Download Work for Alg-Trig Review

Work for all of the Algebra-Trigonometry Review for MAT220
I.
1.
2
x3  x 2  2 x x  x  x  2  x  x  2  x  1  x  2 
 2 2
 2

x4  x2
x  x  1 x  1 x  x  1
x  x  1
x  1
2.
2
 3x  2 
12 x 2  22 x  20 2  6 x  11x  10  2  3x  2  2 x  5



4 x 2  6 x  10 2  2 x2  3x  5 2  2 x  5 x  1
 x  1
x
5
2
3. Since we are given a “hint” that 2 is a zero of the denominator we KNOW (x – 2)
MUST be one of the factors. Using synthetic division we see…..
2
1
1
-3
2
-1
3
-2
1
-2
2
0
So we can see that the denominator
x3  3x 2  3x  2   x  2   x 2  x  1
The numerator is clearly the sum of two cubes so using the formula
a 3  b3   a  b   a 2  ab  b 2 
We factor the numerator
x 3  1   x  1  x 2  x  1
So we have
 x  1  x2  x  1 x  1
x3  1


for all real x
x3  3x 2  3x  2  x  2   x 2  x  1 x  2
II.
1. A) Since the polynomial has integer coefficients and the terms are in descending
order, with a last term of 24 we get the candidates for p to be + or - 1, 2, 3, 4, 6, 8, 12, 24
(that is all of the possible integer divisors of 24). Since the leading coefficient is 4 we get
the candidates for q to be + or – 1, 2, 4. If you put each possible p over each possible q
(without repeating any) you obtain the following list….
p
1 1
3 3
: 1, , ,2,3, , ,4,6,8,12,24
p
2 4
2 4
B) By factoring out a 4 the new polynomial has a last term of 6 and a lead coefficient
of 1. This means my candidates for p are now + or – 1, 2, 3, 6 and for q are now + or – 1.
This yields a MUCH shorter list of possible rational zeros..
p
: 1,2,3,6
q
C) To find the zeros of the polynomial given in part B) we choose a possible rational
zero from our list in B) and see if we get a remainder of zero when we do synthetic
division… let’s try – 1.
-1
1
1
2
-1
1
-5
-1
-6
-6
6
0
Since we obtained a remainder of zero we know that – 1 is a zero for the original
polynomial (which means that (x + 1) MUST be a factor. Also, since the original
polynomial was of degree three (and we are dividing by a linear factor, degree 1) the
coefficients in our bottom line are those of a quadratic (degree 2…..3 – 1 = 2)….
x3  2 x 2  5 x  6   x  1  x 2  x  6 
We can find the remaining zeros by factoring the quadratic (and IF it didn’t factor we
could find them by either completing the square or utilizing the quadratic formula).
 x2  x  6    x  3 x  2 
So our three zeros of the polynomial in B) are -1, 2, and -3
2.
A) For this polynomial (integer coefficients and descending order) the last term is a
6 and the first term is a 2 so doing something similar to what we did in 1 A) we
obtain…
p : 1,2,3,6
q : 1,2
p
1
3
: 1, ,2,3, ,6
q
2
2
B) By factoring out a 2 our new polynomial has a last term of 3 and a lead
coefficient of 1 so we obtain…
p : 1,3
q : 1
p
: 1,3
q
C) Doing synthetic division with 1 is certain to work here because the sum of the
coefficients is zero, I always check that first!
1
1
1
3
1
4
-1
4
3
-3
3
0
As explained in 1 C) above we know know that
x3  3x 2  x  3   x  1  x 2  4 x  3   x  1 x  3 x  1
Thus the zeros for the polynomial given in 2 B) are 1, - 3, - 1
D) For the polynomial x3 + 3x2 – x – 3 we certainly COULD have used a
different method. Anytime you have a four term polynomial you should
always check to see if it can be factored by GROUPING (then obtain the
zeros from the factors)…
x3  3x 2  x  3  x 2  x  3  1 x  3   x  3  x 2  1   x  3 x  1 x  1
E) For the polynomial in 1 B) we could NOT have done the same thing…
the coefficients just don’t have the same relationship that we see in D) above.
3. To simplify this rational expression start by factoring out the GCF (which is
something that you should ALWAYS do as a first step then notice that you can make use
of your work from #1 and #2 above!!!!
3
2
4 x 4  8x3  20 x 2  24 x 4 x  x  2 x  5x  6  4 x  x  1 x  3 x  2  2  x  2 
 2 3
 2

2 x5  6 x 4  2 x3  6 x 2
2 x  x  3x 2  x  3 2 x  x  3 x  1 x  1 x  x  1
x  1, 3
III.
1.
2  4i 2  4i 3  6i 6  12i  12i  24i 2 6  12i  12i  24(1) 18  24i
18 24
2 8




  i  i
2
3  6i 3  6i 3  6i
9  36i
9  36(1)
45
45 45
5 15
2.




 x  11 2  x  7
x  11
x  11 2  x  7  x  11 2  x  7




4   x  7
4 x7
2 x7 2 x7 2 x7
 x  11  2 
11  x
x7
   x  11  2 
x7
  x  11
   2

x7

x  11  2  x  7
x  11
HOWEVER, if you notice, x = 11 is perfectly okay in the original expression, it makes
the fraction zero, AND our new expression is also zero when x = 11 SO we our actual
answer is….
x  11
 .............  2  x  7
2 x7
IV.
1. There’s not much I could put down here to show you how to fill out the unit circle.
HOPEFULLY you already knew how to do this from your prerequisite class OR you paid
really good attention during class when I went through it. You can always go view the
construction of the unit circle on some of the links on our class web page..
Here is one of the links from our class web page…
http://www.libraryofmath.com/unit-circle.html
2. Your ability to QUICKLY evaluate trig functions for various “standard” angles
(particularly in radian measure) WITHOUT the use of a calculator will be EXTREMELY
important later on in this course.
Here is a copy of the unit circle again…I will try to “talk” you through how you use the
unit circle to evaluate the various trig functions at the given value…
Before you start you must recall the fundamental definitions of the trig functions. For
any given angle, A, the Cos A = the “x” coordinate of the point on the unit circle where
the terminal side of the angle passes through. Similarly the Sin A = the “y” coordinate
the point on the unit circle where the terminal side of the angle passes through. Since
Tan A = Sin A / Cos A, then to find the Tan A just take the “y” coordinate and divide it
by the “x” coordinate of the point where the terminal side of the angle A crosses through
the unit circle. The other three trig functions are just reciprocals of these three…
Sec A = 1/Cos A so to find Sec A just calculate 1/”x” similarly Csc A = 1/Sin A so to
find sec A just calculate 1/”y”, lastly Cot A = 1/TanA so just calculate “x”/”y”
A) Cos 1200 … go to 1200 and select the “x” coordinate. Cos 1200= - ½
B) Sec (-2250) …go to – 2250 , remember if the angle is “negative” you travel
clockwise around the unit circle(which is COTERMINAL with 1350) and
calculate 1/”x”.
sec  225   sec 135  
1
2

2
1
2 2 2


2
2
2
2 2 2

 2
2
2
C) Csc (-2100)… go to – 2100 (which is COTERMINAL with 1500) and calculate
1/”y”
csc  210   csc 150  
1
2
1  2
1
1
2
D) Tan 3300…. just go to 3300 and calculate “y”/”x”.
1
1 2
1
1
tan 330  2  


2 3
3
3
3
2

3
3

3
3
E) Sin (-1350)… traveling clockwise (since the angle is negative) we see that -1350 is
coterminal with 2250 and since we are looking for the Sin of the angle, all we care about
is the “y” coordinate!
sin  135   sin  225   
2
2
F) Cot 1800….. just travel counterclockwise (since the angle is positive) to 1800. Cot A
is defined to be “x”/”y” so…
cot180 
1
 undefined
0
Remember that when the denominator of a fraction is 0 and the top isn’t then we say the
fraction is “undefined”. If the numerator were zero and the denominator not zero then the
fraction value would be zero. If both the numerator AND denominator are zero we say
the value of the fraction is “indeterminate”…we will study these more in Calculus.
G)
cos
5
5
3
 the x coordinate at

6
6
2
H)
sin
2
2
3
 the y coordinate at

3
3
2
I) Tan A = “y”/”x” and we need to find the angle on the unit circle that is coterminal
with 11 pi/4.
11
3
tan
 tan

4
4
2
2  1
2

2
J) Sec A = 1/”x” so
sec

2

1
 undefined
0
K) Csc A = 1/”y” and we will need to find the coterminal angle to -3pi/4
 3
csc  
 4

 5
  csc 

 4



1

2
2
1
2
2
2


2
2
2
2
2 2

 2
2
2
L) Cot A = “x”/”y” and we must find the coterminal angle for 5pi/2
cot
5
 0
 cot   0
2
2 1
V. Simplifying trig expressions often involves the use of trigonometric identities. There
are a few that SHOULD be part of your mathematical vocabulary. If you go to our class
web page you can print out a paper that has a bunch (28) of trig identities on it. You
should know all of the basic identities at the top along with #9 (and know how to use 9 to
get 10 and 11) , #20, 21(note: #21 has two “alternate” versions that you can obtain by
substituting a “version” #9 into the version of #21 on the handout), and 12, 13 , 14 and
15. These are ones that will come up frequently in MAT220.
1. Use #21 and then factor “the difference of two squares” in the denominator and then
reduce.
 cos   sin  
cos   sin 
cos   sin 
1



cos 2
cos 2   sin 2   cos   sin   cos   sin   cos   sin 
 ?
Will there be some values of theta for which the ending expression will NOT match the
original expression?
2. Use #20 and then just cancel!
sin 2 2sin  cos 

 2sin    ?
cos 
cos 
Will there be some values of theta for which the ending expression will NOT match the
original expression?
3. This one requires you to rewrite the Pythagorean identity (#9 on the list on your
webpage) to substitute in for the numerator and then factor the difference of two squares.
sin 2 
1  cos 2  1  cos  1  cos  


 1  cos    ?
1  cos  1  cos 
1  cos  
Will there be some values of theta for which the ending expression will NOT match the
original expression?