MAT220 Class Notes:
Algebra and Trigonometry Review.
I. Properties of Fractions
\
How to “split up” a single fraction into two fractions…
1.
x y x y
 
(usually we go the other direction with this…i.e common denominator and write as a single fraction)
z
z z
(Note that when there are terms in the numerator and we split up the fraction, the denominator goes with BOTH terms)
2.
x y
y
 x
z
z
" or "
x y x
 y
z
z
(Note that when there are factors in the numerator and we split up the fraction, the denominator only goes with ONE of the factors)
Common Mistake.
x y x y
 
z
z z
Adding / Subtracting Fractions (requires a common denominator)
a c a d c b ad bc ad  bc
     


b d b d d b bd bd
bd
(note: when you “find common denominators” you actually “multiply by one”…this will be a VERY common thing for us to do in
this class! More on “multiplying by one” on Day 3)
Common Mistake.
a c ac
 
b d bd
II. Factoring:
Factoring (and using factoring to simplify fractions): NOTE: If you are in Calculus YOU should be an expert at factoring!!!!
Here is a general Factoring Strategy that you should use to factor polynomials.
1. Always factor out the GCF(Greatest Common Factor) first.
2. Next check the number of terms in your polynomial.
A. Two terms
i. Factor the difference of two squares
a2  b2   a  b  a  b 
ii. Factor the difference of two cubes
a 3  b3   a  b   a 2  ab  b 2 
iii. Factor the sum of two cubes
a 3  b3   a  b   a 2  ab  b 2 
B. Three terms ---- try reverse foil
(although sometimes a three term polynomial will factor into the product of two trinomials)
C. Four terms ---- try factor by grouping
D. If none of these work you could try to use the rational roots theorem from
College Algebra / Precalculus to find a zero of the polynomial (which in turn
will give you one of the factors) and you may be able to go from there.
Note: We will review the Rational Roots Theorem on Day 2.
3. Repeat step 2. until all factors are prime.
Factor the GCF:
1. x  2e  e
2
2.
x
 2 x  2 xe x  x  e x 
2x
3 2
1
1
1
1
2
x  x  3  x3  x  3  x 2  x  3 6  x  3  5 x   x 2  x  3  6 x  18  5 x   x 2  x  3 x  18 
5
2
10
10
10
Factor:

1. x3  3 x 2  9 x  27  x 2  x  3  9  x  3   x  3  x 2  9    x  3  x    3
2
2

=  x  3 x  3 x  3
  x  3
2. 6 x2  5x  6   3x  2 2 x  3

2
 x  3
remember that there are many different methods that can be used to factor this!


3. 16 x3  54  2 8 x3  27   2  2 x    3  2  2 x  3  2 x    2 x  3   3
3
3
2
2

 2  2 x  3  4 x 2  6 x  9 
III. Simplifying Fractions: (This MEANS Factor THEN Cancel)
2
1.
x
2
4 x  x  4
4x  x  4
4x  x  4
4 x  16 x
2x




x  0, 4
3
2
2
2 x  6 x  8 x 2 x  x  3x  4  2 x  x  4  x  1 2 x  x  4   x  1 x  1
3
2
2
2
(why do we need to exclude those two values?)
Note: A common error..
x
2
1
x2  4 x2  4 x  2 x  2 x 1




 x  1 THIS IS WRONG!!!! You may NEVER cancel TERMS
x2
2
1
x 2
2
1
1
1
This problem SHOULD be done as follows…
x 2  4  x  2  x  2   x  2   x  2 


 x2 x  2
x2
 x  2
 x  2
3x  4 x 2  9 
3x  2 x  3 2 x  3 3x  2 x  3  2 x  3 3x  2 x  3
12 x3  27 x
2.




36 x 2  6 x  90 6  6 x 2  x  15 6  3x  5 2 x  3
6  3x  5 
6  3 x  5   2 x  3
x
3
2
IV. Finding Zeros of Polynomials
Rational Roots Theorem:
Let
f  x   an xn  an1 xn1  an2 xn2  .....  a2 x2  a1x  a0 be a polynomial with integer coefficients.
If the polynomial has any rational zeros (roots), p/q, then p must be an integer factor of a 0 and q must be a factor of an.
Example: List the possible rational zeros for
f  x   3x4 11x3  10x  4
.
p : 1,  2,  4
q : 1,  3
p
1
2
4
: 1,  , 2,  ,  4, 
q
3
3
3
Other important polynomial theorems for College Algebra / Precalculus.
Conjugate Pairs Theorems.
i. If your polynomial has rational coefficients and
ii. If your polynomial has real coefficients and
a  b c is a zero then so is it’s conjugate a  b c
a  bi is a complex zero then so is it’s conjugate a  bi
A) The Remainder Theorem.
If you wish to evaluate a polynomial at a number “c” just do synthetic division using
“c” and whatever remainder you get will be f (c). Note: This works for ANY number,
integer, irrational or imaginary.
B) The Factor Theorem.
If doing synthetic division with “c” yields a remainder of zero then we say that “c” is
a zero (or root) of f (x) AND it means that ( x – c ) is a factor of f (x).
C) The Upper Bound Theorem
If doing synthetic division with a positive number yields a whole row of non-negatives
then there is no zero greater than the one that you just tried.
D) The Lower Bound Theorem
If doing synthetic division with a negative number yields a whole row of alternating signs
then there is no zero smaller than the one that you just tried.
E)
The Intermediate Value Theorem.
For any polynomial P(x), with real coefficients, if a is not equal to b and if P(a) and
P(b) have opposite sings (one negative and one positive) then P(x) MUST have at least
one zero in the interval (a , b).
Note: The Intermediate Value Theorem holds for any CONTINUOUS function.
We will study the idea of continuity in MAT220.
f  x   3x4 11x3  10x  4 to find ALL
Example: Use your “list” of possible rational zeros of
of the zeros for the polynomial. Write the polynomial in factored form.
p
1
2
4
: 1,  , 2,  ,  4, 
q
3
3
3
1
(Always “consider” trying “1” first…..do the coefficients sum to zero?)
3
 11
0
10
4
3
 14
14
4
0
So  1 is a zero thus  x  1 is a factor!
Now go off of the NEW numbers!!!!
2
3
3
 14
14
4
3
 12
6
0
So
2
is a zero thus  3 x  2  is a factor!
3
NOW your new numbers are the coefficients of a quadratic so find the last two zeros a different way.
3 x  12 x  6  0  x  4 x  2  0  x 
2
2
x

2 2 2
2
  2
2
Thus
  4  
 4 
2 1
2
 4 1 2 

 x   2  2  and  x   2  2  are factors!!!
2
So the four zeros of the polynomial are  1, , 2  2 and the polynomial factors as follows...
3

f  x   3x 4  11x 3  10 x  4   x  1 3x  2  x  2  2
 x  2  2 

4 8 42 2 2 2 2


2
2
2

Let’s Try…
f  x   6 x5 19 x4  23x3  82x2  4x  24
p : 1,  2,  3,  4,  6,  8,  12, 24
q : 1,  2,  3,  6
p
1 1 1
2
3
4
8
: 1,  ,  ,  ,  2,  ,  3,  ,  4,  , 6,  8,  ,  12,  24
q
2 3 6
3
2
3
3
Note: we know 1 is not a zero as the sum of the coefficients is not zero.
6
 19
 23
82
6
3
6
6
17
 37
79
88
556
 182
3
6
1
 26
4
3332
542
4
 24
19968
 1650
8
0
Note: 6 is a UB
Note:  3 is a LB
So 3 is a zero!
ALSO note that the “new a0” (8) has less factors than the original a0 (24) so we can cross more off of our list!
Go off of the “new” numbers now!!!! Typically we always immediately see if a found zero is a multiple zero BUT in this case since
our “new a0” (8) would not have 3 in the list we won’t bother.
2
6
6
1
13
 26
0
4
4
8
0
So  2 is a zero!!!
At this point we are down to a cubic so we always check to see if we can factor by grouping to find the remaining zeros but we can’t
ALSO note that the “new a0” (4) has less factors than the previous a0 (8) so we may be able to cross more off of our list!
Go off of the “newest” numbers!
2
2
6
6
6
 13
 25
1
0
50
2
4
 96
0
So  2 is NOT a multiple zero
So 2 is a zero AND we are NOW down to a quadratic
Now that we are down to a quadratic we can solve
6 x 2  x  2  0 by factoring.
 3x  2  2 x  1  0
 x
2
3
x
1
2
2 1
,  and our polynomial factors as follows
3 2
f  x   6 x5 19 x4  23x3  82x2  4x  24 =  x  3 x  2 x  2 3x  2 2 x  1
So the five zeros of f  x   6 x  19 x  23x  82 x  4 x  24 are 3,  2, 2,
5
4
3
2
V. Multiplying by one, fractions, conjugation and adding zero
Dividing Fractions: (Multiply by the reciprocal…this means YOU are actually “multiplying by one”)
a a d a d

a c b b c b c a d
Note:
    
 
b d c c d
1
b c
d d c
(THIS is WHY dividing by a fraction is THE SAME AS multiplying by the reciprocal)
1.
 x  4  x  3   x  3 x  3
x 2  7 x  12 x 2  x  12 x 2  7 x  12
x2  9




x 3
x2  9
x 3
x 2  x  12
 x  3
 x  4  x  3

 x  4   x  3  x  3  x  3

 x  3
 x  4   x  3
 x  3
=
 x  3
2
(Do YOU see WHY
x  4
x  3 is unnecessary?)
Adding and Subtracting Fractions: (need a common denominator AND check to see if you can simplify at the end)
2.
 x  4 
 x  6
x
5
x
5
x
5






x 2  11x  30 x 2  9 x  20  x  6  x  5   x  5  x  4   x  6  x  5   x  4   x  5  x  4   x  6 
=
x2  4 x
5 x  30

 x  6  x  5 x  4   x  6  x  5 x  4 
=
x 2  4 x  5 x  30
 x  6  x  5 x  4 
=
x 2  x  30
 x  6  x  5 x  4 
 x  6  x  5   x  6   x  5
 x  6  x  5 x  4   x  6   x  5  x  4 
 x  6
=
x  5
 x  6  x  4 
2
(adding zero) Complete the square to transform the quadratic function into the “graphing form” f  x   a  x  h   k
f  x   3x 2  12 x  2  f  x   3x 2  12 x
2
 f  x   3  x 2  6 x
  2  f  x   3  x 2  6 x   2  f  x   3  x 2  6 x  9   2  27
=
6
3
9
f  x   3  x  3  29
2
Conjugation: (multiplying by one)
Simplify
2  5i 2  5i 3  2i 6  4i  15i  10i 2 6  4i  15i  10 4  19i
4 19





  i
2
3  2i 3  2i 3  2i
9  4i
94
13
13 13
Rationalize the denominator
1.

2.





 x  4 3  5  x  x  4 3  5  x
x4
x4
3  5  x  x  4 3  5  x





9  5  x 
95 x
x4
3 5 x 3 5 x 3 5 x
 x  4 3  5  x 
 3
 x  4
5 x

x  4





 x  27  5  x  2  x  27  5  x  2
x  27
x  27 5  x  2  x  27  5  x  2





25   x  2 
25  x  2
 x  27
5 x2 5 x 2 5 x 2

 x  27   5  x  2 
  x  27 
 x  27   5  x  2 

  x  27 

 5  x  2
Note: It would be WRONG if you wrote
IF TIME
3.
x  27 on this particular problem! Do YOU understand why?





 x  11 4  5  x  x  11 4  5  x
x  11
x  11 4  5  x  x  11 4  5  x





16   5  x 
16  5  x
x  11
4 5 x 4 5 x 4 5 x

4.




 x  11  4  5  x 
 x  11
 4 5 x

x  11
 x  11 3  x  2  x  11 3  x  2
x  11
x  11 3  x  2  x  11 3  x  2





9   x  2
9 x2
 x  11
3 x  2 3 x  2 3 x  2

x  11 on this particular problem! Do YOU understand why?

 x  11  3  x  2 
  x  11
 3 x  2
Note: It would be WRONG if you wrote

VI. Lines: (A quick review…we will be utilizing a lot of what you see here in Calculus)
Slope Intercept form of the equation of a line
Point-Slope form of the equation of a line
Slope Formula
m
y  mx  b (m is the slope and  0,b  is the y-intercept)
y  y1  m  x  x1  (m is the slope and  x1 , y1  is ANY point on the line)
y2  y1
(  x1 , y1  and
x2  x1
 x2 , y2  are ANY two points on the line)
Note: Do YOU see HOW to obtain the slope formula from the Point-Slope form of the equation of a line?
y  y1  m  x  x1  
m  x  x1 
y  y1


 x  x1   x  x1 
There are four (4) different “types” of slope:
positive
“uphill”-increasing
m  x  x1 
y  y1

 x  x1   x  x1 
m
negative
“downhill”-decreasing
zero
undefined
“horizontal”-constant
“vertical”-undefined
and
9
9
18
y   x  b  5    2   b  5   b 
5
5
5
9
7
y  x
5
5
25 18
 b 
5 5
VII. Absolute Value: This is the most “basic” piecewise defined function.
Examples:
Write
3, 4
y2  y1 4  5 9
9



x2  x1 3  2 5
5
Official Mathematical Definition:
5 5
3  3
y  y1
x  x1
positive, negative, zero and undefined.
Example: Find the equation of the line that passes through the points  2,5
Solution:
 m
 w if w  0
w 
 w if w  0
0 0
x2  1  x2  1
f  x   4 x  2 as a piecewise defined function and graph the result.
Solution: Just use the definition of absolute value where YOUR “w” is 4x – 2 .
1

4x  2 x 
 4 x  2 4 x  2  0 
2
f  x  4x  2  

  4 x  2  4 x  2  0 4 x  2 x  1

2
7
b
5
f  x   6  2 x as a piecewise defined function and graph the result.
Write
Solution: Just use the definition of absolute value where YOUR “w” is 6 – 2x .
 6  2 x 6  2 x  0  6  2 x
f  x  6  2x  

  6  2 x  6  2 x  0 6  2 x
x3
x3
VIII. Other types of piecewise defined functions.
Graph:
  x  2 2  1 x  1

g  x    x  3  1  x  2
 x 1
x2

Note: Sometimes the pieces matchup and the function is “continuous” at that x value and other times the pieces do NOT matchup and
the function is discontinuous at that value of x. We will study the concept of continuity in Calculus.
Write a piecewise defined function that represents the given graph below.
3
 1
 2 x  2 x  1

f  x  
1
1  x  2
 x 1
x2


Note: The “pieces” of a piecewise defined function could be from any function that you studied in College Algebra and Trigonometry
or Precalculus (the prerequisite class(es) that you should have if you are in Calculus).
Here are some basic functions that YOU should know the graphs for.
yx
y  x2
y  x3
y x
y x
y  ex
y  ln x
y  log b x
y  sin x
y  cos x
1
x
y  tan x
y
IX. Evaluating trigonometric functions using the unit circle.
Here is a copy of The Unit Circle…YOU should know this completely!!!!!
As YOU know from your Trigonometry background, the “x” coordinate of a point on the unit circle is the cosine of the given angle
and the “y” coordinate of a point on the unit circle is the sine of the given angle. In terms of the coordinates on the unit circle we
know….
cos   x
sec  
1
x
sin   y
csc 
1
y
x
y
y
cot  
x
tan  
Note: the equation of the unit circle is
x 2  y 2  1 and as is the case with every graph, if a point lies on the graph then the
coordinates of the point must make the equation true. So, if you take any of the coordinates shown on the graph and substitute them
into the equation
x 2  y 2  1 you will get a true statement.
Example: Use the unit circle to evaluate the following…
A) sin 2100 = 
1
2
3
2
 5
C ) tan  
 6
 2
E ) cot 
 3
1
3

or 

3
3

B) cos  300  =
D) csc  7200   undefined



1
or
3
3
3
When you studied Trigonometry you restricted the domain on your trigonometric functions so they became 1-to-1 and consequently
would have an inverse that was also a function.
f  x   sin x 

2
x

f 1  x   sin 1 x  1  x  1

2
1  y  1

f  x   cos x 0  x  

2
x
2
 y

2
f 1  x   cos 1 x  1  x  1

0 y 
1  y  1
f  x   tan x 


2

f 1  x   tan 1 x    y  
  y  


2
 y

2
Knowing the RANGE for the inverse trigonometric functions will be very important later in this class.
Example: Use your knowledge of the range of the inverse trigonometric functions along with your unit circle (or drawing a triangle)
to evaluate the following.

3
A) cos 1  
 
 2 
 3
B) sin 1 
 
 2 
  7  
D) sin 1  sin 
 
  6 

 3 
C ) cos  sin 1     
 5 

5
-3
x 2   3  52
2

3
,
 
2





3  5
cos 1  
 
2

 6



,
3

2 
 3 
sin 1 
 
2

 3
x4

 3  x 4
cos  sin 1      
 5  r 5

7
6
11
6
we must find an angle in
  
 , 
 2 2
7
. As you
6
11
know from the unit circle,
has the
6
7
the same sine as
.
6
that has the same sine as
11    

   ,  so we pick 
6  2 2
6
which is coterminal and in the range
Here are 28 trigonometric identities that you studied in your Trigonometry or Precalculus class. You do NOT need to memorize all 28
of them BUT there are several that you should know because they come up frequently in Calculus.
Trigonometric Identities
1. Sin x 
1
Csc x
2. Cos x 
4. Sin x  Sin x
7. Tan x 
Sin x
Cos x
9. Sin 2 
 Cos 2   1
1
Sec x
3. Tan x 
5. Cos x   Cos x
8. Cot x 
1
Cot x
6. Tan x   Tan x
Cos x
Sin x
10. Cot 2 
 1  Csc 2 
11. 1  Tan 2   Sec 2 
12. Cos A  B  CosACosB  SinASinB
13. Cos A  B  CosACosB  SinASinB
14. Sin A  B  SinACosB  CosASinB
15. Sin A  B  SinACosB  CosASinB
16. Tan A  B  
TanA  TanB
1  TanATanB
17. Tan A  B  


 x   Cos x
2

TanA  TanB
1  TanATanB


 x   Sin x
2

18. Sin 
19. Cos
note: #’s 18 and 19 also hold for the function pairs….tan x, cot x AND sec x, csc x
Cos 2 x  Cos 2 x  Sin 2 x
20. Sin 2 x  2SinxCosx
21.
22. Tan 2 x 
 2Cos 2 x  1
=1  2 Sin 2 x
23. Sin 2 x 
1  Cos 2 x
2
24. Cos 2 x 
1  Cos 2 x
2
26. Sin
x
1  Cosx

2
2
28. Tan
Sinx
1  Cosx
x
1  Cosx
=
=

1  Cosx
Sinx
2
1  Cosx
27. Cos
25. Tan 2 x 
x
1  Cosx

2
2
The identities that come up often in calculus are #’s 1 – 3, 7, 9 – 11, 12 – 15, 20, 21.
Note: #’s 4 – 6 discuss “odd” and “even” functions!!!!!
A function is “even” if
f   x   f  x  for all x in the domain of the function.
A function is “odd” if
f   x    f  x  for all x in the domain of the function.
2Tan x
1  Tan 2 x
1  Cos2 x
1  Cos2 x
X. Simplifying Trigonometric Fractions
Examples: Simplify the following fractional expressions that involve trigonometric functions.
1.
cos 
cos 
cos 
1
1



 csc 
sin 2 2sin  cos  2sin  cos  2sin 
2
cos   0 



2
 2k where k is any integer
3
 2k where k is any integer
2
Note: Identities # 20 and # 1 were used in my work.
2.



1  sin 
1  sin  1  sin 
1  sin 




cos 2  1  sin 2  1  sin  1  sin   1  sin   1  sin  
1
3
1  sin   0   
 2k where k is any integer
2
1  sin  
Note: A version of identity # 9 was used in my work.
3.
cos 2
cos 2   sin 2   cos   sin   cos   sin    cos   sin   cos   sin  



sin   cos 
sin   cos 
sin   cos 
  cos   sin  
=
 cos   sin    cos   sin  
  cos   sin  
=  cos   sin 
cos   sin   0   


4
 2k where k is any integer
5
 2k where k is any integer
4