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. " " xy y xy x x or y z 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. xy 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 2 a b a b a b - = + - ii. Factor the difference of two cubes ( ) ( ) 3 3 2 2 a b a b a ab b - = - + + iii. Factor the sum of two cubes ( ) ( ) 3 3 2 2 a 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.
13
Embed
MAT220 Class Notes: Algebra and Trigonometry Review
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
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= − − + − −