MAT220 Class Notes: Algebra and Trigonometry Review.

Properties of Fractions, Factoring, Simplifying Fractional Expressions (factor and cancel)

How to “split up” a single fraction into two fractions…

1. x y x y

z z z

±= ± (usually we go the other direction with this…i.e common denominator and write as a single fraction)

(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 y x

x or yz z 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)

d b

d b

a c a c ad bc ad bc

b d b d 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

±± =

±

Factoring:

I. 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

( )( )2 2a b a b a b− = + −

ii. Factor the difference of two cubes

( )( )3 3 2 2a b a b a ab b− = − + +

iii. Factor the sum of two cubes

( )( )3 3 2 2a b a b a ab b+ = + − +

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. ( )2 22 2 2x xx xx e e x xe x e⋅ + ⋅ = +

2. ( ) ( ) ( ) ( ) ( )[ ] ( ) ( )2 2 22 23 1 1 1

3 6 3 5 3 63 1

3 18 5 3 1810 10

305 12

x x x x x x x x xx x x xx x + + − = + + − = + + + − + =

Factor:

( ) ( ) ( )( ) ( ) ( ) ( )( )( )( )( )

2 23 2 2 23 9 3 3 9 3 3

= 3 3 3

1.

3 9 27 x x x x x x x

x

x x

x x

x − − − − − = − − = − −

+

+

−

=

−

( ) ( )2

3 3x x= − +

( )( )22. 6 5 6 3 2 2 3xx x x− = +− − remember that there are many different methods that can be used to factor this!

( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( )( )( ) ( )

3 3 3 2 23

2

2 8 27 2 2 3 2 2 3 2 2 3 3

3. 16 5

2 2 3 4 6 9

4 x x x x x

x x

x

x

− = − = − + +

= − +

−

+

=

Simplifying Fractions: (This MEANS Factor THEN Cancel)

( )

( )( )

( ) ( )

3

2

2

2

2 2

3

44 44 1 4 4

2 4 12 31

2 8 4

6.

6

x x x x

x x xx

x x

x x x x x

+ += =

+ −+ −

+=

+ −

22

x ( )4

x

x +

2 x ( )4x + ( )

20, 4

11

xx

xx= ≠ −

−−

(why do we need to exclude those two values?)

Note: A common error..

22 4

2

x

x

x−=

+

4

x

−

2

x1

2+1

2 2

2

x x− −= =

1

21

11 THIS IS WRONG!!!! You may NEVER cancel TERMS

1

xx

−= = −

This problem SHOULD be done as follows… ( )( )

( )

( ) ( )2 2 24 2

22

2 x xx xx

x x

+ −+ −=

−

−=

+ ( )2x −2 2x x= + ≠

( )( )

( )( )

( )( )

( )2

3

2

2 3 2 33 4 9 3 2 3 2 3

6 3 5 2 36 6

12 272.

36 56 90 1

x x x xx x x x x

xx xx xx

+− + −= =

− +− −

−=

− −

( )

( ) ( )

2 3

6 3 5 2 3

x

x x

−

− +

( )

( )

3 2 3 3

6 3 5 2

x xx

x

−= ≠ −

−

Rational Roots Theorem:

Let ( ) 1 2 2

1 2 2 1 0.....n n n

n n nf x a x a x a x a x a x a− −

− −= + + + + + + be a polynomial with integer coefficients.

If the polynomial has any rational zeros (roots), p/q, then p must be an integer factor of a0 and q must be a factor of an.

Example: List the possible rational zeros for ( ) 4 33 11 10 4f x x x x= − + − .

: 1, 2, 4

: 1, 3

1 2 4: 1, , 2, , 4,

3 3 3

p

q

p

q

± ± ±

± ±

± ± ± ± ± ±

Other important polynomial theorems for College Algebra / Precalculus.

Conjugate Pairs Theorems.

i. If your polynomial has rational coefficients and a b c+ is a zero then so is it’s conjugate a b c−

ii. If your polynomial has real coefficients and 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.

Example: Use your “list” of possible rational zeros of ( ) 4 33 11 10 4f x x x x= − + − to find ALL

of the zeros for the polynomial. Write the polynomial in factored form.

1 2 4: 1, , 2, , 4,

3 3 3

p

q± ± ± ± ± ± (Always “consider” trying “1” first…..do the coefficients sum to zero?)

( )

3 11 0 10 4

1 3 14 14 4 0 So 1 is a zero thus 1 is a factor!

x

− −

− − − − +

Now go off of the NEW numbers!!!!

3 14 14 4

2 3 12

3

− −

− ( )2

6 0 So is a zero thus 3 2 is a factor!3

NOW your new numbers are the coefficients of a quadratic so find the last two zeros a different way

x −

( ) ( ) ( )( )

( )

( )2

2 2

.

2 2 24 4 4 1 2 4 8 4 2 23 12 6 0 4 2 0

2 1 2 2 2

2

x x x x x

x

±− − ± − − ± ±− + = → − + = → = = = =

=( )2 2

2

±

( )( ) ( )( )

( ) ( )( )( )( )4 3

2 2 Thus 2 2 and 2 2 are factors!!!

2So the four zeros of the polynomial are 1, , 2 2 and the polynomial factors as follows...

3

1 3 2 2 2 2 23 11 10 4

x

f x x x x

x

x x x x

= ± − + − −

−

= − + − =

±

+ − − − − +

Let’s Try… ( ) 5 4 3 26 19 23 82 4 24f x x x x x x= − − + − −

: 1, 2, 3, 4, 6, 8, 12, 24

: 1, 2, 3, 6

1 1 1 2 3 4 8: 1, , , , 2, , 3, , 4, , 6, 8, , 12, 24

2 3 6 3 2 3 3

p

q

p

q

± ± ± ± ± ± ± ±

± ± ± ±

± ± ± ± ± ± ± ± ± ± ± ± ± ± ±

Note: we know 1 is not a zero as the sum of the coefficients is not zero.

6 19 23 82 4 24

6 6 17 79 556 3332

1996

− − − −

8 Note: 6 is a UB

3 6 37 88 182 542 1650 Note: 3 is a LB

3 6 1 26

− − − − −

− − 4 8 0 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.

6 1 26 4 8

2 6 13 0 4 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!

6 13 0 4

2 6 25 50 96 So 2 is NOT a multiple zero

2 6 1

−

− − − −

− 2 0 So 2 is a zero AND we are NOW down to a quadratic −

Now that we are down to a quadratic we can solve

( )( )

26 2 0 by factoring.

2 13 2 2 1 0

3 2

x x

x x x x

− − =

− + = → = = −

So the five zeros of ( ) 5 4 3 26 19 23 82 4 24f x x x x x x= − − + − − are 2 1

3, 2, 2, ,3 2

− − and our polynomial factors as follows

( ) 5 4 3 26 19 23 82 4 24f x x x x x x= − − + − − = ( )( )( ) ( )( )3 2 2 3 2 2 1x x x x x− + − − +

Multiplying by one, fractions, conjugation and adding zero

Dividing Fractions: (Multiply by the reciprocal…this means YOU are actually “multiplying by one”)

Note: 1

d

cd

a a a d

a c a db

c

b b cc cb d b c

d d

⋅

÷ = = ⋅ = = ⋅

(THIS is WHY dividing by a fraction is THE SAME AS multiplying by the reciprocal)

( ) ( )

( )

( )( )

( ) ( )

( )

2 2

2

2 2

2

4 3 3 37 12 9

3 12 3 4 3

4

7 12 121.

3

9

x x x xx x x

x x x x x x

x

x x x x

x x

+ + + −+ + −⋅ = ⋅

− + −

+ + + −÷ =

− − + −

+=

−

( )

( )

3

3

x

x

+

−

( ) ( )3 3x x+ −⋅

( )4x + ( )

( )

( )

2

3

3 = 4

3

x

xx

x

−

+≠ −

−

(Do YOU see WHY 3x ≠ is unnecessary?)

Adding and Subtracting Fractions: (need a common denominator AND check to see if you can simplify at the end)

( )( ) ( ) ( ) ( ) ( )

( )

( ) ( ) ( )

( )

( )

( )( )( ) ( ) ( ) ( )

2

2 2

5 5

6 5 5 4 6 5 5 4

4 5 30 =

6 5 4 6

4

5

6

4

52.

11 30 9 20

4 6

x

x x

x xx x

x x x x x x x x

x x x

x x x x x x

x xx x− = −

+ + + + + + + +

+

− =+ + + +

+

+−

+

+⋅ ⋅

+

+ +

+

+ + +

( )( )( )

2 4 5 30 =

6 5 4

x x x

x x x

+ −−

+ + +

( )( )( )

( )( )

( )( )( )

( ) ( )

2 30 =

6 5 4

6 56 5 =

6 5 4

x x

x x x

x xx x

x x x

− −

+ + +

− +− +=

+ + + ( ) ( )6 5x x+ + ( )

( )

( )( )

4

6 = 5

6 4

x

xx

x x

+

−≠ −

+ +

Conjugation: (multiplying by one)

Simplify

2

2

2 5 6 4 15 10 6 4 15 10 4 19 4 19

3 2 9

2 5

3 2 4 9 4 13 13

2

1

3

3 2 3

i i i i i i i

i ii i

iii

− −⋅

− − + − − − − −= = = =

−= − −

− +−++

Rationalize the denominator

( ) ( )( )

( )( ) ( )( )

( )

3 5 4 3 5 4 3 5 4 3 54

9 5

41.

3 9 53 5 45 3 5

4

x x x x x xx

x x

x

x

x

x xx

x

+ + − + + − + + −+= = =

− − − + +− −

+ −⋅

+ −=

−

=

−

+

+

( )( )

3 5

4

x

x

+ −

+3 5 4x x= + − ≠ −

( )( )( )

( )( ) ( )( )27 5 2 27 5 2 27 5 227

25 2 25 2 275 2

272.

5 2

5 2

5 2

xx x x x x xx

x x x

x

xx x

− − − − − − − − −−= = =

− − − + − ++

−=

+

−

−⋅

−

−

− −

( )( )( )

( )

27 5 2

27

27

x x

x

x

− − −=

− −

−=

( )( )

5 2

27

x

x

− −

− −

5 2x= − + −

Note: It would be WRONG if you wrote 27x ≠ on this particular problem! Do YOU understand why?

IF TIME

( ) ( )( )

( ) ( ) ( ) ( )11 4 5 11 4 5 11 4 511

16 5 16 5 114 5

113.

4 5

4

5

4 5

x x x x x xx

x

x

x xx

x

x x

+ − − + − − + − −+= = =

− − − + +− −

+ −⋅

+ −

+=

− −

( )11

x +=

( )( )

4 5

11

x

x

− −

+

4 5 11x x= − − ≠ −

( )( )( )

( ) ( ) ( ) ( )11 3 2 11 3 2 11 3 211

9 2 9 2 113 2

114.

3 2

3 2

3 2

x x x x x xx

x x xx

x x

xx

− − − − − − − − −− −⋅

−

−= = =

− − − + − ++ −

−=

+ − −

( )( )( )

11 3 2

11

3 2

x x

x

x

− − −=

− −

= − −

Note: It would be WRONG if you wrote 11x ≠ on this particular problem! Do YOU understand why?

Lines, Absolute Value, Piecewise defined functions and Trigonometry (a very quick review)

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 y mx b= + (m is the slope and ( )0,b is the y-intercept)

Point-Slope form of the equation of a line ( )1 1y y m x x− = − (m is the slope and ( )1 1,x y is ANY point on the line)

Slope Formula 2 1

2 1

y ym

x x

−=

− ( ( ) ( )1 1 2 2, ,x y and x y 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?

( )( )

( )

( ) ( )

( )111 11 1

1 1 1

m x xm x xy y y yy y m x x

x x x x x x

−−− −− = − → = → =

− − − ( )1x x−

1

1

y ym

x x

−→ =

−

There are four (4) different “types” of slope: positive, negative, zero and undefined.

positive negative zero undefined

“uphill”-increasing “downhill”-decreasing “horizontal”-constant “vertical”-undefined

Example: Find the equation of the line that passes through the points ( ) ( )2,5 3, 4and− −

Solution: 2 1

2 1

4 5 9 9

3 2 5 5

y ym

x x

− − − −= = = = −

− − −

( )9 9 18 25 18 7

5 2 55 5 5 5 5 5

9 7

5 5

y x b b b b b

y x

= − + → = − − + → = + → − = → =

= − +

Absolute Value: This is the most “basic” piecewise defined function.

Official Mathematical Definition: 0

0

w if ww

w if w

≥=

− <

Examples: 2 25 3 5 3 0 0 1 1x x= − = + = +=

Write ( ) 4 2f x x= − as a piecewise defined function and graph the result.

Solution: Just use the definition of absolute value where YOUR “w” is 4x – 2 .

( )( )

14 24 2 4 2 0 2

4 24 2 4 2 0 1

4 22

x xx x

f x xx x

x x

− ≥− − ≥

= − = = − − − ≤ − + <

Write ( ) 6 2f x x= − as a piecewise defined function and graph the result.

Solution: Just use the definition of absolute value where YOUR “w” is 6 – 2x .

( )( )

6 2 6 2 0 6 2 36 2

6 2 6 2 0 6 2 3

x x x xf x x

x x x x

− − ≥ − ≤= − = =

− − − ≤ − + >

Other types of piecewise defined functions.

Graph:

( )

( )2

3 1 2

2 1 1

1 2

x

x

xg

x

x

x

x − + − ≤

+ + < −

=

+

≥

<

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.

( )

1 31

2 2

1 2

1 2

1

x x

x

x xf

x

− ≤

=

− − <

−

<

≥

−

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.

2 3 1

ln log sin cos tanx

b

y x y x y x y x y x yx

y e y x y x y x y x y x

= = = = = =

= = = = = =

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 sin tan

sec csc cot1 1

xx y

y

y

x y x

θ θ θ

θ θ θ

= = =

= = =

Note: the equation of the unit circle is 2 2 1x y+ = 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 2 2 1x y+ = you will get a true statement.

Example: Use the unit circle to evaluate the following…

( )

( )

0 0

0

5) sin 210 = ) cos 30 = ) tan

1 3 1 3or

2 2 33

1 3unde

6

2) csc 720 ) cotfined

333

A B C

oD rE

π

π

− − =

= =

−

− −

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.

( ) ( )

( ) ( )

1 1

1 1

sin sin 1 12 2

1 1 2 2

cos 0 cos 1 1

1 1

f x x x f x x x

y y

f x x x f x x x

y

π π

π π

π

− −

− −

= − ≤ ≤ → = − ≤ ≤

− ≤ ≤ − ≤ ≤

= ≤ ≤ → = − ≤ ≤

− ≤ ≤

( ) ( )1 1

0

tan tan 2 2

2 2

y

f x x x f x x y

y y

π

π π

π π

− −

≤ ≤

= − < < → = − ∞ ≤ ≤ ∞

− ∞ ≤ ≤ ∞ − < <

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.

1 1 1 13 3 3 7) cos ) sin ) cos sin ) sin sin

2 2 5 6A B C D

π− − − −

− = = − = =

5 -3

3

,2

−

3

,2

( )22 2

1

3 5

4

3 4cos sin

5 5

x

x

x

r

−

+ − =

=

− = =

7

6

π

11

6

π

1 3 5

cos2 6

π−

− =

1 3

sin2 3

π−

=

we must find an angle in ,2 2

π π −

that has the same sine as 7

6

π. As you

know from the unit circle, 11

6

πhas the

the same sine as 7

6

π.

11,

6 2 2

π π π ∉ −

so we pick 6

π−

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. xCsc

xSin1

= 2. xSec

xCos1

= 3. xCot

xTan1

=

4. ( ) xSinxSin −=− 5. ( ) xCosxCos =− 6. ( ) xTanxTan −=−

7. xCos

xSinxTan = 8.

xSin

xCosxCot =

9. 122=+ θθ CosSin 10. θθ 22 1 CscCot =+ 11. θθ 221 SecTan =+

12. ( ) SinASinBCosACosBBACos +=− 13. ( ) SinASinBCosACosBBACos −=+

14. ( ) CosASinBSinACosBBASin +=+ 15. ( ) CosASinBSinACosBBASin −=−

16. ( )TanATanB

TanBTanABATan

−

+=+

1 17. ( )

TanATanB

TanBTanABATan

+

−=−

1

18. xCosxSin =

−

2

π 19. xSinxCos =

−

2

π

note: #’s 18 and 19 also hold for the function pairs….tan x, cot x AND sec x, csc x

20. SinxCosxxSin 22 = 21.

2 2

2

2

2

2 1

=1 2

Cos x Cos x Sin x

Cos x

Sin x

= −

= −

−

22. xTan

xTanxTan

21

22

−=

23. 2

212 xCosxSin

−= 24.

2

212 xCosxCos

+= 25.

xCos

xCosxTan

21

212

+

−=

26. 2

1

2

CosxxSin

−±= 27.

2

1

2

CosxxCos

+±=

28. Cosx

CosxxTan

+

−±=

1

1

2 =

Cosx

Sinx

+1 =

Sinx

Cosx−1

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.

Examples: Simplify the following fractional expressions that involve trigonometric functions.

cos ccos1.

sin

os

2s2 in cos

θ

θ θ

θ θ

θ==

2sin cosθ θ

1 1csc cos 0 2 where k is any integer

2sin 2 2

3 2 where k is any int

2

k

k

πθ θ θ π

θ

πθ π

= = ≠ → ≠ +

≠ + eger

Note: Identities # 20 and # 1 were used in my work.

( ) ( )2 2

1 sin1 sin 1 sin

1 si

1 sin2

n 1 sin 1 sin.

cos

θθ θ

θ θ

θ

θ θ

−− −= =

−=

+ −

− ( )( ) ( )1 sin 1 sinθ θ+ −

( )

1 31 sin 0 2 where k is any integer

1 sin 2k

πθ θ π

θ

=

= + ≠ → ≠ ++

Note: A version of identity # 9 was used in my work.

( )( ) ( ) ( )

( )

2 2 cos sin cos sin cos sin cos sincos sin

sin cos sin cos cos sin

cos

cos 23.

sin

cos

=

θ θ θ θ θ θ θ θθ θθ

θ θ θ θ θ θθ θ

+ − + −−=

− −=

−=

− −

( ) ( )sin cos sinθ θ θ θ+ −

( )cos sinθ θ− −

= cos sin

cos sin 0

θ θ

θ θ

− −

− ≠ 2 where k is any integer4

5 2 where k is any integer

4

k

k

πθ π

πθ π

→ ≠ +

≠ +