1. Finals’ Lessons in
ADVANCE ALGEBRA
a powerpoint compilation

2. THE REMAINDER THEOREM

3.  When we divide a polynomial f(x) by x-c we
get:
P(x) = (x-c) · q(x) + r(x)
dividend divisor quotient remainder

4. Now see what happens when we have x equal to c:
Let x=c
f(c) = (x-c)● q(c) + r
f(c) = (0)·q(c) + r
f(c) = r

5. So we get this:The Remainder Theorem:
When we divide a polynomial f(x) by x-c the
remainder r equals f(c).
So when we want to know the remainder after
dividing by x-c we don't need to do any division. Just
calculate f(c).

6. Example:
1. 2x2-5x-1 divided by x-3
(x-c)=x-3
3-c=-3-(-3)
3-c=0
c=3
P(C)= 3
P(x)=2-5x-1
P(3)=2(3)2-
5(3)-1
P (3)= 2(9)-
5(3)-1
P(3)=18-15-1
P(3)= 2 ----
REMAINDER

7. 2.) 2x2-5x-1 divided by x-4
(x-c)=x-4
4-c=4-(4)
4-c=0
c=4
P(C)= 4
P(x)=2-5x-1
P(4)=2(4)2-5(4)-1
P (4)= 2(16)-(20)-1
P(4)=32-20-1
P(4)= 11 -----
REMAINDER

8. 3.) x2+ 5x -3 / x+1
(x-c)=x+1
-1-c=-1-(-1)
-1-c=0
C=-1
P(C) = -1
P(x) = x2+ 5x -3
P(-1)=(-1)2 + 5(-1) -3
P(-1)=1-5-3
P(-1)=-6 -----
REMAINDER

9. THE FACTOR THEOREM

10. When f(c)=0 then x-c is a factor of the polynomial
And the other way around, too:
When x-c is a factor of the polynomial then f(c) =0.

11. Example:
f (x) = 2x4 + 3x2 – 5x + 7 ; x – 1
1. First set x – 1 equal to zero and solve to find the
proposed zero, x = 1.
2. Then use synthetic division to divide f (x) by x = 1.
Since there is no cubed term, careful to remember
to insert a "0" into the first line of the synthetic
division to represent the omitted power
of x in 2x4 + 3x2 – 5x + 7.

12. 2 0 3 -5 7
2 2 5 0

13. remainder
Since the remainder is not zero, then the Factor
Theorem says that:
x – 1 is not a factor of f (x).

14. Exercise:
1. f (x) = 5x4 + 16x3 – 15x2 + 8x + 16 / x + 4
remainder
The remainder is zero, so the Factor Theorem says
that:
x + 4 is a factor of 5x4 + 16x3 – 15x2 + 8x + 16.

15. • f (x) = 3x4 + 5x3 + x2 + 5x – 2 / x+2
remainder
x+2 is a factor of f (x) = 3x4 + 5x3 + x2 + 5x – 2

16. SYNTHETIC DIVISION

17. Synthetic division is a shorthand method of dividing
polynomials where you divide the coefficients of the
polynomials, removing the variables and exponents.
It allows you to add throughout the process instead
of subtract, as you would do in traditional long
division.

18. STEPS IN SYNTHETIC DIVISION OF
POLYNOMIALS
Step 1 :
To set up the problem, first, set the denominator
equal to zero to find the number to put in the
division box. Next, make sure the numerator is
written in descending order and if any terms are
missing you must use a zero to fill in the missing
term, finally list only the coefficient in the division
problem.

19. Step 2 :
Once the problem is set up correctly, bring the
leading coefficient (first number) straight down.
Step 3 :
Multiply the number in the division box with the
number you brought down and put the result in the
next column.

20. Step 4 :
Add the two numbers together and write the result
in the bottom of the row.
Step 5 :
Repeat steps 3 and 4 until you reach the end of the
problem.

21. Step 6 :
Write the final answer. The final answer is made up
of the numbers in the bottom row with the last
number being the remainder and the remainder
must be written as a fraction. The variables or x’s
start off one power less than the original
denominator and go down one with each term.

22. Examples:
1. Divide:
Solution:
Step 1:
To set up the problem, first, set the denominator equal to
zero to find the number to put in the division box. Next,
make sure the numerator is written in descending order
and if any terms are missing you must use a zero to fill
in the missing term, finally list only the coefficient in the
division problem.

23. Step 2:
Once the problem is set up correctly, bring the
leading coefficient (first number) straight down.

24. • Step 3: Multiply the number in the division box
with the number you brought down and put the
result in the next column.

25. • Step 4: Add the two numbers together and write
the result in the bottom of the row.

26. Step 5:
Multiply the number in the division box with the
number you brought down and put the result in the
next column.

27. Step 6:
Add the two numbers together and write the result in
the bottom of the row.

28. Step 7:
Multiply the number in the division box with the
number you brought down and put the result in the
next column.

29. Step 8:
Add the two numbers together and write the result
in the bottom of the row.

30. Step 9:
Write the final answer. The final answer is made up
of the numbers in the bottom row with the last
number being the remainder and the remainder
must be written as a fraction. The variables or x’s
start off one power less than the original
denominator and go down one with each term.

31. 1. Divide:
Solution:
Step 1:

32. Step 2:
Step 3:
Step 4:

33. Step 5:
Repeat Steps 3 and 4 until you reach the end of the
problem.
Step 6:

34. RATIONAL ZEROS OF POLYNOMIAL
FUNCTION

35. The Rational Zeros Theorem states:
If P(x) is a polynomial with integer coefficients and
if is a zero of P(x) ( P() = 0 ), then p is a factor of the
constant term of P(x) and q is a factor of the leading
coefficient of P(x) .

36. We can use the Rational Zeros Theorem to find all
the rational zeros of a polynomial. Here are the
steps:
Arrange the polynomial in descending order
1. Write down all the factors of the constant term.
These are all the possible values of p .
2. Write down all the factors of the leading coefficient.
These are all the possible values of q .

37. 1. Write down all the possible values of . Remember
that since factors can be negative, and - must both
be included. Simplify each value and cross out any
duplicates.
2. Use synthetic division to determine the values of
for which P() = 0 . These are all the rational roots of
P(x) .

38. Example:
Find all the rational zeros of P(x) = x 3 -9x + 9 +
2x 4-19x 2 .
1. P(x) = 2x 4 + x 3 -19x 2 - 9x + 9
2. Factors of constant term: ±1 , ±3 , ±9 .
3. Factors of leading coefficient: ±1 , ±2 .
4. Possible values of : ± , ± , ± , ± , ± , ± . These can be
simplified to: ±1 , ± , ±3 , ± , ±9 , ± .
5. Use synthetic division:

39. Thus, the rational roots of P(x) are x = - 3 , -1 , , and 3 .

40. REAL ROOTS OF A POLYNOMIAL EQUATION

41. Here are three important theorems relating to
the roots of a polynomial:
A polynomial of n-th degree can be factored into n
linear factors.
A polynomial equation of degree n has exactly n
roots.
If (x−r) display style {left({x}-{r}right)}(x−r) is a
factor of a polynomial, then x=r display style
{x}={r}x=r is a root of the associated polynomial
equation.

42. The cubic polynomial f(x) = 4x3 − 3x2 − 25x −
6 has degree 3 (since the highest power of x that
appears is 3. This polynomial can be factored
and written as
4x3 − 3x2 − 25x − 6 = (x − 3)(4x + 1)(x +
2)
So we see that a 3rd degree polynomial has 3
roots.

43.  The associated polynomial equation is formed by
setting the polynomial equal to zero:
f(x) = 4x3 − 3x2 − 25x − 6 = 0
 In factored form, this is:
(x−3)(4x+1)(x+2)=0.

44. In this example, all 3 roots of our polynomial
equation of degree 3 are real.
 Since (x−3) is a factor, then x=3 is a root.
 Since (4x+1is a factor, then x=−​4​​1​​ is a root.
 Since (x+2) is a factor, then x=−2 is a root.

45. Solving Polynomials

46. The first step is to apply the Rational Roots Test to the
polynomial to get a list of values that might possibly be
solutions to the polynomial equation. You can follow this
up with an application of Descartes' Rule of Signs, if you
like, to narrow down which possible zeroes might be best
to check.
Of course, if you've got a graphing calculator, it's a good
idea to do a quick graph, since x-intercepts of the graph
are the same as zeroes of the equation.
Seeing where the graph looks like it crosses the axis can
quickly narrow down your list of possible zeroes.

47. Once you've found a value you want to test, you use
synthetic division to see if you can get a zero remainder.
If you get a zero remainder, you've not only found a zero,
but you've also reduced your polynomial by one degree.
 Remember that synthetic division is, among other
things, division, so checking if x = a is a solution is the
same as dividing out the linear factor x – a.
This means that you should not return to the original
polynomial for your next computation (for finding the
other zeroes); you should instead work with the output
of the synthetic division.

49. EXPONENTIAL FUNCTIONS AND THEIR
GRAPH

50. Let’s start off this section with the definition of an
exponential function.
If b is any number such that
and
then an exponential function is a function in the
form,
where b is called the base and x can be any
real number.

51. Notice that the x is now in the exponent and the
base is a fixed number. This is exactly the opposite
from what we’ve seen to this point. To this point the
base has been the variable, x in most cases, and the
exponent was a fixed number. However, despite
these differences these functions evaluate in exactly
the same way as those that we are used to. We will
see some examples of exponential functions shortly.

52. Before we get too far into this section we should
address the restrictions on b. We avoid one and zero
because in this case the function would be,
and these are constant functions and won’t have
many of the same properties that general
exponential functions have.

53. Next, we avoid negative numbers so that we don’t
get any complex values out of the function
evaluation. For instance if we allowed the
function would be,
and as you can see there are some function
evaluations that will give complex numbers. We
only want real numbers to arise from function
evaluation and so to make sure of this we require
that b not be a negative number.

54. Now, let’s take a look at a couple of graphs. We will
be able to get most of the properties of exponential
functions from these graphs.
Example 1: Sketch the graph of and
and
on the same axis system.

55. Solution
Okay, since we don’t have any knowledge on
what these graphs look like we’re going to have to
pick some values of x and do some function
evaluations. Function evaluation with exponential
functions works in exactly the same manner that all
function evaluation has worked to this
point. Whatever is in the parenthesis on the left we
substitute into all the x’s on the right side.

56. Here are some evaluations for these two functions,
x
-2
-1
0
1
2

57. Here is the sketch of the two graphs.

58. Note as well that we could have written in
the following way,
--- Sometimes we’ll see this kind of exponential
function and so it’s important to be able to go
between these two forms.

59. Properties of Exponential Functions.
1. The graph of will always contain the point .
Or put another way, regardless of the value of b.
2. For every possible b . Note that this implies
that .
3. If then the graph of will decrease as we
move from left to right. Check out the graph
of above for verification of this property.

60. 4. If then the graph of will increase as we
move from left to right. Check out the graph
of above for verification of this property.
5. If then .
--- All of these properties except the final one can be
verified easily from the graphs in the first
example. We will hold off discussing the final
property for a couple of sections where we will
actually be using it.

61. Example 2
Sketch the graph of
Solution
Let’s first build up a table of values for this function.
• To get these evaluation you will need to use a
calculator. In fact, that is part of the point of this
example. Make sure that you can run your
calculator and verify these numbers.
x -2 -1 0 1 2
f(x) 0.135 0.3679 1 2.178 7.389

62. Here is a sketch of this graph.

63. Notice that this is an increasing graph as we should
expect since
There is one final example that we need to work
before moving onto the next section. This example
is more about the evaluation process for
exponential functions than the graphing
process. We need to be very careful with the
evaluation of exponential functions.

64. Example 3
Sketch the graph of
Solution
Here is a quick table of values for this function.
Now, as we stated above this example was more
about the evaluation process than the graph so
let’s go through the first one to make sure that you
can do these.
x -1 0 1 2 3
g(x) 32.945… 9.951…. 1 -2.161… -3.323…

65.  Notice that when evaluating exponential
functions we first need to actually do the
exponentiation before we multiply by any
coefficients (5 in this case). Also, we used only 3
decimal places here since we are only graphing. In
many applications we will want to use far more
decimal places in these computations.

66. Here is a sketch of the graph.

67. Notice that this graph violates all the properties
we listed above. That is okay. Those properties
are only valid for functions in the
form or . We’ve got a lot more
going on in this function and so the properties, as
written above, won’t hold for this function.

68. Logarithmic Functions & Their Graphs

69. For x  0 and 0  a  1,
y = loga x if and only if x = a y.
The function given by f (x) = loga x is called the
logarithmic function with base a.
Every logarithmic equation has an equivalent
exponential form:
y = loga x is equivalent to x = a y
A logarithm is an exponent!

70.  A logarithmic function is the inverse function of an
exponential function.
Exponential function: y = ax
Logarithmic function: y = logax is equivalent to x = ay

71. Examples:
Write the equivalent exponential equation
and solve for y.
Logarithmic
Functions & Their
Graphs
Equivalent
Exponential
Equation
Solution
y = log216
y = log2( )
y = log416
y = log51
16 = 2y
= 2 y
16 = 4y
1 = 5 y
16 = 24  y = 4
= 2-1 y = –1
16 = 42  y = 2
1 = 50  y = 0

72. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 5
Properties of Logarithms
Examples: Solve for x: log6 6 = x
log6 6 = 1 property 2 x = 1
Simplify: log3 35
log3 35 = 5 property 3
Simplify: 7log
7
9
7log
7
9 = 9 property 3
Properties of Logarithms
1. loga 1 = 0 since a0 = 1.
2. loga a = 1 since a1 = a.
4. If loga x = loga y, then x = y. one-to-one property
3. loga ax = x and alogax = x inverse property

73. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 7
Example: Graph the common logarithm function f(x) = log10 x.
by calculator
1
10
10.6020.3010–1–2f(x) = log10 x
10421x 1
100
y
x
5
–5
f(x) = log10 x
x = 0
vertical
asymptote
(0, 1) x-intercept

74. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 8
The graphs of logarithmic functions are similar for different
values of a.
f(x) = loga x (a  1)
3. x-intercept (1, 0)
5. increasing
6. continuous
7. one-to-one
8. reflection of y = ax in y = x
1. domain ),0( 
2. range ),( 
4. vertical asymptote
 
)(0as0 xfxx
Graph of f(x) = loga x (a  1)
x
y y = x
y = log2 x
y = ax
domain
range
y-axis
vertical
asymptote
x-intercept
(1, 0)

75. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 9
The function defined by
f(x) = loge x = ln x
is called the natural
logarithm function.
Use a calculator to evaluate: ln 3, ln –2, ln 100
ln 3
ln –2
ln 100
Function Value Keystrokes Display
LN 3 ENTER 1.0986122
ERRORLN –2 ENTER
LN 100 ENTER 4.6051701
y = ln x
(x  0, e 2.718281)
y
x
5
–5
y = ln x is equivalent to ey = x

76. PROPERTIES OF LOGARITHMS
loga 1 = 0 because a0 = 1
No matter what the base is, as long as it is legal, the
log of 1 is always 0. That's because logarithmic
curves always pass through (1,0)
loga a = 1 because a1 = a
Any value raised to the first power is that same
value.

77. loga ax = x
The log base a of x and a to the x power are inverse
functions. Whenever inverse functions are applied
to each other, they inverse out, and you're left with
the argument, in this case, x.
loga x = loga y implies that x = y
If two logs with the same base are equal, then the
arguments must be equal.
loga x = logb x implies that a = b
If two logarithms with the same argument are equal,
then the bases must be equal.

78. LOGARITHMS

79. When we are given the base 2, for example, and exponent
3, then we can evaluate 23.
23 = 8.
Inversely, if we are given the base 2 and its power 8 – (2?=
8) then what is the exponent that will produce 8?
That exponent is called a logarithm. We call the exponent
3 the logarithm of 8 with base 2.
We write 3 = log28
We write the base 2 as a subscript.
3 is the exponent to which 2 must be raised to produce 8.
A logarithm is an exponent.

80. Since 104 = 10,000 then log1010,000 = 4.
"The logarithm of 10,000 with base 10 is 4."
4 is the exponent to which 10 must be raised to
produce 10,000.
"104 = 10,000" is called the exponential form.
"log1010,000 = 4" is called the logarithmic form.
logbx = n means bn = x.

81. Example 1.
Write in exponential form: log232 = 5.
Answer. 25 = 32.
Example 2.
Write in logarithmic form: 4−2 = 1
16
Answer. log4 1 = −2.
16

82. THE LAW OF LOGARITHMS:
The first law of logarithms
--- relates multiplication to addition and states that
the logarithm of a product of two terms A and B is
the sum of the logarithms of those terms so that:
logAB logAlogB
--- This law is true for any base of logarithm so
long as you use the same base throughout a
calculation. You could re-write law 1 in base a as:
ABAB as a log loglog

83. You should remember that, because of the equals
sign, this law also works from right-to-left and so
can be used to combine logarithms of the same base
that are added together. When quoting this law you
should try to say it in full as it is often
misremembered as “the log of A plus B”. If you make
the effort to quote it in full then this (very common)
error can be avoided: There is no law of logarithms
which relates to logAB.

84. Example:
Rewrite log15 in terms of log3 and log5. As 15 
35 you can use law 1 with A  3 and B  5 So, for any
base:
log(15)  log35 log3log5
Example: Express lnxln(3) as a single logarithm.
As lnxln(3) is the sum of two logarithms both in base e
(natural logarithm), you can use law 1 to show that:
lnxln3ln3 xln3x
Where the answer also has to be a natural logarithm

85.  The second law of logarithms
--- In the same way that law 1 relates multiplication to
addition, law 2 relates division to subtraction. The law
states that the logarithm of a quotient of two terms A
and B is the logarithm of the numerator minus the
logarithm of the denominator so that:
log)= log (A)- log(B)

86.  Similar to law 1, the equals sign means that this
law also works from right-to-left. When quoting this
law you should try to say it in full as it is often
misremembered as “the log of A minus B”. If you
make the effort to quote it in full then this (very
common) error can be avoided: There is no law of
logarithms which relates to logAB or B A log log .

87. Example:
 Express log () in terms of logx and log4.
This is a logarithm of a fraction with A x and B  4 so
you can use law 2 to show that, for any base:
Log= log (x)- log (4)

88. The third law of logarithms
--- relates exponentiation (raising to a power) to
multiplication and states that the logarithm of A
raised to the power of n is n multiplied by the
logarithm of A:
log(An) = nlog(A)

89. The logarithm of A divided by B is the logarithm of
A minus the logarithm of B As with the other two
laws, you need to use the same base throughout a
calculation:
loga(An) = nloga(A)

90. This law also works from right-to-left and can be
used to change a number that is multiplied by a
logarithm into exponentiation; law 3 is often
thought of as the most important because of this.
It is especially useful when changing relationships
described by curves into straight lines. The final
section of this guide shows you how to change an
exponential function into a straight line.

91. Example:
 Calculate log3(9).
This may not seem like a question where
you can use law 3. However, as 9 = 32 you can
use law 3 with A  3 and n 2 to find that:
log3(9) = log3 (32) = 2log3(3)

92. SOLVING EXPONENTIAL EQUATIONS

93. An exponential equation is one in which a variable
occurs in the exponent.
An exponential equation in which each side can be
expressed in
terms of the same base can be solved using the
property:
If the bases are the same, set the exponents equal.

94. Rule:
To solve an exponential equation, take the log of both sides,
and solve for the variable.
Example 1:
Solve for x in the equation
Solution:
Step 1:
Take the natural log of both sides:

95. Step 2:
Simplify the left side of the above equation
using Logarithmic Rule 3:
Step 3:
Simplify the left side of the above equation:
Since Ln(e)=1, the equation reads
Ln(80) is the exact answer
and x=4.38202663467 is an approximate answer
because we have rounded the value of Ln(80)..

96. Check:
Check your answer in the original equation.
Example 2:
Solve for x in the equation
Solution:
Step 1:
Isolate the exponential term before you take the
common log of both sides. Therefore, add 8 to both
sides:

97. Step 2:
Take the common log of both sides:
Step 3:
Simplify the left side of the above equation using
Logarithmic Rule 3:
Step 4:
Simplify the left side of the above equation: Since
Log(10) = 1, the above equation can be written

98. Step 5:
Subtract 5 from both sides of the above
equation:
is the exact answer. x = -3.16749108729 is an
approximate answer..
Check:
Check your answer in the original equation

99. SOLVING LOGARITHMIC EQUATIONS

100. • Logarithmic equations contain logarithmic
expressions and constants. A logarithm is another
way to write an exponent and is defined
by if and only if . .
• When one side of the equation contains a single
logarithm and the other side contains a constant,
the equation can be solved by rewriting the
equation as an equivalent exponential equation
using the definition of logarithm from above.

101. Rule:
To solve a logarithmic equation, rewrite the equation in
exponential form and solve for the variable.
Example 1:
Solve for x in the equation Ln(x)=8
Solution:
Step 1:
Let both sides be exponents of the base e. The
equation Ln(x)=8 can be rewritten .
• Step 2: By now you should know that when the base
of the exponent and the base of the logarithm are the
same, the left side can be written x. The
equation can now be written .

102. Step 3:
The exact answer is
and the approximate answer is

103. Example 2:
Solve for x in the equation
Answer:
is the exact answer
and x=104.142857143 is an approximate answer.
Solution:
Step 1: Since you cannot take the log of a negative
number, we have to restrict the domain so that 7x >0
or x > 0.
Step 2: Isolate the Log term in the original equation by
subtracting 4 from each side of the equation:

104. Step 3:
Convert the above logarithmic equation to an
exponential equation with base 3 and exponent 6:
Step 4: Divide both sides of the above equation
by 7:
is the exact answer and is an approximate
answer.

105. Check:
Let's substitute the approximate value in the
answer and determine whether the left side of the
equation equals the right side of the equation after
the substitution. Remember we rounded the
number and the answer is only a close
approximation, so the left and right side of the
equation will most likely be very close but not equal;
it depends on the number of decimals were rounded
in your answer
Or

106. EXPONENTIAL EQUATION AND
LOGARITHMIC MODEL

107. An exponential equation is one in which a variable
occurs in the exponent.
An exponential equation in which each side can be
expressed in
terms of the same base can be solved using the
property:
If the bases are the same, set the exponents equal

108. Examples:Solve for x. Answer
1.
Since the bases are the same, set the
exponents equal to one another:
2x + 1 = 3x - 2
3 = x
2.
27 can be expressed as a power of 3:
2x - 1 = 3x
-1 = x
3.
25 can be expressed as a power of 5:
3x - 8 = 4x
-8 = x

109. Exponential Growth

110. FUNCTIONy = C ekt, k > 0
Features
Asymptotic to y = 0 to left
Passes through (0,C)
C is the initial value
Increases without bound to right

111. Exponential Decay (decreasing form)

112. FUNCTIONy = C e-kt, k > 0
Features
Asymptotic to y = 0 to right
Passes through (0,C)
C is the initial value
Decreasing, but bounded below by y=0

113. LOGARITHMIC MODEL

114. FUNCTIONy = a + b ln x
Features
Increases without bound to right
Passes through (1,a),
Very rapid growth, followed by slower growth,
Common log will grow slower than natural log
b controls the rate of growth

115. The logarithmic model has a period of rapid
increase, followed by a period where the growth
slows, but the growth continues to increase without
bound.
 This makes the model inappropriate where there
needs to be an upper bound.
The main difference between this model and the
exponential growth model is that the exponential
growth model begins slowly and then increases very
rapidly as time increases.

116. Submitted by:
Allado, Swedenia
Artizona, Anamae
Bolivar, Ma. Cris
Annabelle
Callocallo, April
Mae
Gabilagon, Liezl
Lerado, Aira
Grace
Matalubos, Lyra
Sorenio, Shiela
Mae
Tabuga, Natalie
T0rda, Marjocel
Submitted to:
Prof. Danilo Parreño
ADVANCE ALGEBRA