 2  16)
 3( x  7)( x  3)
 5( x  4)( x  4)
Solving quadratic equations by factoring is an application
of the Zero Product Property.
Examples:
1.
Solve 3x 2  9 x .
2. Solve  2 x 2  12 x  16  0 .
3
and 1.
4
2
Whenever you factor a polynomial, always look for a common
factor first. Then determine whether the resulting polynomial
factor can be factored again using one or more of the methods
listed above.
1. Factor 3 x 2  12 x  63 .
* GCF & Magic X
Now that you know that a quadratic equation of the form
( x  p)( x  q)  0 has roots (or solutions) p and q, you
can use this pattern to help you find a quadratic equation
for a given pair of roots.
(x – p)(x – q) = 0
  3   (x – 1) = 0
 x    4 

 
3

 x   (x – 1) = 0
4

1
3
2
x –
x–
4
=0
Write the pattern
p=–
3
and q = 1
4
Simplify
Use FOIL
4
4x2 – x – 3 = 0
Multiply equation by 4
4
5  3i

5  3i 5  3i
20  12i

25  15i  15i  9i 2
20  12i

25  9( 1)
20 12

 i
34 34
10 6

 i
17 17
Section 5.4 ~ Complex Numbers (pg. 259-266)


Simplified square root expressions:
o
o
Do not have radicals in the denominator
Do not have any number remaining under
the square root that has no perfect square
factor other than 1
Examples: Simplify the following expressions.
1.
2.
40

 4  10
32
81
16  2
9
4 2

9
 4  10

 2 10

32
81

You can solve some quadratic equations by using the
Square Root Property.
Examples: Solve the following equations.
1. 4 x 2  64  0
2. 6 x 2  72  0
4 x 2  64
6 x 2  72
x 2  16
x 2  12
These properties also hold true with square roots of
negative numbers, except you also have to simplify
1  i (an imaginary number).
x 2   12
x 2   16
x  4i
x  2i 3
Examples: Simplify the following expressions.
1.
 72
2.
 108b
7
  1  36  2
  1  36  b6  3b
  1  36  2
  1  36  b6  3b
 6i 2
 6b3i 3b
Just remember to pull out the largest perfect square when
simplifying square root expressions.

Operations with Complex Numbers
o
o
o
Adding/Subtracting  Combine like terms
Multiplying  Use FOIL & simplify
Dividing  Multiply entire fraction by the conjugate
Examples: Simplify the following expressions.
Remember: When you take the square root of any number
while solving a quadratic equation, you always end up with two
answers.
Section 5.5 ~ Completing the Square (pg. 268-275)

The Square Root Property can only be used to solve
quadratic equations when the quadratic expression is a
perfect square. However, few quadratic expressions are
perfect squares. To make a quadratic expression a perfect
square, completing the square can be used.
Examples: Solve the following by completing the square.
1. x 2  8 x  15  0
x  8 x  15
x 2  8 x  16  15  16
1. (5  2i )  (4  4i )  9  2i
2. (15  3i )  (9  3i )  6  6i
( x  4) 2  1
3. (1  4i )( 2  i )
x  4  1
 2  i  8i  4i 2
 2  7i  4(1)
 2  7i  4
 6  7i
4.
4
5  3i
-Original equation
-Move coefficient
-Find new constant
using magic X
-Factor (using magic X)
2
& simplify
-Take the square root of
both sides
-Solve for x
x  4 1
x  5 or x  3
ac
new constant (c)
factors  b 
16
-4
-4
2
-8
value of b from original equation
b
Follow the same process when solving any equation by
completing the square. (2 more examples below)
2x2  7 x  6  0
7
x2  x  3  0
2
7
2
x  x  3
2
7
49
48 49
2
x  x


2
16
16 16
2 x 2  7 x  12  0
7
x2  x  6  0
2
7
2
x  x  6
2
7
49
96 49
2
x  x
 
2
16
16 16
2
7
1

x   
4
16


7
1
3
x     x  2 or x  
4
4
2
o

Solve for x!
 b  b2  4ac
2a
The expression under the square root, b 2  4ac , is called
x
the discriminant.
o
o

Consider ax 2  bx  c  0 , where a, b, and c are rational #s.
Example of Graph
Value of
Type & Number
of Related Function
Discriminant
of Roots
perfect square
b 2  4ac > 0;
not perfect square
a represents the direction of the opening

a > 0 (positive)  graph opens up

a < 0 (negative)  graph opens down
(h, k) = vertex
x = h represents the equation for the axis of symmetry
Given a function of the form y  ax 2  bx  c , you can
complete the square to write the function in vertex form. If
the coefficient of the quadratic term is not 1, the first step
is to factor that coefficient from the quadratic and linear
terms.
Examples: Write each equation in vertex form.
1. y  x 2  4 x  6
y  ( x 2  4 x)  6
-Separate the constant
y  ( x  4 x  4)  6  4
-Find new constant by
using magic X
-Factor trinomial & simplify
2
The value of the discriminant can be used to determine
the number and type of roots (or solutions) of a
quadratic equation. The following table summarizes the
possible types of roots (or solutions).
b 2  4ac > 0;
The vertex form of a quadratic equations is
y  a( x  h)2  k
The solutions of a quadratic equation of the form
ax 2  bx  c  0 , where a  0 , are given by the following
formula.
Quadratic Formula


2
7
47

x    
4
16


7
i 47
x 
4
4
Section 5.6 ~ The Quadratic Formula & the Discriminant
(pg. 276-283)

Section 5.7 ~ Analyzing Graphs of Quadratic Functions
(pg. 286-292)
y  ( x  2)2  2
Therefore, the axis of symmetry is x = -2, the vertex is the point
(-2, 2), and the direction of the opening is up since a=1
(positive).
2. y  2 x 2  12 x  17
y  (2 x 2  12 x)  17
y  2( x 2  6 x)  17
y  2( x 2  6 x  9)  17  2(9)
2 real, rational
roots
2 real, irrational
roots
y  2( x  3)2  1
-Separate the constant
-Factor out 2
-Find new constant by
using magic X
-Factor trinomial & simplify
Therefore, the axis of symmetry is x = -3, the vertex is the point
(-3, -1), and the direction of the opening is up since a=2
(positive).
b 2  4ac = 0
1 real, rational
root
Remember: Writing equations in vertex form is exactly like
completing the square, except instead of adding the new
constant on both sides, you add the new constant inside the
parentheses and subtract the constant outside the
parentheses. It is just like adding zero!

If the vertex and one other point on the graph of a
parabola are known, you can write the equation of the
parabola in vertex form.
Example: Write an equation for a parabola with vertex at (3,-2)
and another point at (5,3).
b 2  4ac < 0
2 complex roots
This means that (h, k) = (3, -2) & (x, y) = (5, 3). Use these
values to plug it into the vertex form of the equation and solve
for a.
Example: Solve x 2  4 x  5 .
y  a( x  h)2  k
3  a(5  3)2  2
Now, plug in a, h, and k
3  a(2)  2
3  4a  2
5  4a
5
a
4
2
x2  4x  5
x2  4x  5  0
( x  5)( x  1)  0
to get an equation.
y  a( x  h)2  k
5
y  ( x  3) 2  2
4
x 5  0
x5
x 1  0
x  1
Test x = 0:
x2  4x  5
0  5
Since the statement is true,
shade the region that
includes x = 0.
The solutions are
1  x  5 .
-1
5
Section 5.8 ~ Graphing & Solving Quadratic Inequalities
(pg. 294-301)
Chapter 6: Polynomial Functions

Section 6.1 ~ Properties of Exponents (pg. 312-318)
You can graph quadratic inequalities in two variables
using the same techniques you used to graph linear
inequalities in two variables.

Example: Graph y  x 2  12 x  31 .
a = 1, b = 12, c = 31

y-intercept: c = 31

axis of symmetry (& x-coordinate of vertex):
x
The following table summarizes all the properties of
exponents needed to simplify an expression containing
powers.
Name of Property
an 
Negative Exponents
b
12

 6
2a
2(1)
x2 + 12x + 31
16 – 48 + 31
25 – 60 + 31
36 – 72 + 31
49 –84 + 31
64 – 96 + 31
1
an
Flip & switch
am  an  amn
Product of Powers
Now, make a table of values that includes the vertex.
x
-4
-5
-6
-7
-8
Rule
f(x)
-1
-4
-5
-4
-1
(x, f(x))
(-4, -1)
(-5, -4)
(-6, -5)
(-7, -4)
(-8, -1)
To multiply powers of the
same variable, add the
exponents.
am
 amn
an
Quotient of Powers
To divide powers of the same
base, you subtract exponents.
Power of a Power
(a m ) n  a m  n
Power of a Product
(ab) m  a mb m
n
an
a
   n
b
b
Since the inequality symbol is <, the parabola should be solid.
Power of a Quotient
n
Then, test a point outside the
parabola. The easiest point to test
would be (0,0).
Test (0,0):
Zero Power
Remember: To simplify an expression containing powers
means to rewrite the expression without parentheses or
negative exponents.
y  x 2  12 x  31
0  (0)2  12(0)  31
0  31 
ALSO, WHEN IN DOUBT, WRITE IT OUT!
Since the statement is true, shade the region that includes
(0,0). So, the region outside the parabola represents the
solutions of the inequality.


You can solve quadratic equations the same way.
n
bn
a
b
     n
a
b
a
a 0  1 , except when a = 0
Section 6.2 ~ Operations with Polynomials (pg. 320-324)
The degree of a polynomial is the degree of the
monomial with the greatest degree.
Examples: Determine whether each expression is a
polynomial. If it is a polynomial, state the degree of the
polynomial.
m5
1.
2.
5x y  x 3
2
The degree of the first term is 2 + 4 or 6, and the degree of the
second term is 1. The degree of the polynomial is 6.
This is not a polynomial because there is
a variable under the root.
4
Adding/Subtracting Polynomials  Combine like terms
Multiplying Polynomials  Distributive Property & FOIL


This is a polynomial because
each term is a monomial.
If you know how to do both of these, you should be okay
with adding, subtracting, and multiplying polynomials.
Section 6.3 ~ Dividing Polynomials (pg. 325-330)

In Lesson 6.1, you learned to divide monomials. You can divide a polynomial by a monomial by using those same skills.
4 2
5 3
Example: Simplify
3
25a b – 20a b + 10a b
2
.
5a b
5 3
4
2
3
25a b – 20 a b  10 a b
2
5 3
=
5a b
=
25a b
2
4
5a b
25
5–2
a
20a b
–
2
2
5a b
b3 – 1 –
5
3
+
10 a b
Sum of quotients.
2
5a b
20
 a4 – 2 b2 – 1 +
5
10
 a 3 – 2 b1 - 1
Divide.
5
b1 – 1 = b0 or 1
= 5a3b2 – 4a2b + 2a

You can use a process similar to long division to divide a polynomial by a polynomial with more than one term. The process is
known as the division algorithm.
Example: Use long division to find (x2 – 7x – 18) ÷ (x + 2).
x
x-9
x  2 x  7 x  18
x  2 x  7 x  18
2
(–) x2 + 2x
–9x – 18
2
x(x + 2) = x2 + 2x
–7x – (2x) = –9x
(–) x2 + 2x
–9x – 18
–9x – 18
0
The quotient is x – 9. The remainder is 0.

Synthetic division is a simpler process for dividing a polynomial by a binomial.
Example 1: Use synthetic division to find (x4 + 3x3 – 2x + 5) ÷ (x + 1).
Step 1
Write the terms of the dividend so that the degrees of
the terms are in descending order. Then write just the
coefficients as shown at the right. Since there is no x2
term, you must include a coefficient of 0 for x2.
x4 + 3x3 + 0x2 – 2x + 5
 



1
3
0 –2
5
Step 2
Write the constant r of the divisor x – r to the left. In
this case, r = –1. Bring the first coefficient, 1, down as
shown.
–1 |
Multiply the first coefficient by r: 1 • –1 = –1. Write the
product under the second coefficient. Then add the
product and the second coefficient: 3 + (–1) = 2.
–1 |
Step 4
Multiply the sum, 2, by r: –1(2) = –2. Write the product
under the next coefficient and add:
0 + (–2) = –2.
–1 |
3
0 –2 5
–1 –2
1 2 –2
|
Step 5
Multiply the sum, –2, by r: –1(–2) = 2. Write the
product under the next coefficient and add:
–2 + 2 = 0.
–1 |
1
Step 3
1
3
0
–2
1
1
3
–1
1 2
5
|
0
–2
5
|
1
3
–1
1 2
0
–2
–2
–2
2
0
5
|
Multiply the sum, 0, by r: –1(0) = 0. Write the product
under the next coefficient and add:
5 + 0 = 5. The remainder is 5.
Step 6
–1 |
3
0 –2 5
–1 –2
2 0
1 2 –2
0 |5
1
The numbers along the bottom row are the coefficients of the quotient. Start with the power that is one less than the degree of the
dividend. Thus, the quotient is x3 + 2x2 – 2x +
5
x  1
.
Example 2: Use synthetic division to find (12x3 – 7x2 + 4x – 3) ÷ (3x – 1).
it has a first coefficient of 1.
Use division to rewrite the divisor so
3
12 x – 7 x
2
3
 4x – 3
=
3x – 1
(12 x – 7x
3
=
1
3
, so r =
1
 4x – 3)  3
(3x – 1)  3
4x –
x–r=x–
2
1
.
3
Simplify numerator and denominator.
7 2
4
x 
x– 1
3
3
1
x –
3
4
3
–
7
4
3
4
3
1
3
4
–1
–1
1
–
3
1
Divide the numerator and denominator
by 3.
–
3
2
3
2
3
The result is 4x2 – x + 1 –
x–
1
.
Now simplify the fraction.
3
= 4x2 – x + 1 –
2
3x – 1
.
Rewrite the remainder so that there are no fractions in the denominator.
That’s the semester in a nutshell. Have fun studying!!!