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Unformatted text preview: PARTS 1 & 2 COMBINED COVER PRINCIPLES FOR BASIC, INTERMEDIATE AND COLLEGE COURSES SET THEORY NOTATION . { } braces indicate the beginning and end of a set notation; when listed, el
ements or members must be separated by commas. EX: A = {4, 8, 16}; sets
are finite (ending, or having a last element) unless otherwise indicated.  indicates continuation of a pattern. EX: B = {5, 10, 15, , 85, 90} 0 at the end indicates an infinite set, that is, a set with no last element.
EX: C = {3, 6, 9, 12, ...} 0  is a symbol which literally means "such that." 0 E means “is a member of” OR “is an element of.” EX: IfA = {4, 8, 12}
then 12 E A because 12 is in set A. 0 65 means “is not a member of” OR “is not an element of.” EX: If
B = {2, 4, 6, 8} then 365B because 3 is not in set B. 0% means empty set OR null set; a set containing no elements or mem
bers, but which is a subset of all sets; also written as { }. 0 C means “is a subset of;” also may be written as Q 0 ¢ means “is not a subset of;” also may be written as @  AC B indicates that every element of set A is also an element of set B.
EX: IfA = {3, 6} and B = {1, 3, 5, 6, 7, 9} then ACB because the 3 and 6
which are in set A are also in set B. 0 2“ is the number of subsets of a set when n equals the number of elements
in that set. EX: If A = {4, 5, 6} then set A has 8 subsets because A has 3 elements and 23 = 8.
. OPERATIONS  U means unlon. 0 AU B indicates the union of set A with set B; every element of this set is
either an element of set A OR an element of set B; that is, to form the
union of two sets, put all of the elements of both sets together into one set
making sure not to write any element more than once. EX: If A = {2,4} and B = {4, 8, 16} then A U B = {2, 4, 8,16}.
0 0 means intersection. 0 AﬂB indicates the intersection of set A with set B; every element of this
set is also an element of BOTH set A and set B; that is, to form the in tersection of two sets, list only those elements which are found in BOTH O P E RAT I O N S O F R EA L N U M B E RS _ ofthe two sets. EX: IfA = {2,4} and B = {4, 8, 16} then A n B = {4}. ' A indicates the complement of set A; that is, all elements in the univer
sal set which are NOT in set A. EX: If the Universal set is the set of
Integers and A = {0, 1, 2, 3, ...} then A {1, 2, 3, 4,...}. A I) A = Q. PROPERTIES A = B means all of the elements in set A are also in set B and all ele
ments in set B are also in set A, although they do not have to be in the
same order. EX: IfA = {5, 10} and B = {10, 5} then A = B. 0 n(A) indicates the number of elements in set A. EX: If A = {2, 4, 6} then
n(A) = 3. 0 ~ means “is equivalent to”; that is, set A and set B have the same number of el
ements although the elements themselves may or may not be the same. EX: If
A = {2, 4, 6} and B = {6, 12, 18} then A ~ B because n(A) = 3 and n(B) = 3. 0 A (I B = Q indicates disjoint sets which have no elements in common. SETS OF NUMBERS ' Natural or Counting numbers = {1, 2, 3, 4, 5, ..., 11, 12, ...} 'Whole numbers = {0, 1, 2, 3,..., 10, 11, 12, 13, ...} ' Integers = {... , 4, 3, 2, 1, 0, 1, 2, 3, 4, ...} ' Rational numbers = {p/q  p and q are integers, q at 0}; the sets of Nat
ural numbers, Whole numbers, and Integers, as well as numbers which
can be written as proper or improper fractions, are all subsets of the set of
Rational numbers. ° Irrational numbers : { x  x is a Real number but is not a Rational num
ber}; the sets of Rational numbers and Irrational numbers have no ele
ments in common and are therefore disjoint sets. 'Real numbers = {x  x is the coordinate of a point on a number
line}; the union of the set of Rational numbers with the set of Irra
tional numbers equals the set of Real numbers. ° Imaginary numbers = {ai  a is a Real number and i is the number
whose square is 1}; i2 = 1; the sets of Real numbers and Imaginary
numbers have no elements in common and are therefore disjoint sets. ° Complex numbers = {a + bi  a and b are Real numbers and i is the number
whose square is 1}; the set of Real numbers and the set of Imaginary num
bers are both subsets of the set of Complex numbers. EXs: 4 + 7i ; 3  2i ~___ I W  Part 1_ PROPERTIES OF REAL NUMBERS ... I mudelﬂ:puLﬁJﬂﬁlm.fm FOR ANY REAL NUMBERS a, b, AND G FOR ADDITION FOR MULTIPLICATION
a+bisaReal number ab isaReal number
a+b=b+a ab=ba
(a+b)+c=a+(b+c) (ab)c=a(bc)
0+a=aanda+0=a a‘1=aandl'a=a
a+(a)=0and a1/,=1and
(a)+a=0 ‘/a°a=1ifa¢0 Distributive Property a(b + c) = ab + ac; a(b  c) = ab  ac PROPERTIES OF EQUALITY
FOR ANY REAL NUMBERS a, b, AND c Reﬂexive: a = a
Symmetric: If a = b then b = a
Transitive: If a = b and b = c then a = c
Addition Property: If a = b then a + c = b + c
Multiplication Property: If a = b then ac = bc
Multiplication Property of Zero: a 0 0 = 0 and 0 0 a = 0
Double Negative Property: (a) = a PROPERTIES OF INEQUALITY
FOR ANY REAL NUMBERS a, b, AND c Trichotomy: Either a>b, or a=b, or a<b
Transitive: If a<b, andb<cthena<c Ifa<bthenac<b+c Ifa>bthenac>b+c Ifc¢0andc>0,anda>bthenac>bc;
also, if a<bthen ac<bc lfc¢0andc<0,anda>bthenac<bc;
also, ifa<bthen ac>bc PROPERTY Closure
Commutative
Associative Identity VNBd Inverse Addition Property of Inequalities: Multiplication Property of Inequalities: ABSOLUTE VALUE
x = x if X is zero or a positive number; x = X if x is a negative number;
that is, the distance (which is always positive) of a number from zero on the
number line is the absolute value of that number. EXs:   4 =  ( 4) = 4;
29=29; 0=0;43=(43)=43
ADDITION
If the signs of the numbers are the same: add the absolute values of the numbers; the sign of the answer is the same as the signs of the original two
numbers. EXs: 11 + 5 = 16 and 16 + 10 = 26
If the signs of the numbers are different: subtract the absolute values of the numbers; the answer has the same sign as the number with the larger
absolute value. EXs: 16 + 4 = 12 and 3 + 10 = 7 S U BTRACTI O N a  b = a + (b); subtraction is changed to addition of the opposite number:
that is, change the sign of the second number and follow the rules of addition
(never change the sign of the ﬁrst number since it is the number in back of the
subtraction sign which is being subtracted; 14  6 ¢ 14 +  6).
EXs:1542= 15+(42)=27; 245=24+(5)=29; 13 (45)=
13 + (+45) = 32;  62  (20) =  62 + (+20) =  42 MULTIPLICATION
The product of two numbers which have the same signs is positive;
EXs: (55)(3) = 165; ( 30)( 4) = 120; ( 5)( 12) = 60
The product of two numbers which have different signs is negative no matter
which number is larger. EXs: ( 3)(70) =  210; (21)( 40) =  840; (50)(3) =  150 DIVISION
(DIVIsons no NOT EQUAL ZERO) The quotient of two numbers which have the same sign is positive.
EXs: ( 14) / (7) = 2; (44) / (11) = 4; ( 4) / ( 8) = .5
The quotient of two numbers which have different signs is negative no
matter which number is larger.
EXs: (24) / (6) = 4; (40) / (8) =  5; (14) / (56) =  .25 DOUBLE NEGATIVE  ( a) = a; that is, the negative sign changes the sign of the contents of the
parentheses. EXs:  (4) = 4;  (17) = 17 VNSdov vuadov ALGEBRAIC TERMS COMBINING LIKE TERMS
ADDING on SUBTRAGTING a + a = 2a; when adding or subtracting terms, they must have exactly
the same variables and exponents, although not necessarily in the
same order; these are called like terms. The coefficients
(numbers in the front) may or may not be the same. ° RULE: combine (add or subtract) only the coefficients of like terms and
never change the exponents during addition or subtraction. EXs: 4xy3 and
7y3x are like terms and may be combined in this manner: 4xy3 + 7y3x =
3xy3. Notice only the coefficients were combined and no exponent changed. 15a2bc and 3bca4 are not like terms because the exponents of the a are not the same in both terms, so they may not be combined. MULTIPLYING
PRODUCT RULE FOR EXPONENTS (a"’)(a") = am"; any terms may be multiplied, not just like terms.
The coefficients m the variables are multiplied which means the
exponents also change. ' RULE: multiply the coefﬁcients and multiply the variables (this means you have to add the exponents of the same variable).
EX: (4a2c)(12a3b2c) = 48a5b2c2, notice that 4 times 12 became 48, a2
times a3 became as, c times c became c2, and the b2 was written down.
DISTRIBUTIVE PROPERTY FOR POLYNOMIALS
° Type 1:a(c + d) = ac + ad. EX: 4x3(2xy + y2) = 8x4y + 4x3y2
'Type2:(a+b)(c+d)=a(c+d)+b(c+d)=ac+ad+bc+bd
EX: (2x + y)(3x  5y) = 2x(3x  5y) + y(3x  5y) =
6x2  10xy + 3xy  5y2 = 6x2  7xy  5yz
' Type 3:(a + b)(c2 + cd + d2) = a(c2 + cd + d2) + b(c2 + cd + d2) = ac2 +
acd + ad2 + be2 + bcd + bd2
EX: (5x + 3y)(x2  6xy + 4y2) = 5x(x2 6xy+4y2) +3y(x2  6xy + 4y2) =
5x330x2y+20xy2+3x2y18xy2+12y3=5x327x2y+2xy2+12y3
“FOIL” METHOD FOR PnODucTs 0F BINOMIALs ' This is a popular method for multiplying 2 terms by 2 terms only. FOIL
means ﬁrst times ﬁrst, outer times outer: inner times inner: and last times last.
EX: (2x + 3y)(x + 5y) would be multiplied by multiplying first term
times first term, 2x times x = 2x2; outer term times outer term, 2x times
5y = 10xy; inner term times inner term, 3y times x = 3xy; and last term
times last term, 3y times 5y = 15y2; then combining the like terms of 10xy
and 3xy gives 13xy with the final answer equaling 2x2 + 13xy + 15y2. SPECIAL PRODUCTS 'Type 1: (a+b)2=(a+b)(a+b)=a2+2ab+b2 'Type2: (ab)2=(ab)(ab)=a22ab+b2 'Type3: (a+b)(ab)=a2+0abb2=azb2 EXPONENT RULES ' RULE 1: (a“')11 = a‘"'11 ; (am)n means the parentheses contents are multi
plied n times and when you multiply you add exponents; EX: (2m“n2)3 = (2m4n2) (2m4n2) (2m4n2) = 8m12n6, notice the
parentheses were multiplied 3 times and then the rules of regular multi
plication of terms were used; ' SHORTCUT RULE: when raising a term to a power, just multiply expo
nents; EX: (2m4n2)3 = 23m12n6, notice the exponents of the 2, m4 and n2
were all multiplied by the exponent 3, and that the answer was the same as the
example above. CAUTION: am at (a)“; these two expressions are different.
EXs: 4yz2 at (4yz)2 because (4yz)2 = (4yz) (4yz) = 16yzz2 while 4yz2
means 4  y ' z2 and the exponent 2 applies only to the z in this situation. ° RULE 2: (ab)m = am b’“. EX: 6x3 y)2 = 62 x6 y2 = 36x6 y2
BUT (6x3 + y)2 = (6x3 + y) (6x + y) = 36x6 + 12x3y + yz; because there
is more than one term in the parentheses the distributive property for
polynomials must be used in this situation. m m 2
'RULE 3: 3 =Lwhen b at 0; EX: ‘4X2y =16x4y2
b 1"” 5z 25z2 ' RULE 4: Zero Power a0 = 1 when a at 0 DIVIDING ' QUOTIENT RULE: 27:11:?“ ; any terms may be divided, not just like terms; the coefficients ﬂi the variables are divided which means the
exponents also change. RULE: Divide coefficients and divide variables (this means you have
to subtract the exponents of matching variables). EX: (20x5yzz) / (5x22) = 4x3y2, notice that 20 divided by 5 became 4,
x5 divided by x2 became x3, and 1 divided by z became one and therefore
did not have to be written because 1 times 4x3y2 equals 4x3y . ' NEGATIVE EXPONENT: a' '1 = 1 / a'1 when a 1: 0; EXs: 2'1 = 1I2;
(41 '3y2) / (3ab'1) = (4y2b1) / (3az3). Notice that the 4 and the 3 both stayed
where they were because they both had an invisible exponent of positive 1;
the y remained in the numerator and the a remained in the denominator be
cause their exponents were both positive numbers; the z moved down and the
b moved up because their exponents were both negative numbers. STEPS FOR SOLVING A FIRST DEGREE
EQUATION WITH ONE VARIABLE ° FIRST, eliminate any fractions by using the Multiplication Property of
Equality. EX: 1I2 (3a + 5) = 2/3 (7a  5) + 9 would be multiplied on both sides
of the = sign by the lowest common denominator of 1I2 and 2I3, which is 6;
the result would be 3(3a + 5) = 4(7a  5) + 54, notice that only the 1/2, the
2/3, and the 9 were multiplied by 6 and not the contents of the parentheses;
the parentheses will be handled in the next step which is distribution. ° SECOND, simplify the left side of the equation as much as possible by us
ing the Order of Operations, the Distributive Property, and Combining
Like Terms. Do the same to the right side of the equation. EX: Use dis
tribution first, 3(2k  5) + 6k  2 = 5  2(k + 3) would become 6k  15 +
6k  2 = 5  2k  6 and then combine like terms to get 12k  17 = 1  2k. ' THIRD, apply the Addition Property of Equality to simplify and organize all
terms containing the variable on one side of the equation and all terms which do
not contain the variable on the other side. EX: 12k  17= 1  2k would become
2k + 12k  17 + 17= 1 + 17  2k + 2k and then combining like terms, 14k = 16. ' FOURTH, apply the Multiplication Property of Equality to make the co
efficient of the variable 1. EX: 14k = 16 would be multiplied on both sides
by 1/14 (or divided by 14) to get a 1 in front of the k so the equation would
become 1k = 16/14 or simply k = 11/7 or 1.143. ' FIFTH, check the answer by substituting it for the variable in the origi
nal equation to see if it works. NOTE: 1. Some equations have exactly one solution (answer). They are conditional
equations. EX: 2k = 18 2. Some equations work for all real numbers. They are identities. EX: 2k = 2k 3. Some equations have no solutions. They are inconsistent equations.
EX: 2k+3=2k+7 STEPS FOR SOLVING A FIRST DEGREE
INEQUALITY WITH ONE VARIABLE ADDITION PROPERTY OF INEQUALITIES For all real numbers a, b, and c, the inequalities a < b and a + c < b + c
are equivalent; that is, any terms may be added to both sides of an inequal
ity and the inequality remains a true statement. This also applies to a > b
and a + c > b + c. MULTIPLIOATION PROPERTY OF INEOUALITIES ' For all real numbers a, b, and c, with c at 0 and c > 0, the inequalities a >
b and ac > be are equivalent and the inequalities a < b and ac < be are
equivalent; that is, when c is a positive number the inequality symbols stay
the same as they were before the multiplication. EX: If 8 > 3 then multi
plying by 2 would make 16 > 6, which is a true statement. ° For all real numbers a, b, and c, with c at 0 and c < 0, the inequalities a > b and
ac < be are equivalent and the inequalities a < b and ac > be are equivalent; that
is, when c is a negative number the inequality symbols must be reversed from
the way they were before the multiplication for the inequality to remain a true
statement. EX: If 8 > 3 then multiplying by 2 would make 16 > 6, which is
false unless the inequality symbol is reversed to make it true, 16 < 6. STEPS FOR SOLVING ' FIRST, simplify the left side of the inequality in the same manner as an
equation, applying the order of operations, the distributive property, and
combining like terms. Simplify the right side in the same manner. ' SECOND, apply the Addition Property of Inequality to get all terms which
have the variable on one side of the inequality symbol and all terms which
do not have the variable on the other side of the symbol. ' THIRD, apply the Multiplication Property of Inequality to get the
coefficient of the variable to be a 1; (remember to reverse the in
equality symbol when multiplying or dividing by a negative number, this
is NOT done when multiplying or dividing by a positive number). ‘FOURTII, check the solution by substituting some numerical values of
the variable in the original inequality. ORDER OF OPERATIONS  FIRST, simplify any enclosure symbols: parentheses ( ), brackets [ ],
braces { } if present:
1.Work the enclosure symbols from the innermost and work outward.
2. Work separately above and below any fraction bars since the entire top of
a fraction bar is treated as though it has its own invisible enclosure sym
bols around it and the entire bottom is treated the same way.  SECOND, simplify any exponents and roots, working from left to
right; Note: The J— symbol is used only to indicate the positive root,
except that «m = 0.  THIRD, do any multiplication and division in the order in which they oc
cur, working from left to right; Note: If division comes before multiplica
tion then it is done first, if multiplication comes first then it is done first. ° FOURTH, do any addition and subtraction in the order in which they oc
cur,working from left to right; Note: If subtraction comes before addition in
the problem then it is done first, if addition comes first then it is done first. FACTORING Some algebraic polynomials cannot be factored. The following are meth
ods of handling those which can be factored. When the factoring process
is complete the answer can always be checked by multiplying the factors
out to see if the original problem is the result. That will happen if the
factorization is a correct one. A polynomial is factored when it is written as a product of polynomials
with integer coejﬁcients and all of the factors are prime. The order of the
factors does not matter. FIRST STEP  'GCF' Factor out the Greatest Common Factor (GCF), if there is one. The GCF
is the largest number which will divide evenly into every coefficient togeth
er with the lowest exponent of each variable common to all terms. EX: 15a3c3 + 25azc4d2  10a2c3d has a greatest common factor of 5a2c3 be
cause 5 divides evenly into 15, 25, and 10; the lowest degree of a in all three
terms is 2; the lowest degree of c is 3; the GCF is 5a2c3 ; the factorization is
Sazc3 (33 + Scd  2d). SECOND STEP  CATEGORIZE AND FACTOR Identify the problem as belonging in one of the following categories. Be
sure to place the terms in the correct order first: highest degree term to
the lowest degree term. EX: 2A3 + A4 + 1 =A4  2A3 + 1 CATEGORY FORM OF PROBLEM FORM OF FACTORS ax2+bx+c
(aaé0) Ifa= 1: (x+h)(x+k)wherehk=cand
h + k = b; h and k may be either
positive or negative numbers. Ifa ¢ 1: (mx + h)(nx + k) where m  n
=a,hk=c,andhn+mk=b; m,
h, n and k may be either positive or
negative numbers. Trial and error
methods may be needed. TRINOMIALS
(a rEnMs) (see Special Factoring Hints at right)
x2 + 2cx + c2 (x + c) (x + c) = (x + c)2 where c may be @e'fe“ 5‘1"!" 9) either a positive or a negative number a2x2 _ bzyz
(difference of 2 squares) (ax + by)(ax  by) a2x2 + bzyz
(sum of 2 squares)
asxs + bsys
(sum of 2 cubes) asxs _ bays
(difference of 2 cubes) BN0MAL5 PRIME — cannot be factored! (2 TERMS)
(ax + by) (azx2  abxy + bzyz) (ax  by) (azx2 + abxy + bzyz)
(see Special Factoring Hints at right) PERFECT
cUBEs
(4 TERMS) a3);3 + 3a2bx2  3ab2x + b3 (ax + b)3 = (ax + b)(ax + b) (ax + b) a3x3  3a2bx2  3ab2x  b3 (ax  b)3 = (ax  h)(ax  b)(ax  b) ax + ay + bx + by
(2  2 grouping) X2+20X+02y2
(3  I grouping) a(x+y)+b(x+y)=(x+y)(a+b) GROUPING (x + c)2  y2 = (x + c + y)(x + c  y) yzxzZcxc2 y2(x+c)2=(y+x+c)(yxc) (1  3 grouping) SPECIAL FACTORING HINTS
TRI NOMIALS
WHERE THE cOEFFIcIENr or THE HIGHEST DEGREE TERM :5 NOT 1 The first term in each set of parentheses must multiply to equal the first
term (highest degree) of the problem. The second term in each set of
parentheses must multiply to equal the last term in the problem. The
middle term must be checked on a trial and error basis using: outer
times outer plus inner times inner; ax2 + bx + c = (mx + h)(nx + k)
where mx times nx equals axz, h times k equals c, and mx times k plus
h times nx equals bx. EX: To factor 3x2 + 17x  6 all of the following are possible correct factor
izations, (3x + 3)(x  2); (3x + 2)(x  3); (3x + 6)(x  1); (3x + 1)(x  6). How
ever, the only set which results in a 17x for the middle term when applying
“outer times outer plus inner times inner” is the last one, (3x + 1)(x  6). It
results in 17x and +17): is needed, so both signs must be changed to get the
correct middle term. Therefore, the correct factorization is (3x  1)(x + 6). BINOMIALS
'rHE SUM on DIFFERENCE or Two cUBEs This type of problem, a3x3 i b3y3, requires the memorization of the fol lowing procedure: The factors are two sets of parentheses with 2 terms in
the first set and 3 terms in the second. To find the 2 terms in the first set
of parentheses take the cube root of both of the terms in the problem and
join them by the same middle sign found in the problem. The 3 terms in
the second set of parentheses are generated from the 2 terms in the first
set of parentheses. The first term in the second set of parentheses is the
square of the first term in the first set of parentheses; the last term in the
second set is the square of the last term in the first set; the middle term
of the second set of parentheses is found by multiplying the first term
and the second term from the first set of parentheses together and chang
ing the sign. Thus, a3x3 i b3y3 = (ax i by)(a2x2 I abxy + bzyz). EX: To factor 27x3  8 find the first set of parentheses to be (3x  2) be
cause the cube root of 27x3 is 3x and the cube root of 8 is 2. Find the 3
terms in the second set of parentheses by squaring 3x to get 9ng square
the last term 2 to get +4; and to find the middle term multiply 3x times
2 and change the sign to get +6x. Therefore, the final factorization of
27x38 is (3x  2) (9x2 + 6x + 4). PERFECT CUBES
(4 TERMS) Perfect cubes, such as a3x3 i 3a2bx2 + 3ab2x 1 b3, factor into three sets
of parentheses, each containing exactly the same two terms; therefore,
the final factorization is written as one set of parentheses to the third
power, thus a perfect cube; (ax i b)3 = a3):3 1 3a2bx2 + 3ab2x 1 b3. EX: To factor 27x3  54x2 + 36x  8 it must be first observed that the prob
lem is in correct order and that it is a perfect cube; then the answer is sim
ply the cube roots of the first term and the last term placed in a set of
parentheses to the third power; so the answer to this example is (3x  2)3. NOTICE TO STUDENT This chart is the first of 2 charts outlining the major topics taught in
Algebra courses. Keep it handy as a quick reference source in the
classroom, while doing homework, and use it as a memory refresher
when reviewing prior to exams. It is a durable and inexpensive study tool
that can be repeatedly referred to during and well beyond your college
years. Due to its condensed format, however, use it as an Algebra glide and
not as a replacement for assigned course work. @2002 BarCharts, Inc. Boca Raton, FL RATIONAL EXPRESSIONS DEFINITION
The quotient of two polynomials where denominator cannot
(x+4) equal zero is a rational expression. EX: (x_3) where x at 3, since 3 would make the denominator, x  3, equal to zero. BASICS
' DOMAIN: set of all Real numbers which can be used to replace a variable. EX: The domain for the rational expression; w
is {xx E Reals and x at 1 or x at 4}. (X+1)(4—X) 1. That is, x can be any Real number except 1 or 4 because 1 makes
(x + 1) equal to zero and 4 makes (4  x) equal to zero; therefore,
the denominator would equal zero, which it must not. 2. Notice that numbers which make numerator equal to zero, 5 and 2,
are members of the domain since fractions may have zero in
numerator but not in denominator. ° RULE 1:
1. If x/y is a rational expression then x/y = xa/ya when a at 0. a. That is, you may multiply a rational expression (or fraction) by any
nonzero value as long as you multiply both numerator and denom
inator by the same value. i. Equivalent to multiplying by 1 since a/a=1.
EX: (x/y)(l)=(x/y)(a/a) = xa/ya
ii.Note: 1 is equal to any fraction which has the same numera tor and denominator. RULE 2:
xa _ x l.If La is a rational expression,i_iwhen a 7* 0.
ya ya 3’ a. That is, you may write a rational expression in lowest term because E =[EIEJ=(£)(1)=£ since 3:1
ya Y a y y a
I LOWEST TERMS:
l. Rational expressions are in lowest terms when they have no com
mon factors other than 1.
2. STEP 1: Completely factor both numerator & denominator.
3. STEP 2: Divide both the numerator and the denominator by the
greatest common factor or by the common factors until no common (1:2 +8x+15) _(x+5)(x+3) _(x+3) (x2 +3x—10) (x+5)(x—2) (x—2)
because the common factor of (x + 5) was divided into the
(x +5) _1
(x +5)
4.NOTE: Only factors can be divided into both numerators and denominators, never terms.
OPERATIONS ADDITION
(DENOMINATORS MUST BE THE SAME) factors remain. EX:
numerator and the denominator since 0 RULE 1: (a + c)
1. If a/b and c/b are rational expressions and b at 0, then:% +%= b . a. If denominators are already the same simply add numerators and write this sum over common denominator.
° RULE 2: 1. If a/b and c/d are rational expressions and b at: 0 and d at 0, then:
a + c = (ad) + (cb) =(ad+cb)
b d (bd) (bd) (bd)
a. If denominators are not the same they must be made the same
before numerators can be added.
' ADDITION STEPS
1. If the denominators are the same, then: a. Add the numerators. b. Write answer over common denominator. .
c.Wr1te final answer 1n lowest terms, maklng sure to follow directions for finding lowest terms as indicated above.
Ex: (x+2) + (x—l) _ (2x+1)
(x—6) (x—6) (X—6)
2. If the denominators are not the same, then:
a. Find the least common denominator.
b. Change all of the rational expressions so they have the same
common denominator.
0. Add numerators.
(1. Write the sum over the common denominator.
e. Write the final answer in lowest terms. EX; x+3+ x+1 _(x+3)(x—1)+(x+1)(x+5) _ 2X2 +8x+2
x+5 x—l (X+5)(x—1) (x—l)(x+5) x2 +4x—5
f. NOTE: If denominators are of a degree greater than one, try to factor all denominators first so the least common denominator
will be the product of all different factors from each denominator. SUBTRACTION o RULE 1: (DENOMINATORS MUST BE THE SAME) 1. If a/b and c/b are rational expressions and b at 0, then: a_£=a+;c=(ac)
b b b b b
a. If denominators are the same be sure to change all signs of the
terms in numerator of rational expression which is behind (to the right of) subtraction sign; then add numerators and write
result over common denominator. EX_ x—3 _x+7 : x—3+(—X)+(—7) : —10
' x+1 x+1 x+1 X+1 ' RULE 2: l.lf a/b and c/d are rational expressions and b at 0 and d at 0, then: 3_£_@_@_ (ad—ch) b (1 (bd) (bd) bd a. If denominators are not the same they must be made the same
before numerators can be subtracted. Be sure to change signs of
all terms in numerator of rational expression which follows sub
traction sign after rational expressions have been made to have
a common denominator. Combine numerator terms and write
result over common denominator. b. Note: When denominators of rational expressions are additive
inverses (opposite signs), then signs of all terms in denominator
of expression behind subtraction sign should be changed. This
will make denominators the same and terms of numerators can
be combined as they are. Subtraction of rational expressions is
changed to addition of opposite of either numerator (most of
time) or denominator (most useful when denominators are addi
tive inverses) but never both. ' SUBTRACTION STEPS:
1.lf the denominators are the same, then: a. Change signs of all terms in numerator of a rational expression
which follows any subtraction sign. b. Add the numerators. c. Write answer to this addition over common denominator. d. Write final answer in lowest terms, making sure to follow
directions for finding lowest terms as indicated above. EX.(x+2)_(x—1)_[x+2+(—x)+1]_ 3
(216) (X6) (X6) (X6)
2. If the denominators are not the same, then:
a. Find the least common denominator. b. Change all of the rational expressions so they have the same
common denominator. c. Multiply factors in the numerators if there are any. d. Change the signs of all of the terms in the numerator of any
rational expressions which are behind subtraction signs. e. Add numerators.
f. Write the sum over the common denominator. g. Write the ﬁnal answer in lowest terms. EX:
(x+3) _ (x+1) _ (x+3)(x—1) _(x +1)(x +5) _ —4x—8
(x+5) (x—l) (x+5)(x—1) (x—1)(x+5) x2 +4x—5 h. NOTE: If denominators are of a degree greater than one,.try to fac tor all denominators first, so the least common denominator will
be the product of all different factors from each denominator. MULTIPLICATION
(DENOMINATORS DO NOT HAVE TO BE THE SAME) I RULE:
1. If a, b, c, & d are Real numbers and b and d are nonzero numbers, then: (£I£)= (ac) ; [top times top and bottom times bottom]
b d (bd) OMULTIPLICATION STEPS: l.Complete1y factor all numerators and denominators. 2. Write problem as one big fraction with all numerators written as
factors (multiplication indicated) on top and all denominators
written as factors (multiplication indicated) on bottom. 3. Divide both numerator and denominator by all of the common
factors; that is, write in lowest terms. 4.Multiply the remaining factors in the numerators together and
write the result as the final numerator. 5. Multiply the remaining factors in the denominators together and
write the result as the final denominator. EX: ( x+3 Ixz—Zx—3)_
x2+2x+1 x2—9 ) DIVISION
' DEFINITION . . . x .
1. Recrprocal of a rational expressron — ml because 5
x [reciprocal may be found by inverting the expression].
EX: The reciprocal of is (£7) . X+7 x— 3
l.lf a, b, c, and d are Real numbers a, b, c, and d are nonzero 3+:=(%)(%)=% ' DIVISION STEPS
l. Reciprocate (ﬂip) rational expression found behind division sign
(immediately to right of division sign).
2.Multiply resulting rational expressions, making sure to follow
steps for multiplication as listed above.
Ex: x2 —2x—15 +(x+2) = x2 —2x—15 .(x—S)
x2 —10x+25 (X5) x2 —10x+25 (X+2)
Numerators and denominators would then be factored, written in
(x + 3) (x+2)' numbers, then: lowest terms, and yield a final answer of COMPLEX FRACTIONS An understanding of the Operations section of Rational Expressions is
required to work “complex fractions.”
° DEFINITION: A rational expression having a fraction in the 1 numerator or denominator or both is a complex fraction. EX: X x ' TWO AVAILABLE METHODS: 1. Simplify the numerator (combine rational expressions found
only on top of the complex fraction) and denominator (combine
rational expressions found only on bottom of the complex frac
tion) then divide numerator by denominator; that is, multiply
numerator by reciprocal (flip) of denominator. OR 2. Multiply the complex fraction (both in numerator & denomi
nator) by least common denominator of all individual fractions
which appear anywhere in the complex fraction. This will elimi
nate the fractions on top & bottom of the complex fraction and
result in one simple rational expression. Follow steps listed for
simplifying rational expressions. SYNTHETIC DIVISION ' DEFINITION: A process used to divide a polynomial by a binomial
in the form of x + h where h is an integer. ' STEPS:
1.Write the polynomial in descending order [from highest to low
est power of variable]. EX: 3x3  6x + 2 2.Write all coefﬁcients of dividend under long division symbol, making sure to write zeros which are coefficients of powers of variable which are not in polynomial. EX: Writing coeﬁ‘icients of polynomial in example above, write 3 0 6 2
because a zero is needed for the x2 since this power of x does not
appear in polynomial and therefore has a coefficient of zero. 3. Write the binomial in descending order. EX: x  2 4. Write additive inverse of constant term of binomial in front of long division sign as divisor. EX: The additive inverse of the 2 in
the binomial x  2 is simply +2; that is, change the sign of this term. 5. Bring up first number in dividend so it will become first num
ber in quotient (the answer). 6. Multiply number just placed in quotient by divisor, 2.
a. Add result of multiplication to next number in dividend. b. Result of this addition is next number coefficient in quotient; so
write it over next coefficient in dividend. 7. Repeat step 6 until all coefficients in dividend have been used.
a. Last number in the quotient is the numerator of a remainder which
is written as a fraction with the binomial as the denominator. EX: 2 3 0 —6 2 results in a quotient of 3 6 6 with remainder 14;
14
(X  2) 8. First exponent in answer (quotient) is one less than highest power
of dividend because division was by a variable to first degree. therefore; (3x3 —6x+2)+(x—2) =3x2 +6x+6+ BASICS
° DEFINITION: The real number b is thelnth root of a if P“ = a.
 RADICAL NOTATION: Ifn 96 0 then ai =35 andi‘B =ai. The symbol J—is the radical or root symbol. The a is the radicand.
The n is the index or order. 'SPECIAL NOTE: Equation a2 = 4 has two solutions, 2 and 2.
However, the radical «5 represents only the nonnegative square root of a. 'DEFINITION OF SQUARE ROOT: For any Real number a,
w/aT =la , that is, the nonnegative numerical value of a only. EX: 4] =+2 only, by definition of the square root.
R U L ES  FOR ANY REAL NUMBERS, m and n, with m/n in lowest terms
In 1 m l
and n =16 0, aF =(am)i =«n/a'“; OR a.F = (a3)“1 = (3/3)?
' FOR ANY REAL NUMBERS, m and n, with m and n, with m/n
"' _ 1 % in lowest terms and n at 0, aT —§ ' FOR ANY NONZERO REAL NUMBER 11,
(an)i =21 =a; OR (aﬁ)n =31 =a
' FOR REAL NUMBERS a and b and natural number n,
('ng)=m; OR mqfuﬁ
i.e., as long as the radical expressions have the same index 11, they
may be multiplied together and written as one radical expression of
a product OR they may be separated and written as the product of two or more radical expressions; the radicands do not have to be the
same for multiplication. 'FOR REAL NUIMBERS a and b, and natural number n,
ﬁZVE.OR:/E=E
3/5 b’ b 3/5 i.e., as long as the radical expressions have the same index they may
be written as one quotient under one radical symbol OR they may be
separated and written as one radical expression over another radical
expression; the radicands do not have to be the same for division. ' TERMS CONTAINING RADICAL EXPRESSIONS cannot be
combined unless they are like or similar terms and the radical expres
sions which they contain are the same; the indices and radicands
must be the same for addition and subtraction. EX: 3x42 +5xﬁ =8X'\/§ BUT 3y\/§ +7y«/§ cannot be combined
because the radical expressions they contain are not the same. The
terms 7mx/2 and 8m§/2 cannot be combined because the indices (plural
of index) are not the same.
SIMPLIFYING RADICAL EXPRESSIONS
' WHEN THE RADICAL EXPRESSION CONTAINS ONE
TERM AND NO FRACTIONS (EX: \112m2) then: 1. Take the greatest root of the coefficient. EX: Form usedﬁﬁ,NOTf«/§, becauseJSis not in
simplest form.
2. Take the greatest root of each variable in the term. Remember 3/217“ :31; that is, the power of the variable is divided b the index. a. This is accomplished by first noting if the power 0 the variable
in the radicand is less than the index. If it is, the radical expres
sion is in its simplest form. b. If the power of the variable is not less than the index, divide the power
by the index. The quotient is the new power of the variable to be writ
ten outside of the radical symbol. The remainder is the new power of
the variable still written inside of the radical symbol. EX: §/a7 =a2 SUE; il8ab5 =2b §/ab2
'WHEN THE RADICAL EXPRESSION CONTAINS MORE THAN ONE TERM AND NO FRACTIONS (V X2 +6x+9I then:
1. Factor, if possible, and take the root of the factors. Never take the
root of individual terms of a radicand. EX; \[Xz +4 ¢X+2, \IXz +4X+4 = (X+2)2 because the root of the factors (x + 2)2 was taken to get x + 2 as the answer.
2. If the radicand is not factorable, then the radical expression cannot be simpliﬁed because you cannot take the root of the terms of a radicand. ' WHEN THE RADICAL EXPRESSION CONTAINS FRACTIONS 1. If the fraction(s) is part of one radicand (under the radical symbol.
EXn/g) then: a. Simplify the radicand as much as possible to make the radicand one Rational expression so it can be separated into the root of the numer ator over the root of the denominator. Simplify the radical expression in
the numerator. Simplify the radical expression in the denominator. ROOTS AND RADICALS CONTINUED RADICAL EXPRESSIONS IN EQUATIONS b. Never leave a radical expression in the denominator. It is not con  RULE: Both sides of an equation may be raised to same power. Cau
sidered completely simplified until the fraction is in lowest terms. tion: Since both entire sides must be raised to the same power, place each
Rationalize the expression to remove the radical expression from the side in a separate set of parentheses first.
denominator as follows: ' STEPS: i. Step 1: Multiply the numerator and the denominatorby the radi 1. If the equation has only one radical expression EX: 4% +5 =x then:
cal expressron needed to elirninate the radical expressron from the a. Isolate the radical expression on one Side ofthe equal Sign. denominator. A radical expression in the numerator is acceptable. b. Raise both Sides ofthe equation to the same power as the index.
(Sm/2) («5) c. Solve the resulting equation. Exmmust be multiplied by (.550 the denominator becomes d. Check the solution(s) in the original equation because extraneous so
lutions are possible. EX:\/E+5=x becomes JE=x—5,then squaring both sides gives 3x = x2  10x + 25 because when the entire right side, which is a
binomial, x  5, is squared, (x  5)2, the result is x2  10x+ 25. This is AOPENA 21 with no radical symbols in it. The numerator becomes 5 “E:
ii. Step 2: Write the answer in lowest terms. (X) 2. If the fraction contains monomial radical expressions. EX: (ﬁﬁhen: a. If the radical expression is in the numerator only, simplify it and
write the fraction in lowest terms. b. If the radical expression is in the denominator only, rationalize
the fraction so no radical symbols remain there. Simplify the result
ing fraction to lowest terms. c. If radical expressions are in both numerator and denominator, either:
i. Simplify each separately, rationalize the denominator and write the answer in lowest terms, OR
ii. Make the indices on all radical symbols the same, put the numer now a second degree equation. The steps for solving a quadratic
equation should now be followed. 2. With equations containing two radical expressions; EX: «5 + 2X =4 a. Change the radical expressions to have the same index. b. Separate the radical expressions, placing one on each side of the
equal sign. c. Raise both sides of the equation to the same power as the index. d. Repeat steps b and c above until all radical expressions are eliminated. e. Solve the resulting equation and check the solution(s). . If the equation has more than two radicals; ator and the denominator under one common radical symbol, write
in lowest terms, separate again into a radical expression in the nu
merator and a radical expression in the denominator, rationalize
the denominator, and write the answer in lowest terms. 3. IF THE RADICAL EXPRESSIONS ARE PART OF POLYNOMIALS
(x he)
IN A RATIONAL EXPRESSION. EX: (3x+J5) then:
a. If the radical expressions are not in the denominator, then simpli
fy the fraction and write the answer in lowest terms. a. Change the radical expression to the same index. b. Separate as many radical expressions as possible on different sides
of the equal sign. c. Raise both sides of the equation to the power of the index. (1. Repeat steps b and (3 above until all radical expressions are eliminated. e. Solve the resulting equation and check the solution(s). QUADRATIO EOUATIONS ' DEFINITION: Seconddegree equations in one variable which can be writ
ten in the form ax2 + bx + c = 0 where a, b, and c are Real numbers and a at 0. b. If the radical expressions are in the denominator rationalize it by mul . .
   ’  _ PROPERTY: If a and b are Real numbers and (a)(b) = 0 then either a = 0 or
hplymg the numerator and the denommator by the conjugate 0f the dc b = 0 or both equal zero. At least one of the numbers has to be equal to zero. nominator. Conjugates are expressions with the middle sign changed. , ST E PS_
EX: The conjugate of 3x +542 is 3x —SJ2 because when they are
multiplied, the radical symbol is eliminated. o RATIONAL EXPRESSIONS IN EQUATIONS {  DEFINITION: Equations containing rational expressions are algebraic fractions.
 STEPS:
1. Determine least common denominator for all rational expressions in equation. 2. Use the Multi lication Pro er 0 E ualit to multi 1 all terms on both A  . —bi\/b2—4ac sides of the eguality sign b1; tIlJC) cjdmijnon dzenominatgryand thereby elimi 3" The quadratlc formUIa 15' x zT Date an algebraic erCti0n5 . t b. a, b, and c come from the seconddegree equation which is to be
3.Solveresu1t1,ng equation usmg apprepnate steps, depending on degree of solved. After the seconddegree equation has been set equal to zero, equatlon Whmh resulted from followmg Step , a is the coefficient (number in front of) of the seconddegree term, b
4. Check answers because numerical values which cause denominators of ra— is the coefficient (number in from of) of the first_degree term (if no tiona} expreSSions in original equation to be.equa1 to Zero are exuaneous firstdegree term is present then b is zero), and c is the constant term
solutions, not true solutions of original equation. (no variable showing) Note. aX2 + bx + c = 0 1. Set the equation equal to zero. Combine like terms. Write in
descending order.
2. Factor; (if factoring is not possible then go to step 3)
a. Set each factor equal to zero. See above: "If a product is equal
to zero at least one of the factors must be zero."
b. Solve each resulting equation and check the solution(s).
3. Use the quadratic formula if factoring is not possible. FENA 0. Substitute the numerical values for a, b, and c into the quadratic formula.
RADICAL OPERATIONS  ADDITION AND SUBTRACTION: Only radical expressions which have the
same index and the same radicand may be added.
 MULTIPLICATION AND DIVISION: (1. Simplify completely. e. Write the two answers, one with + in front of the radical expression, and
one with — in front of the radical expression in the formula. Complete any
additional simplification to get the answers in the required form. 1. Monomials be multi lied when the indices are the same even thou the radican
dsmo,E§f‘{3xg)&ﬁ)=6x.F35 gh COMPLEX NUMBERS 2. Binomials may be multiplied using any method for multiplying regular bi ' DEFINITION: The set of an numbers, a + bi, where a and b are Real numbers
nomial CXPYCSSiOHS ifindices are the same (eg FOILipage 2) and i2 = 1; that is, i is the number whose square equals 1 or i = J——1.
EX: (9111+ZN/g )(31‘1—5\/7 )=27m2 ‘45nh/7 “mil/g ‘10“! 35 NOTE: 1'2 = 1 will be used in multiplication and division. The complex 3. Other polynomials are multiplied using the distributive property for multi number 3 + 2i at 3  21' because the numbers must be identical to be equal.
plication if the indices are the same. 0 OPERATIONS; 4. Division may be simplification of radical expressions or multiplication by 1_ Addition and Subtraction:
the reciprocal of the divisor. Rationalize the answer so it is in lowest terms a_ Combine complex numbers as though the ' Were a variable_ without a radical expression in the denominator. EX, (4 + 5 i) + (7 _ 30 = 11 + 2i. (_3 + 7i) _ (5 _ 8i) = _8 + 15'
. ’
b. The sum or diﬂ‘erence of complex numbers is another complex number. Even
the number 21 is a complex number of the form a+ bi where a = 21 and b = 0.
001°“? 2°02 2. Multiplication and Division: 5 O 5 9 5
PRICE: "5 3595 CAN $835 a. Multiply complex numbers using the methods for multiplying two
 biriomials. Remember that i2 = 1, so the answer is not complete un ISBN l5?EEE?35 Li 9 781572 227354 Customer Hotline # 1.800.230.9522. We welcome your feed
back so we can maintain and exceed your expectations. mitigate. y.com .. Author: S. Kizlik til i2 has been replaced with 1 and simpliﬁed. EX: (3 + 51') (1  i) =
3+3i+5i5i2=3+3i+5i5(—1)=3+3i+5i+5=2+8i b. Divide complex numbers by rationalizing the denominator. The answer
is complete when there is no radical expression or i in the denominator and the answer is in simplest form. EX: The conjugate of the complex
number 3 + 12i is 312i. AOPENA 54614 20735 7 ...
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This note was uploaded on 05/23/2010 for the course IB 1405 taught by Professor Joane during the Spring '10 term at GwyneddMercy.
 Spring '10
 Joane

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