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Chapter 1 Review Topic in Algebra 1 
Chapter I 
Review Topics in Algebra 1 
 Sets of Real Numbers 
 Exponents and Radicals 
 Polynomials 
 Factoring Polynomials 
 Rational Expressions 
 Rational Exponents 
 CHAPTER TEST
1. Set of real numbers 
1.1 Real number is a set of rational numbers and the set of irrational numbers make 
up. 
If the numbers are repeating or terminating decimal they called rational number. The 
square roots of perfect squares also name rational number. 
Examples: 
1) √0.16 
2) 0.666
3) 
1 
3 
4) 
10 
9 
5) 
9 
6 
If the numbers are not repeating or terminating decimals. They called irrational number. 
For examples: 
1) π 
2) √2 
3) 0.61351 
4) √8 
5) √11 
Exercise 1.1 
Direction: Determine whether each statement is true or false. 
1. Every integer is also a real number. 
2. Every irrational number is also an irrational number. 
3. Every natural number is also a whole number. 
4. Every real number is also a rational number. 
State whether each decimal represents a rational o irrational number. 
5. √4 
6. √5 
7. 0 
8. 3 
9. 0.63586358 
10. √866 
1.1.1 Properties of real numbers 
Let us denote the set of real numbers by 푅. These properties are statement derived from 
the basic axioms of the real numbers system. Axioms are assumptions on operation with 
numbers. 
Axioms of Equality 
Let a, b, c, d ∈ R 
1. Reflexive Law 
If a=a 
2. Symmetric Law 
If b=c then c=b 
3. Transitive Law 
If b=c and c=d then b=d 
4. Additional Law of Equality 
If a=b then a+c=b+c
5. Multiplication Law of Equality 
If a=b then a.c=b.c 
Axioms for Addition and Multiplication 
Let a, b, c, d, ∈ R 
1) A. Closure property for addition 
a+b ∈ R 
Examples: 
1) 3+3=6 
2) 7+(-4)=3 
3) -8+4=-4 
B. Closure property for multiplication 
a.b ∈ R 
Examples: 
1) 3(7)=21 
2) -8(3)=-24 
3) 0.11=0 
2) A. Commutative prroperty for addition 
a+b=b+a 
Examples: 
1) 
1 
2 
+ 7 = 7 + 1 
2 
2) 0.3 + (− 5 
6 
) = − 5 
6 
+ 0.3 
3) 
1 
3 
+ 21 = 21 + 1 
3 
B. Commutative prroperty for multiplication 
a.b=b.a 
Examples: 
1) 
4 
5 
(22) = 22 (4 
5 
) 
2) 6.3=3.6 
3) 
10 
9 
(−25) = −25 (10 
9 
) 
3) A. Associative property for addition 
(a+b)+c=a+(b+c) 
Examples: 
1) (3+7)+0.4=3+(7+0.4) 
2) (0.36+89)+ 
1 
2 
= 0.36 + (89 + 1 
2 
) 
3) (3 
5 
+ 0.8) + 3 
8 
= 3 
5 
+ (0.8 + 3 
8 
) 
B. Associative property for multiplication
(a.b).c=a.(b.c) 
Examples: 
1) (3.x).y=3.(x.y) 
2) [5(7)] 1 
4 
= 5 [7 (1 
4 
)] 
3) [3푥(6푥)]]5 = 3푥[6푥(5)] 
4) Identity property for multiplication 
a.1=a 
Examples: 
1) 1.a3=a3 
2) 
3 
7 
(1) = 3 
7 
3) 3.1=3 
5) A. Inverse property for addition 
a+(-a)=0 
Examples: 
1) 6+(-6)=0 
2) 10+(-10)=0 
3) -3+3=0 
B. Inverse property for multiplication 
푎. 1 
푎 
= 1 
Examples: 
1) -2(− 1 
)=1 
2 
1 
8 
2) 8( 
)=1 
3) -6(- 
1 
6 
)=1 
6) Distributive property of multiplication over addition 
a(b+c)=ab+ac 
Examples: 
1) 3(4+6)=3(4)+3(6) 
2) -6(7+1)=-6(7)+[-6(1)] 
3) a(7+5)=7a+5 
Exercise 1.1.1 
Determine which real number property is shown by each of the following. 
1. − 1 
4 
+ 1 
4 
= 0 
2. 2(1)=2 
3. 
1 
4 
(4)=1 
4. -7+(-4)=-4+(-7) 
5. 0.3(0)=0.3
6. 5[3+(-1)]=5(3-1) 
7. (8+ 
9 
8 
9 
8 
)+0.45=8+( 
+0.45) 
8. 5(8+8)=5(8)+5(8) 
9. 6x+(8x+10)=(6x+8x)+10 
10. 5a+2b=2b+5a 
1.2 Exponents and Radicals 
In the expression 훼푛 , α is the base and 혯 is the exponent. The expression 훼푛 
means that the value α is multiplied 혯 times by itself. 
Examples: 
1) 63= 6.6.6 
=216 
2) 56= 5.5.5.5.5 
=15625 
3) 42= 4.4 
=16 
1.2.1 Integral and zero exponents 
Laws of Integral and Zero Exponents 
Theorem 1: 
For any real number α, (α≠ 0) 
푎0 = 1 
Examples: 
1) (6푎0 + 3)0=1 
2) 6α0+70=6(1)+1=7 
3) 2α0+70=2(1)+1=3 
Theorem 2: 
For any real numbers α, 
αm. α혯= αm+n 
where m and n are integers. 
Examples: 
1) α5.α4=푎5+4 = 푎9
2) 4푥푦2(2푥2푦2) = 8푥 1+2푦2+2 = 8푥 3푦4 
3) 푥 푎+3. 푥 푎+4 = 푥 2푎 +7 
Theorem 3: 
For any real numbers a+b, 
(ab)n=anbn, 
where n is any integer. 
Examples: 
1) (5x)2=55x2=25x2 
2) (-2x)3=-23x3=-8x3 
3) [x(x-3)]2=x2(x-3)2 
=x2(x2-6x+9) 
=x4-6x3+9x2 
Theorem 4: 
For any real numbers a 
(am)n=amn 
where m and n are integers. 
Examples: 
1) (-x2)3=-x2(3)=-x6 
2) [(3x+4)2]3=(3x+4)6 
3) (-x2y3z)4=-x8y12z4 
Theorem 5: 
For any real numbers a and b (b≠0), 
( 
푎 
푏 
)푛 = 
푎푛 
푏푛 
where n is any integer. 
Examples: 
1) ( 
푎2 
푏3 )2 = 푎4 
푏6 
3 
4 
2) ( 
)3 = 33 
43 = 27 
64 
푥 
푦 +2 
3) ( 
)2= 
푥2 
2 
= 푥(푦 +2)2 푦2 +4푦 +4 
Theorem 6: 
For any real numbers a(a≠0), 
푎푚 
푎푛 = 푎푚−푛 
where m and n are integers. 
Examples:
1) 
푎7 
푎5 = 푎7−5=푎2 
2) 
푥3푦4 푧5 
푥푦푧 
= 푥 3−1푦4−1 푧5−1 = 푥 2푦3푧4 
3) 
푥4푦4 
푥4푦4 = 푥 4−4푦4− 4 = 푥 0푦0 = 1(1) = 1 
Theorem 7: 
For any real numbers a(a≠0), 
푎−푛 = 
1 
푎푛 
Where n is any positive integer. 
Examples: 
1) 3푥 3푦−2= 
3푥3 
푦2 
2) (4푥 2푦)−2 = 1 
(4푥2푦)2 = 1 
8푥4 푦2 
3) (푥 2 + 푦)−2 = 1 
(푥2+푦)2 = 1 
푥4+푦2 
Exercises 1.2.1 
Simplify and express the following expressions with positive and negative 
integrals only. 
1. 50 
2. 
10푚4 
30m 
3. 
16푏4 푐 
−4푏푐3 
4. 푦3. 푦4 
5. (5푥푦)6 
6. (푎푏)3 
7. (푥 3푦2)3 
8. [(−5)2]2 
9. 
푥5푦6 
푥푦 
= 
10. 
푎7 
푎3 
1.2.2 Fractional Exponents: Radicals
Since not all numbers are integers, we can’t expect exponents to always 
whole number or zero. Exponents can be form fractional. Fractional exponents 
may seem unfamilliar for they are usually expressed as radicals. 
For expression 푥 
1 
2 is the same as √2 (read as square root of 2), and 푥 
2 
3 is 
the same as 3√푥2 
(read as cube root of x squared). The expression 푛√푎푚 is called a 
radical. The symbol √ is called a radical sign, where n is the index, a is the 
radicand and m is the power of the radicand. 
푎 
푚 
푛 
=푛√푎푚 
Laws of Radicals 
Theorem 1: 
For any real numbers a, 
√푎푛 = 푎 푛 
Examples: 
1) √42 = 4 
2) 3√(푥2푦)3=푥 2푦 
3 
3) √33 =3 
Theorem 2: 
For any real numbers a,and b. 
√푎 푛 . √푏 푛 = √푎푏 푛 
Examples: 
1) √3. √3 = √3.3 = √9=3 
2) √4. √3 = √4.3 = √12 
3) √푎. √푏 = √푎. 푏 
Theorem 3: 
For any real numbers a,and b, (b≠0) 
√푎 푛 
√푏 푛 = √ 
푎 
푏 
푛 
Examples: 
1) 3√푎 
√푏 3 = √ 
푎 
푏 
3 
2) √4 
√5 
= √4 
5 
3) 4√푥 
√푦 4 = √ 
푥 
푦 
4 
Theorem 4:
For any real numbers a , 
푚 
√푎 푚푛 = √ √푎 푛 
푛 
= √ √푎 푚 
Examples: 
3 
1) √64 6 = √√64 
= √8 3 = 2 
2 
2) √16 4 = √√16 
= √4 2 =2 
3) √100 3 = √100 3 =√100 = 10 
Theorem 5: 
For any real numbers a 
k푛√푎푘푚 = 푛√푎푚 
Examples: 
1) √24 6 
= √22.2 2.3 = 3√22 = √4 3 
2) √93 6 = 3.2√93.1 = √9 2 =3 
3) 
1.2.1 Addition and Sutraction of Radicals 
To add and subtract radicals, first we need to combine the like terms with 
similar radicals. 
Examples: 
1) √2 + 3√2 − 2√2 = 2√2 
2) √8 + √18 + √32 = √4.2 + √9.2+√16.2 = 2√2 + 3√2 + 4√2 = 9√2 
3) 푦√푥 3푦 − √푥 3푦3 + 푥√푥푦3 = 푦√푥 2. 푥푦 − √푥 2. 푥. 푦2. 푦 + 푥√푥. 푦2. 푦 = 
푥푦√푥푦 − 푥푦√푥푦 + 푥푦√푥푦 = 푥푦√푥푦 
1.2.2 Multiplication and Division of Radicals 
To multiply and divide radicals with the same index, multiply, or divide 
the radicals and copy the common index. 
Examples: 
1) √3.√3 = √32 = 3 
2) 3√푥푦. √푥 2푦 3 . √푥푧 3 =√푥푦. 푥 2푦. 푥푧 3 = √푥 4푦2푧 3 = 푥 √푥푦2푧 3 
3) √16 3 ÷ √−2 3 =√16 ÷ (−2 3 )= √−8 3 = −2 
Exercise: 1.2.2 
Simplify and solve. 
1. (5√2)(3√6)
2. (3푎3√4푥2)(4 √3푥푦 3 ) 
3. 4√ 9 
16 
4. √2(3+√3) 
5. 5√2+3√2 
6. √18 − 2√27 + 3√3 − 6√8 
7. √16푏 + √4푏 
8. 
−12√24 
3√2 
9. √8 + √50 
10. 4√푥 7푦10 
1.3 polynomials 
Polynomials was used to describe any algebraic expression. The algebraic 
expression, 5x+4 and x3+x2+1 are examples of polynomials in variable x. A polynomial 
with just one term 2x is called a monomial. If the polynomial is the sum or difference of 
two terms as in -9x+7, then it is called a binomial. If it has three terms like x2+2x+1, then 
it is called a trinomial. In general a polynomial consisting of a sum of any numbers of 
terms is called a multinomial. 
In the binomial, 5x+4 the number 5 is called the numerical coefficient of x while x 
is the literal coefficient and the numbers 4 is the constant term. 
1.3.1 Addition and Sutraction of Polynomials 
To determined the sums and differences of polynomials, only the coefficients are 
combined. By similar terms are refer to the terms with the same coefficients. Those with 
different literal coefficient are called dissimilar or unlike terms. 
Examples: 
1. Find the sum of 2x-3y+5 and x+2y-1, 
=(2x-3y+5)+( x+2y-1) 
=2x+x-3y+2y+5-1 
=3x-y++4 
2. Find the differences between 2x-3y+5 and x+2y-1
=(2x-3y+5)-( x+2y-1) 
=2x-3y+5+(-x-2y+1) 
=2x-x-3y-2y+5+1 
=x-5y+6 
3. Subtract 2(4x+2y+3) from 5(2x-3y+1) 
=5(2x-3y+1)- 2(4x+2y+3) 
=10x-15y+5-8x+4y+6 
=2x-11y+11 
1.3.2 Multiplication of Polynomials 
Examples: 
1) 푥 푚.푥 푛 = 푥 푚+푛 
2) 푥 −2.푥 2=푥 0 = 1 
3) Multiply a+2b+3c by 5m. 
= a+2b+3c(5m) in multiplication, we apply the 
=5am+10bm+15cm distributive property 
1.3.3 Division of Polynomials 
To divide a polynomial by a monomial, divide each term of the polynomial by the 
monomial. 
푥 푚 
푥 푛 = 푥 푚−푛 푎푛푑 푥 −푛 = 
1 
푥 푛 
Examples: 
1) 
푥5 
푥2 = 푥 3 
1 
푥5 
2) 푥 −5= 
3) Divide 7푥 2 − 5푥 푏푦 푥 
푥 is the divisor and 7푥 2 − 5푥 as the dividend, we have 
7푥2 −5푥 
푥 
= 
7푥2 
푥 
- 
5푥 
푥 
=7푥 − 5
Exercise: 1.3 
Simplify. 
1. (5푥 − 1) + (10푥 2 + 7푥) 
2. (20푥 2 + 2) + (15푥2 − 8) + (3푥 2 − 4) 
3. (푥 2 + 푦2 + 8) + (4푥 2 − 2푦2 − 9) 
4. (−3푥 2 + 5푦 − 4푥푦 + 푦2) 푓푟표푚(2푥2 − 4푦 + 7푥푦 − 6푦2) 
5. 2푥 2 + 6푥 + 5 푎푛푑 3푥2 − 2푥 − 1 
6. (푥 + 2)(푥2 − 2푥 + 3) 
7. 푎푏(2푎 + 1) 
푥2−3푥 −10 
8. 
푥+2 
9. 
푥6+2푥4 +6푥−9 
푥3+3 
10. (3푥 3 − 11푥 2푦 + 11푥푦2 − 2푦3) ÷ (푥 − 2푦) 
1.4 Factoring 
1.4.1. Factors and Greatest Common Denominator 
If the two of more numbers are multiplied, each number is a factor of the product. 
In the example above, 18 is expressed as the product of different pair of whole numbers. 
18=2.9 
18=3.6 
18=18.1 
A prime number is a whole number, greater than1, whose only factors are 1 and 
itself. A composite number is a whole number greater than 1, that is not prime. 
Examples: 
1) Find the prime factorization of 84. 
84=2.42 the least prime factor of 84 is 2
=2.2.24 the least prime factor of 42 is 2 
=2.2.3.7 the least prime factor of 21 is 3 
All of the factors in tha last row are prime. Thus, the prime factorization of 84 is 
2.2.3.7 or 22.3.7. 
2) Factor 20a2b 
20a2b=2.10.a.a.b 
=2.2.5.a.a.b 
The greatest common factor of two or more integer is the product of the prime 
factors common to the integers. 
Examples: 
1) Find the GCF of 54, 63, and 180. 
54=2.③.③.3 factor each number 
63③.③7 
180=2.2.③.③.5 then circle the common factors 
The GCF of 54, 63, and 180 is 3.3 or 9. 
2) 8푎2푏 푎푛푑 18푎2푏2푐 
8푎2푏=②. 2.2. 푎 ⃝ . 푎 ⃝ . 푏 ⃝ 
18푎2푏2 푐=②.3.3. 푎 ⃝ . 푎 ⃝ . 푏 ⃝. 푏. 푐 
= 2푎2푏 
The GCF of 8푎2 푏 푎푛푑 18푎2푏2푐 is 2푎2푏. 
1.4.2 Factoring Using the Distributive Property 
To multiplied a polynomial by a monomial by using the distributive property. 
Multiplying Polynomials Factoring Polynomials 
3(a+b)=3a+3b 3a+3b=3(a+b) 
x(y-z)=xy-xz xy-xz= x(y-z) 
3y(4x+2)=3y(4x)+3y(2) 12xy+6y=3y(4x)+3y(2) 
=12xy+6y =3y(4x+2) 
Examples: 
1) Use the distributive property to factor 10푦2 + 15푦
10푦2=2.⑤. 푦 ⃝. 푦 
15푦 =3.⑤.푦 
The GCF is 5y 
10푦2 + 15푦=5y(2y)+5y(3) 
=5y(2y+3) distributive property 
2) Factor 21푎푏2 − 33푎2푏푐 
21푎푏2=③.7. a ⃝ . 푏 ⃝. 푏 
33푎2푏푐=③.11. a ⃝ . 푎. 푏 .⃝c the GCF is 3ab 
21푎푏2 − 33푎2푏푐= 3ab(7b)-3ab(11ac) 
=3ab(7b-11ac) distributive property 
Exercise: 1.4.1-2 
Find the factor and GCF. 
1. 21 
2. 4, 12 
3. 63 
4. 304 
5. 18, 35 
6. 12푎푛2, 40푎4 
7. 7푦2, 14푦2 
8. 15, 10 
9. 6푎2, 18푏2, 9푏3 
10. 18푥 2푦2, 6푦2, 42푥 2푦3 
1.4.3 Factoring by Grouping 
Polynomial with four or more terms, like 3xy-21y+5x-35, can sometimes be 
factored by grouping terms of the polynomials. One method is to group the terms into 
binomials that can each be factored using the distributive property. Then use the the 
distributive property again with a binomial as the common factor. 
Examples: 
1) Factor 3xy-21y+5x-35 
3xy-21y+5x-35= (3xy-21y)+(5x-35) 
=3y(x-7)+5(x-7) 
=3y+5(x-7) 
Check by using FOIL ; 
(3y+5)(x-7)=3y(x)+3y(-7)+5(x)-5(7) 
=3xy-21y+5x-35
2) Factor 8푚2 푛 − 5푚 − 24푚푛 + 15 
8푚2 푛 − 5푚 − 24푚푛 + 15=(8푚2푛 − 5푚) + (−24푚푛 + 15 
= 푚(8푚푛 − 5) + (−3)(8푚푛 − 5) 
=푚 − 3(8푚푛 − 5) 
Check: 
푚 − 3(8푚푛 − 5) = 푚(8푚푛) + 푚(−5) + (−3)(8푚푛) + (−3)(-5) 
= 8푚2 푛 − 5푚 − 24푚푛 + 15 
1.4.4 Factoring Trinomials 
When two numbers are multiplied each number is a factor of the product. 
Similarly if two binomials are multiplied, each binomials is factor of the product. 
Consider the binomials 5x+2 and 3x+7. You can use the FOIL method to find their 
product. 
(5x+2)( 3x+7)=(5x)(3x)+(5x)(7)+(2)(3x)+(2)(7) 
=15x2+35x+6x+14 
=15x2+(35+6)x+14 
=15x2+41x+14 
You can be use this pattern to factor quadratic trinomials, such as 2푦2 + 7푦 + 6 
Factors of 12 Sum of Factors 
1.12 1+12=13 no 
2.6 2+6=8 no 
3.4 3+4=7 yes 
2푦2 + (3 + 4)푦 + 6 Select the factors 3 and 4. 
2푦2 + 3푦 + 4푦 + 6 
(2푦2 + 3푦) + (4푦 + 6) Group terms that have a 
푦(2푦 + 3) + 2(2푦 + 3) common monomials factor 
(푦 + 2)(2푦 + 3) Factor (use the distirbutive property) 
Therefore 2푦2 + 7푦 + 6= (푦 + 2)(2푦 + 3) 
Example: 
Factor 5푥 − 6 + 푥 2 
The trinomials 5푥 − 6 + 푥 2 can be written as 푥 2 + 5푥 − 6. For this 
trinomials, the constant terms is -6 and the coefficient of x is 5. Thus, we need ti 
find two factors two factors of -6 whose sum is 5. 
Factors of -6 Sum of factors 
1, -6 1+(-6)=-5 no
-1, 6 -1+6=5 yes 
Select the factors -1 and 6 
Therefore, 푥 2 + 5푥 − 6 = (푥 − 1)(푥 + 6) 
Exercise: 1.4.3-4 
Factor each trinomials. If possible. If the trinomial cannot be factored using integers. 
Write prime. 
1. 6푚푥 − 4푚 + 3푥 − 2푟 
2. 3푚푦 − 푎푏 + 푎푚 − 30푦 
3. 푎2 − 2푎푏 + 푎 − 20 
4. 3푚2 − 5푚2 푝 + 3푝2 − 5푝3 
5. 4푎푥 − 14푏푥 + 35푏푦 − 10푎푦 
6. 푦2 + 12푦 + 27 
7. 푐2 + 2푐 − 3 
8. 푥 2 − 5푥 − 24 
9. 7푎2 + 22푎 + 3 
10. 6푦2 − 11푦 + 4 
1.4.5 Factoring Differences of Square 
The product of the sum and ifference of two expressions is called the differences 
of squares. The process for finding this product can be reversed in order to factor the 
differenceof squres. Factoring the difference of square can also be modeled 
geometrically. 
푎2 − 푏2 = (푎 − 푏)(푎 + 푏) 
Examples: 
1) 푓푎푐푡표푟 푎2 − 64 
푎2 − 64 = (푎)2 − (8)2 
= (푎 − 8)(푎 + 8) 
푎. 푎 = 푎2 푎푛푑 8.8 = 64 use the difference of square 
2) 푓푎푐푡표푟 푎푥 2 − 100푦2 
푎푥 2 − 100푦2 = (3푥)2 − 10푦2 
= (3푥 − 10푦)(3푥 + 10푦) 
3푥. 3푥 = 9푥 2 푎푛푑 10푦. 10푦 = 100푦2
1.4.6 Perfect Square and Factoring 
Numbers such as 1,4,9 and 16 are called perfect squares. Since they can be expressed as 
the square of an integer. Products of the form (푎 + 푏)2 푎푛푑 (푎 − 푏)2 are called perfect 
squares and the expansions of these products are called perfect square trinomials. 
(푎 + 푏)2 = 푎2 + 2푎푏 + 푏2 
(푎 − 푏)2 = 푎2 − 2푎푏 + 푏2 
Finding a Product Factoring 
(푦 + 8)2 = 푦2 + 2(푦)(8) + 82 푦2 + 16푦 + 64 = (푦)2 + 2(푦)(8) + 
(8)2 
= 푦2 + 16푦 + 64 = (푦 + 8)2 
Examples: 
Determine whether 16푎2 + 81 − 72푎 is a perfect square trinomial. 
1) 16푎2 + 81 − 72푎 = 16푎2 − 72푎 + 81 
= (4푎)2 − 2(4푎)(푎) + (푎)2 
= (4푎 − 9)2 
2) 푥 2 + 22푥 + 121 = (푥)2 + 2(푥)(11) + (11)2 
= (푥 + 11)2 
1.4.7 Solving Equations by Factoring Zero Product Property 
For all numbers a and b, if ab=0, then a=0, b=0 or both a and b equal 0. 
Example: 
1) Solve 16t(9-t)=0 
16t(9-t)=0, then 16t=0 or 9-t zero product property 
16t=0 or 9-t=0 solve each equation 
t=0 9=t 
check: Substitute 0and 9 for t in the original.
16t(9-t)=0 
16(0)(9-0)=0 or 16(9)(9-9)=0 
0(9)=0 144(0)=0 
0=0 0=0 
SOLUTION SET: (0,9) 
2) (y+2)(3y+5)=0 
If (y+2)(3y+5)=0, then y+2=0 or 3y+5=0 
y+2=0 or 3y+5=0 
y=-2 3y=-5 
푦 = − 5 
3 
3) Check: (y+2)(3y+5)=0 
(-2+2)[(3)(-2)+5]=0 or (− 5 
3 
+ 2) [(3) (− 5 
3 
) + 5] = 0 
0(-1)=0 
1 
3 
(0) = 0 
0=0 0=0 
SOLUTION SET: (-2, − 5 
3 
) 
퐄퐱퐞퐫퐜퐢퐬퐞 ퟏ. ퟒ. ퟓ − ퟔ 
Factor. 
1. 푥 2 − 49 
2. 푥 2 − 36푦2 
3. 16푎2 − 9푏2 
4. 2푎2 − 25 
5. 2푧2 − 98 
6. 푛2 − 8푛 + 16 
7. 4푘2 − 4푘 + 1 
8. 푥 2 + 6푥 − 9 
9. 1 − 10푧 + 25푧2 
10. 50푥 240푥 + 8
1.5 Rational Expressions 
A fraction where the numerator and denominator are polynomials, and is defined 
for all values of the variable that do not make the denominator zero. 
1.5.1 Reducing Rational Expression to Lowest Terms 
We need to lowest term the fraction, if the numerator and denominator have no 
common factor. 
Examples: 
1) 
4푎2 푏푐3 
6푎푏3 푐4 = 2.2.푎.푎 .푏.푐.푐.푐 
2.3.푎.푏.푏.푏.푐.푐.푐 .푐 
= 2푎 
3푏2푐 
2) 
푥2+2푥푦 +푦2 
푥2 −푦2 = (푥+푦)(푥+푦 ) 
= 
(푥+푦)−(푥−푦 ) 
푥+ 푦 
푥− 푦 
3) 
푥3+8푦 3 
4푥+8푦 
= 푥+2푦 (푥2−2푥푦 +4푦2 
4(푥+2푦 ) 
푥2−2푥푦+4푦2 
= 
4 
1.5.2 Multiplying and Dividing Rational Expressions 
In multiplication if 
푝 
푞 
푎푛푑 푟 
푠 
are rational expressions and q and s are real numbers 
not equal to 0, then 
푝 
푞 
. 푟 
푠 
= 푝푟 
푞푠 
. 
Examples: 
1) 
4 
3 
. 1 
5 
= 4 
15
2) 
푐 
. (푎 + 2푏)(푎 − 푏) 
푎2−푏2 푐 
= 
(푎 + 푏)(푎 − 푏) 
. (푎 + 2푏)(푎 − 푏) 
푐(푎+2푏) 
= 
푎+푏 
In dividing algebraic fractions, multiply the dividend by the reciprocal of the 
divisor. The reciprocal of a fraction is its multiplicative inverse. 
Examples: 
1) 
4 
3 
÷ 
6 
5 
= 4 
3 
. 5 
6 
= 20 
18 
표푟 10 
9 
2) 
8 
7 
÷ 3 = 8 
7 
. 1 
2 
= 8 
14 
표푟 4 
7 
3) 
푦2 −16 
푦 −5 
÷ 2푦 −8 
푥푦−5푥 
(푦−4)(푦+4) 
= 
푦 −5 
. 푥(푦−5) 
2(푦−4) 
= 푥푦 +4푥 
2 
1.5.3 Adding and Subtracting Rational Expressions. 
To add and subtract rational expressions, it is the important that the least common 
denominator is accurately determined. 
Examples: 
1) 
5 
6 
− 2 
3 
+ 1 
8 
= 20−16+3 
24 
= 7 
24 
2) 
4 
5 
+ 3 
5 
+ 2 
5 
= 4+3+2 
5 
= 9 
5 
3) 3푥 − 2푦 + 2푥2 −푦2 
푥 +푦 
3푥(푥+푦)−2푦 (푥+푦)+2푥2 −푦2 
= 
푥+푦 
= 
3푥2 +3푥푦 −2푥푦 +2푦2 +2푥2 −푦2 
푥 +푦 
= 
5푥2 +푥푦− 3푦2 
푥+푦 
1.5.4 Simplifying Complex Rational Expressions 
A factor which contains one or more fractions either in the numerator or 
denominator or in both. 
Examples:
1) 
4 
31 
3 
= 4 
3 
. 3 
1 
= 12 
3 
표푟 4 
2) 
3 
2+ 
1 
3 
= 3 
6+1 
3 
= 3 
7 
3 
= 3. 3 
7 
= 9 
7 
Exercise: 1.5 
Solve and simplify. 
1. 
푎+1 
푎3 − 푎 +2 
푎2 + 푎 +3 
푎 
2. 
5푥3 
7푦4 . 21푦 2 
10푥2 
3. 
9푥5 
36푥2 
4. 
5−푎 
푎2− 25 
5. 
10푎2 −29푎 +10 
6푎2 −29푎 +10 
÷ 10푎2 −19푎 +6 
12푎 2 −28푎 +15 
6. 
1 
푥+ℎ 
− 
1 
푥 
ℎ 
7. 
푥6−7푥3 − 8 
4푥2−4푥−8 ÷ (2푥 2 + 4푥 + 8) 
8. 
푎 
푏 
− 
푏 
푎 
푎 
푏 
+ 
푏 
푎 
9. 
푡2 −2푡 −15 
푡2 −9 
. 푡2− 6푡 +9 
12−4푡 
10. 
푎−1+푏−1 
푎−2−푏−2
1.6 Rational Exponents 
We defined 푎푛 if n is any integer (positive, negative or zero). To define a 
power of a where the exponent is any rational number, not specifically an integer. 
1 
That is, we wish to attach a meaning to 푎 
⁄푛 푎푛푑 푎 
푚 
⁄푛, where the exponents are 
fractions. Before discussing fractional exponents, we give the following 
definition. 
Definition 
The 푛푡ℎ root of a real number 
If n is a positive integer greater than 
1 푎푛푑 푎 푎푛푑 푏 are real number such that 
푏푛 = 푎, then b is an 푛푡ℎ root of a. 
Examples 1: 
1) 2 is a square root of 4 because 22 = 4 
2) 3 is a fourth root of 81 because 34 = 81 
3) 4 is a cube root of 64 because 43 = 64 
Definition 
The principal 푛푡ℎ root of a real number. If n is a 
positive integer greater than 1, a is a real number, 
and √푎 푛 denotes the princial 푛푡ℎ root of a, then 
If a>0, √푎 푛 is the positive 푛푡ℎ root of a.
. 
The symbol √ is called a radical sign. The entire expression √푎 푛 is 
called a radical, where the number a is the radicand and the number n is 
the index that indicates the order of the radical. 
Examples 2: 
1) √4 = 2 
2) √81 4 = 3 
3) √64 3 = 4 
Definition 
If n is a positive integer greater than 1, and a is 
a real number, then if √푎 푛 is a real number 
⁄ 
Examples 3: 
1) 25 
1 
⁄2 = √25 = 5 
2) −8 
1 
⁄3= √−8 3 = −2 
3) ( 1 
81 
)1/4=√ 1 
81 
4 
= 1 
3 
Definition 
푎 
1 
⁄푛 = √푎 푛 
If m and n are positive integers that are 
relatively prime, and a is a real number, 
then if √푎 푛 is a real number 
푎 
푚 
⁄푛 = (√푎 푛 )m ⇔ 푎 
푚 
⁄푛 = (푎 
1 
⁄푛)m
Examples 4: 
1) 9 
3 
⁄2=(√9)3=33=27 
2) 8 
2 
⁄3 = (√8 3 )2=22 = 4 
3) −27 
4 
⁄3 = (√−27 3 )4=(-3)4=81 
It can be shown that the commutative law holds for rational 
exponents, and therefore 
(푎푚)1/n=(푎 
1 
⁄푛)m 
From which it follows that 푛√푎푚 = ( √푎 푛 )m 
The next theorem follows from this equality and the definition of 푎 
푚 
⁄푛 
Theorem 1 
If m and n are positive integrers that are 
relatively prime, and a is a real number, 
then if √푎 푛 is a real number 
Examples 5: 
Theorem 1 is applied in the following: 
1) 9 
3 
⁄2=√93=729 =27 
2) 8 
2 
⁄3 = √8 3 2=√64 3 = 4 
3) −27 
4 
⁄3 = (√−27 3 )4=√531441 3 =81 
푎 
푚 
⁄푛 = 푛√푎푚 ⇔ 푎 
푚 
⁄푛 = (푎푚)1/n
Observe that 푎 
푚 
⁄푛 can be evaluated by finding either ( √푎 푛 )m or 푛√푎푚. Compare 
example 4 and 5 and you will see the computation of ( √푎 푛 )m in example 4 is simpler than 
that for 푛√푎푚 in example 5. 
The laws of positive-integer exponents are satisfied by positive-rational exponents 
with one exception: For certain values of p and q, (ap)q≠apq for a<0. This situation arises 
in the following example. 
Examples 6: 
1) [(-9)2]1/2=811/2=9 and (-9)2(1/2)=(-9)1=-9 
Therefore [(-9)2]1/2≠(-9)2(1/2). 
2) [(-9)2]1/4=811/4=3 and (-9)2(1/4)=(-9)1/2 (not a real number) 
Therefore [(-9)2]1/4≠(-9)2(1/4). 
The problems that arise in example 6 are avoided by adopting the following rule: 
If m and n are positive even integers and a is a real number, then (푎푚)1/n=│a│m/n 
A particular case of this equality occurs when m=n. We then have (푎푛)1/n=│a│ 
(if n is a positive even integer) or, equivalently, 푛√푎푛 = │a│ (if n is even) 
If n is 2, we have √푎2 = │a│ 
Examples 7: 
1) [(-9)2]1/2=│-9│=9 
2) [(-9)2]1/4=│-9│2/4=91/2=3 
Definition 
Example:8 
1) 8 
−2 
⁄3 = 1 
8 
2 
⁄3= 
1 
( 3√8)2 = 
1 
22= 
1 
4 
2) 8 
−2 
⁄3 = (8 
−1 
⁄3)2=( 1 
1 
⁄3)2=(1 
8 
2 
1 
4 
)2= 
3) 
1 
푥 
⁄3 
푥 
1 
⁄4 =푥 
1 
⁄3. 
1 
푥 
1 
⁄4=푥 
1 
⁄3.푥 
−1 
⁄4 = 푥(1 
⁄3)−1 
⁄4 = 푥 
1 
⁄12 
If m and n positiv e integer that are 
relatively prime and a is a real number and 
a≠0, then if √푎 푛 is a real number. 
푎 
−푚 
⁄푛 = 
1 
푎 
푚 
⁄푛
Exercise 1.6 
Find the value. 
1. 푎)81 
1 
⁄2;푏)27 
1 
⁄3;푐)625 
1 
⁄4;푑)32 
1 
⁄5 
2. 푎)16 
1 
⁄2;푏)125 
1 
⁄3;푐)16 
1 
⁄4;푑)100000 
1 
⁄5 
3. 푎)푥 
−3 
⁄4.푥 
5 
⁄6.푥 
−1 
⁄3;푏) 푦 
−3 
⁄4 
푦 
3 
⁄2 )-1/9 
4. 푎)푦 
1 
⁄4.푦 
−3 
⁄2.푦 
−5 
⁄8;푏) 푥 
−3 
⁄5 
푥 
−7 
⁄10)-1/4 
5. (푥 
1 
⁄3 − 푥 
−2 
⁄3)(푥 
2 
⁄3 − 푥 
−1 
⁄3) 
6. (푎 
1 
⁄4 − 푎 
1 
⁄2)(푎 
−1 
⁄4+ 푎 
−1 
⁄2) 
7. 푎)2푦 
3 
⁄2− 3푦 
5 
⁄2;푏)5푥 
−4 
⁄3+ 4푥 
5 
⁄3 
8. 푎)6푡 
3 
⁄4 + 푡 
7 
⁄4; 푏)4푤 
4 
⁄5 − 3푤 
−6 
⁄5 
9. a)(푎3)n/3(푎3푛 )3/n; b)( 푥 
푛 
⁄2)-1/2(푥 
−1 
⁄2)-n 
10. a)(푦4)n/4(푦2푛)2/n; b)( 푡 
푛 
⁄3)-2/3(푡 
−1 
⁄3)-n
Review exercise 
Name the property that justifies each step. 
1. Simple 6a+(8b+2a) 
a. 6a+(8b+2a)=6a+(2a+8b) 
b. =(6a+2a)+8b 
c. =(6+2)a+8b 
d. =8a+8b 
2. Simplify 6푎2 + (6푎 + 푎2 ) + 9푎 
a. 6푎2 + (6푎 + 푎2 ) + 9푎 = 6푎2 + (푎2 + 6푎)+9a 
b. = (6푎2 + 푎2 ) + (6푎+9a) 
c. = (6푎2 + 1푎2) + (6푎+9a) 
d. =6 + 1(푎2) + (6 + 9)푎 
e. = 7푎2 + 15푎 
Simplify and express the following. 
1. 
푎푚+5 
푎푚−2 
2. [ 
(푥+푦)0+푎0+ 푏0 
푎+푏+푐 
]-2 
3. (푎−2 + 푦)-2 
4. (37푥+5 )(34푥−4) 
5. (9푥푦2)(4푥 3푦) 
Rational Expression (simplify) 
1. 
9푛 
63푛 
÷ 9푛
2. 
−15푚3 푛2 푝2 
−35푚2 푛5 푝 
3. 
푥+푦 
푥2−푦2 
4. 
3푚− 1 
9(푚−1)2−4 
5. 
4푚푛+6 
10푚 +8푛 
6. 
2푥2 +3푥 −5 
10푥 +25 
7. 
푥2−5푥 −24 
4푥2 −27푥 −40 
8. 
25푎2 +70푎 +49 
25푎 2 −49 
Factor each polynomial into two binomials 
1. a2+ 12a+ 27 
2. y2+ 21y+ 110 
3. n2-4n+ 4 
4.x2-12x + 20 
5. x2+ 11x -12 
Answer the following word problemsand multiple choice questions 
1. The area of a rectangle is (x2–12x + 35). If the length is (x-5),find the 
width.(hint:“x5” times“something” will give you “x2–12x + 35.”). 
2. The area of a rectangle is 3a2+ 5a–28. If the length is (a+ 4), find the width. 
3. A rectangle has an area of 3x2+ 5x –12. What factors are the length and width of 
the rectangle? 
a. (3x + 4)(x –3) 
b. (3x –4)(x + 3) 
c. (3x + 3)(x –4) 
d. (3x –3)(x + 4) 
4. The area of a certain rectangle is 5n2–6n–27. Which factors are the width and 
length of the rectangle? 
a. (5n + 3)(n –9) 
b. (5n –3)(n + 9) 
c. (5n + 9)(n –3) 
d. (5n –9)(n + 3) 
5. If the area of a certain rectangle is 6m2–2a –28, and the length is (2m + 4), what is 
the width? 
a. (3m + 7) 
b. (4m –7) 
c. (4m + 7) 
d. (3m –7)
Chapter II Equations and Inequalities
2.1 Equations 
2.1.1 Equations by Using Addition 
To sove an equation means to isolate the variable having a coefficient of 1 
on one side of the equation. By using Addition Property of Equality. 
Examples: 
1) solve r+16=-7 
r+16=-7 
r+16+(-16)=-7+(-16) add -16 to each side 
r=-23 the sum of -16 and 16 is 0 
check: 
r+16=-7 
-23+16=-7 
-7=-7 
2) x+(3.28)=-17.56 
x+(3.28)=-17.56 
x+(3.28)+(3.28)=-17.56+3.28 
x=-14.28 
check: x+(3.28)=-17.56 
-14.28+(-3.28)=-17.56 
-17.56=-17.56 
3) y+21=-7 
y+21+(-21)=-7+(-21) 
y=-28
check: y+21=-7 
-28+21=-7 
-7=-7 
2.1.2 Equations by Using Subtraction 
The property that used to subtract the same number from each side of an 
equation is called the subtraction property of equality. 
Examples: 
1) x+15=-6 
x+15-15=-6-15 
x=-21 
check: 
x+15=-6 
-21+15=-6 
-6=-6 
2) b-(-8)=23 
b+8=23 
b+8-8=23-8 
b=15 
check: 
b-(-8)=23 
15-(-8)=23 
23=23 
2.1.3. Equations by Using Multiplication and Division 
To solve the equation by using multiplication, we use the multiplication property of 
equality. 
For any numbers a,b, and c, if a=b, then ac=bc 
Eamples: 
1) 
5 
12 
= 푟 
24 
24( 5 
12 
) = ( 푟 
24 
)24 multiply each side by 24 
10 = 푟 
Check: 
5 
= 푟 
12 
24 
replace r with 10
5 
12 
= 
10 
24 
5 
12 
= 
5 
12 
2) 24=-2a 
24=-2a 
− 
1 
2 
(24) = − 
1 
2 
(2푎) 
−12=a 
푐ℎ푒푐푘: 
24 =−2a 
24 =−2a(-12) 
24 = 24 
To solve the equation by using division, we use the division property of equality. 
For any numbers a,b,c with c≠ 0, 
If a=b, then 
푎 
푐 
= 푏 
푐 
. 
Examples: 
1. -6x=11 
−6푥 
−6 
= 11 
−6 
divide each side by -6 
푥 = − 
11 
6 
Check: -6x=11 
−6(− 11 
6 
)=11 
11 = 11 
2. 4x=24 
4푥 
4 
= 
24 
4 
X=6 
Check: 4x=24 
4(6)=24 
24=24 
Exercise: 2.1 
Solve and check each equation. 
1. a-15=-32 
2. b+(-14)=6 
3. -1.43+w=0.89
4. − 5 
8 
+ 푤 = 5 
8 
5. k+(-13)=21 
6. 16-y=37 
7. t-(-16)=9 
8. (−4 1 
2 
) 푥 = 36 
9. − 3 
5 
푦 = −50 
10. 3푥 = 4 2 
3 
2.2 Appplication of Linear Equations 
In many applications of algebra, the problems are stated in words. They are called 
word problems, and they give relatiomships between known numbers and unknown 
numbers to be determined. In this section we solve word problems by using linear 
equations. There is no specific method to use. However, here are some steps that give a 
possible procedurefor you to follow. As you read through the examples, refer to these 
steps to see how they are applied. 
1. Read the problem carefully so that you understand it. To gain 
understanding, it is often helpful to make a specific axample that involves 
a similar situation in which all the quatities are known. 
2. Determine the quantities that are known and those that are unknown. Use 
a variable to represent one of the unknown quantities inthe equation you 
will obtain. When employing only one equation, as we are in this section, 
any other unknown quantities should be expressed in terms of this one 
variable. Because the variable is a number, its definition should indicate 
this fact. For instance, if the unknown quantity is a length and lengths are 
mesured in feet, then if x is a variable, x should be defined as the number 
of feet in the length or, equivalently, x feet is the length. If the unknown 
quuantity is time, and time is measured in seconds, then if t is the variable, 
t should be defined as the number of seconds in the time or, equivalently, t 
seconds is the time. 
3. Write down any numerical facts known about the variable.
4. From the information in step 3, determined two algebraic expressions for 
the same number and form an equation from them. The use of a table as 
suggested in step 3 will help you to discover equal algebraic expressions. 
5. Solve the equation you obtained in step 4. From the solution set, write a 
conclusion that answers the questions of the problem. 
6. It is important to keep in mind that the variable represents a number and 
the equation involves numbers. The units of measurement do not appear in 
the equation or its solution set. 
7. Check your results by determining whether the condition of the word 
problem are satisfied. This check is to verify the accuracy of the equation 
obtained in step 4 as well as the accuracy of its solution set. 
Example 1 
If a rectangle has a length that is 3cm less than four times its width and its 
perimeter is 9cm, what arethe dimension? 
Solution 
w: the number of centimeters in the width of the rectangle 
4w-3: the number of centimeters in the length of the rectangle 
(4w-3)cm 
w cm w cm 
(4w-3)cm 
w+(4w-3)+ w+(4w-3)=19 
10w-6=19 
10w=25 
w= 
5 
2 
4w-3=4(5 
)-3 
2 
=7 
Example 2 
A man invested part of $15,000 at 12 percent and the remainder at 8 percent. If 
his annual income from the two investments is $1456, how much does he have invested 
at each rate? 
Solution 
x: the number of dollars invested at 12 percent 
15,000-x: the number of dollars invested at 8 percent
Number of Dollar × Rate = 
Number of Dollars 
invested in 
Interest 
12 percent investment x 0.12 
0.12x 
8 percent investment 15,000-x 0.08 
0.08(15,000-x) 
0.12x+0.08(15,000-x)=1456 
0.12x+1200-0.08x=1456 
0.04x=256 
x=6400 15,000-x=15,000-6400 
=8600 
Thus the man has $6400 invested at 12 percent and $8600 at 8 percent. 
Example 3. 
A father and daughter leave home at the same time in separate automobiles. The father 
drives to his office, a distance of 24 km, and the daughter drives to school, a distance of 
28 km. They arrive at their destinations at the same time. What are their average rates, if 
the father’s average rate is 12km/hr less than his daughter’s? 
Solution: 
r: the number of kilometers per hour in the daughter’s average rate 
r-12: the number of kilometers per hour in the father’s average rate 
Number of Kilometers ÷ Number of Kilometers = number 
of hours 
In Distance per hour in rate in 
time 
Daughter 28 r 
28 
푟 
Father 24 r-12 
24 
푟−12
Equation: 
28 
= 
푟 
24 
푟 − 12 
Solve the equation by first multiplying on both sides by the LCD: 
푟(푟 − 12) 
28 
푟 
= 푟(푟 − 12) 
24 
푟 − 12 
(r-12)28=r(24) 
(r-12)7=r(6) 
7r-84=6r 
7r-6r=84 
r=84 r-12=84-12 
=72 
Therefore, the daughter’s average rate is 84km/hr and the father’s average rate is 
72km/hr. 
Exercise: 2.2 
1. The sum of two numbers is 9 and their difference is 6. What are the 
numbers? 
2. Find two numbers whose sum is 7, given that one is 3 times the other. 
3. A woman invested $25,000 in two business ventures. Last year she made 
a profit of 15 percent from the first venture but lost 5 percent from the 
second venture. If last year’s income from the two investments was 
equivalent to a return of 8 percent on the entire amount invested, how 
much had she invested in each venture? 
4. An investor wishes to realize a return of 12 percent on a total of two 
invesments.if he has $10,000 invested at 10 percent, how much additional 
money should be invested at 16 percent? 
5. Two friends, living 39 mi apart, leave their homes at the same time on 
bicycles and travel toward each other. If one person acerages 2mi/hr more 
than the other, and they meet in 1 1 
2 
ℎ푟, what is each person’s average rate 
of cycling?
Annie

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Annie

  • 1. Chapter 1 Review Topic in Algebra 1 Chapter I Review Topics in Algebra 1  Sets of Real Numbers  Exponents and Radicals  Polynomials  Factoring Polynomials  Rational Expressions  Rational Exponents  CHAPTER TEST
  • 2. 1. Set of real numbers 1.1 Real number is a set of rational numbers and the set of irrational numbers make up. If the numbers are repeating or terminating decimal they called rational number. The square roots of perfect squares also name rational number. Examples: 1) √0.16 2) 0.666
  • 3. 3) 1 3 4) 10 9 5) 9 6 If the numbers are not repeating or terminating decimals. They called irrational number. For examples: 1) π 2) √2 3) 0.61351 4) √8 5) √11 Exercise 1.1 Direction: Determine whether each statement is true or false. 1. Every integer is also a real number. 2. Every irrational number is also an irrational number. 3. Every natural number is also a whole number. 4. Every real number is also a rational number. State whether each decimal represents a rational o irrational number. 5. √4 6. √5 7. 0 8. 3 9. 0.63586358 10. √866 1.1.1 Properties of real numbers Let us denote the set of real numbers by 푅. These properties are statement derived from the basic axioms of the real numbers system. Axioms are assumptions on operation with numbers. Axioms of Equality Let a, b, c, d ∈ R 1. Reflexive Law If a=a 2. Symmetric Law If b=c then c=b 3. Transitive Law If b=c and c=d then b=d 4. Additional Law of Equality If a=b then a+c=b+c
  • 4. 5. Multiplication Law of Equality If a=b then a.c=b.c Axioms for Addition and Multiplication Let a, b, c, d, ∈ R 1) A. Closure property for addition a+b ∈ R Examples: 1) 3+3=6 2) 7+(-4)=3 3) -8+4=-4 B. Closure property for multiplication a.b ∈ R Examples: 1) 3(7)=21 2) -8(3)=-24 3) 0.11=0 2) A. Commutative prroperty for addition a+b=b+a Examples: 1) 1 2 + 7 = 7 + 1 2 2) 0.3 + (− 5 6 ) = − 5 6 + 0.3 3) 1 3 + 21 = 21 + 1 3 B. Commutative prroperty for multiplication a.b=b.a Examples: 1) 4 5 (22) = 22 (4 5 ) 2) 6.3=3.6 3) 10 9 (−25) = −25 (10 9 ) 3) A. Associative property for addition (a+b)+c=a+(b+c) Examples: 1) (3+7)+0.4=3+(7+0.4) 2) (0.36+89)+ 1 2 = 0.36 + (89 + 1 2 ) 3) (3 5 + 0.8) + 3 8 = 3 5 + (0.8 + 3 8 ) B. Associative property for multiplication
  • 5. (a.b).c=a.(b.c) Examples: 1) (3.x).y=3.(x.y) 2) [5(7)] 1 4 = 5 [7 (1 4 )] 3) [3푥(6푥)]]5 = 3푥[6푥(5)] 4) Identity property for multiplication a.1=a Examples: 1) 1.a3=a3 2) 3 7 (1) = 3 7 3) 3.1=3 5) A. Inverse property for addition a+(-a)=0 Examples: 1) 6+(-6)=0 2) 10+(-10)=0 3) -3+3=0 B. Inverse property for multiplication 푎. 1 푎 = 1 Examples: 1) -2(− 1 )=1 2 1 8 2) 8( )=1 3) -6(- 1 6 )=1 6) Distributive property of multiplication over addition a(b+c)=ab+ac Examples: 1) 3(4+6)=3(4)+3(6) 2) -6(7+1)=-6(7)+[-6(1)] 3) a(7+5)=7a+5 Exercise 1.1.1 Determine which real number property is shown by each of the following. 1. − 1 4 + 1 4 = 0 2. 2(1)=2 3. 1 4 (4)=1 4. -7+(-4)=-4+(-7) 5. 0.3(0)=0.3
  • 6. 6. 5[3+(-1)]=5(3-1) 7. (8+ 9 8 9 8 )+0.45=8+( +0.45) 8. 5(8+8)=5(8)+5(8) 9. 6x+(8x+10)=(6x+8x)+10 10. 5a+2b=2b+5a 1.2 Exponents and Radicals In the expression 훼푛 , α is the base and 혯 is the exponent. The expression 훼푛 means that the value α is multiplied 혯 times by itself. Examples: 1) 63= 6.6.6 =216 2) 56= 5.5.5.5.5 =15625 3) 42= 4.4 =16 1.2.1 Integral and zero exponents Laws of Integral and Zero Exponents Theorem 1: For any real number α, (α≠ 0) 푎0 = 1 Examples: 1) (6푎0 + 3)0=1 2) 6α0+70=6(1)+1=7 3) 2α0+70=2(1)+1=3 Theorem 2: For any real numbers α, αm. α혯= αm+n where m and n are integers. Examples: 1) α5.α4=푎5+4 = 푎9
  • 7. 2) 4푥푦2(2푥2푦2) = 8푥 1+2푦2+2 = 8푥 3푦4 3) 푥 푎+3. 푥 푎+4 = 푥 2푎 +7 Theorem 3: For any real numbers a+b, (ab)n=anbn, where n is any integer. Examples: 1) (5x)2=55x2=25x2 2) (-2x)3=-23x3=-8x3 3) [x(x-3)]2=x2(x-3)2 =x2(x2-6x+9) =x4-6x3+9x2 Theorem 4: For any real numbers a (am)n=amn where m and n are integers. Examples: 1) (-x2)3=-x2(3)=-x6 2) [(3x+4)2]3=(3x+4)6 3) (-x2y3z)4=-x8y12z4 Theorem 5: For any real numbers a and b (b≠0), ( 푎 푏 )푛 = 푎푛 푏푛 where n is any integer. Examples: 1) ( 푎2 푏3 )2 = 푎4 푏6 3 4 2) ( )3 = 33 43 = 27 64 푥 푦 +2 3) ( )2= 푥2 2 = 푥(푦 +2)2 푦2 +4푦 +4 Theorem 6: For any real numbers a(a≠0), 푎푚 푎푛 = 푎푚−푛 where m and n are integers. Examples:
  • 8. 1) 푎7 푎5 = 푎7−5=푎2 2) 푥3푦4 푧5 푥푦푧 = 푥 3−1푦4−1 푧5−1 = 푥 2푦3푧4 3) 푥4푦4 푥4푦4 = 푥 4−4푦4− 4 = 푥 0푦0 = 1(1) = 1 Theorem 7: For any real numbers a(a≠0), 푎−푛 = 1 푎푛 Where n is any positive integer. Examples: 1) 3푥 3푦−2= 3푥3 푦2 2) (4푥 2푦)−2 = 1 (4푥2푦)2 = 1 8푥4 푦2 3) (푥 2 + 푦)−2 = 1 (푥2+푦)2 = 1 푥4+푦2 Exercises 1.2.1 Simplify and express the following expressions with positive and negative integrals only. 1. 50 2. 10푚4 30m 3. 16푏4 푐 −4푏푐3 4. 푦3. 푦4 5. (5푥푦)6 6. (푎푏)3 7. (푥 3푦2)3 8. [(−5)2]2 9. 푥5푦6 푥푦 = 10. 푎7 푎3 1.2.2 Fractional Exponents: Radicals
  • 9. Since not all numbers are integers, we can’t expect exponents to always whole number or zero. Exponents can be form fractional. Fractional exponents may seem unfamilliar for they are usually expressed as radicals. For expression 푥 1 2 is the same as √2 (read as square root of 2), and 푥 2 3 is the same as 3√푥2 (read as cube root of x squared). The expression 푛√푎푚 is called a radical. The symbol √ is called a radical sign, where n is the index, a is the radicand and m is the power of the radicand. 푎 푚 푛 =푛√푎푚 Laws of Radicals Theorem 1: For any real numbers a, √푎푛 = 푎 푛 Examples: 1) √42 = 4 2) 3√(푥2푦)3=푥 2푦 3 3) √33 =3 Theorem 2: For any real numbers a,and b. √푎 푛 . √푏 푛 = √푎푏 푛 Examples: 1) √3. √3 = √3.3 = √9=3 2) √4. √3 = √4.3 = √12 3) √푎. √푏 = √푎. 푏 Theorem 3: For any real numbers a,and b, (b≠0) √푎 푛 √푏 푛 = √ 푎 푏 푛 Examples: 1) 3√푎 √푏 3 = √ 푎 푏 3 2) √4 √5 = √4 5 3) 4√푥 √푦 4 = √ 푥 푦 4 Theorem 4:
  • 10. For any real numbers a , 푚 √푎 푚푛 = √ √푎 푛 푛 = √ √푎 푚 Examples: 3 1) √64 6 = √√64 = √8 3 = 2 2 2) √16 4 = √√16 = √4 2 =2 3) √100 3 = √100 3 =√100 = 10 Theorem 5: For any real numbers a k푛√푎푘푚 = 푛√푎푚 Examples: 1) √24 6 = √22.2 2.3 = 3√22 = √4 3 2) √93 6 = 3.2√93.1 = √9 2 =3 3) 1.2.1 Addition and Sutraction of Radicals To add and subtract radicals, first we need to combine the like terms with similar radicals. Examples: 1) √2 + 3√2 − 2√2 = 2√2 2) √8 + √18 + √32 = √4.2 + √9.2+√16.2 = 2√2 + 3√2 + 4√2 = 9√2 3) 푦√푥 3푦 − √푥 3푦3 + 푥√푥푦3 = 푦√푥 2. 푥푦 − √푥 2. 푥. 푦2. 푦 + 푥√푥. 푦2. 푦 = 푥푦√푥푦 − 푥푦√푥푦 + 푥푦√푥푦 = 푥푦√푥푦 1.2.2 Multiplication and Division of Radicals To multiply and divide radicals with the same index, multiply, or divide the radicals and copy the common index. Examples: 1) √3.√3 = √32 = 3 2) 3√푥푦. √푥 2푦 3 . √푥푧 3 =√푥푦. 푥 2푦. 푥푧 3 = √푥 4푦2푧 3 = 푥 √푥푦2푧 3 3) √16 3 ÷ √−2 3 =√16 ÷ (−2 3 )= √−8 3 = −2 Exercise: 1.2.2 Simplify and solve. 1. (5√2)(3√6)
  • 11. 2. (3푎3√4푥2)(4 √3푥푦 3 ) 3. 4√ 9 16 4. √2(3+√3) 5. 5√2+3√2 6. √18 − 2√27 + 3√3 − 6√8 7. √16푏 + √4푏 8. −12√24 3√2 9. √8 + √50 10. 4√푥 7푦10 1.3 polynomials Polynomials was used to describe any algebraic expression. The algebraic expression, 5x+4 and x3+x2+1 are examples of polynomials in variable x. A polynomial with just one term 2x is called a monomial. If the polynomial is the sum or difference of two terms as in -9x+7, then it is called a binomial. If it has three terms like x2+2x+1, then it is called a trinomial. In general a polynomial consisting of a sum of any numbers of terms is called a multinomial. In the binomial, 5x+4 the number 5 is called the numerical coefficient of x while x is the literal coefficient and the numbers 4 is the constant term. 1.3.1 Addition and Sutraction of Polynomials To determined the sums and differences of polynomials, only the coefficients are combined. By similar terms are refer to the terms with the same coefficients. Those with different literal coefficient are called dissimilar or unlike terms. Examples: 1. Find the sum of 2x-3y+5 and x+2y-1, =(2x-3y+5)+( x+2y-1) =2x+x-3y+2y+5-1 =3x-y++4 2. Find the differences between 2x-3y+5 and x+2y-1
  • 12. =(2x-3y+5)-( x+2y-1) =2x-3y+5+(-x-2y+1) =2x-x-3y-2y+5+1 =x-5y+6 3. Subtract 2(4x+2y+3) from 5(2x-3y+1) =5(2x-3y+1)- 2(4x+2y+3) =10x-15y+5-8x+4y+6 =2x-11y+11 1.3.2 Multiplication of Polynomials Examples: 1) 푥 푚.푥 푛 = 푥 푚+푛 2) 푥 −2.푥 2=푥 0 = 1 3) Multiply a+2b+3c by 5m. = a+2b+3c(5m) in multiplication, we apply the =5am+10bm+15cm distributive property 1.3.3 Division of Polynomials To divide a polynomial by a monomial, divide each term of the polynomial by the monomial. 푥 푚 푥 푛 = 푥 푚−푛 푎푛푑 푥 −푛 = 1 푥 푛 Examples: 1) 푥5 푥2 = 푥 3 1 푥5 2) 푥 −5= 3) Divide 7푥 2 − 5푥 푏푦 푥 푥 is the divisor and 7푥 2 − 5푥 as the dividend, we have 7푥2 −5푥 푥 = 7푥2 푥 - 5푥 푥 =7푥 − 5
  • 13. Exercise: 1.3 Simplify. 1. (5푥 − 1) + (10푥 2 + 7푥) 2. (20푥 2 + 2) + (15푥2 − 8) + (3푥 2 − 4) 3. (푥 2 + 푦2 + 8) + (4푥 2 − 2푦2 − 9) 4. (−3푥 2 + 5푦 − 4푥푦 + 푦2) 푓푟표푚(2푥2 − 4푦 + 7푥푦 − 6푦2) 5. 2푥 2 + 6푥 + 5 푎푛푑 3푥2 − 2푥 − 1 6. (푥 + 2)(푥2 − 2푥 + 3) 7. 푎푏(2푎 + 1) 푥2−3푥 −10 8. 푥+2 9. 푥6+2푥4 +6푥−9 푥3+3 10. (3푥 3 − 11푥 2푦 + 11푥푦2 − 2푦3) ÷ (푥 − 2푦) 1.4 Factoring 1.4.1. Factors and Greatest Common Denominator If the two of more numbers are multiplied, each number is a factor of the product. In the example above, 18 is expressed as the product of different pair of whole numbers. 18=2.9 18=3.6 18=18.1 A prime number is a whole number, greater than1, whose only factors are 1 and itself. A composite number is a whole number greater than 1, that is not prime. Examples: 1) Find the prime factorization of 84. 84=2.42 the least prime factor of 84 is 2
  • 14. =2.2.24 the least prime factor of 42 is 2 =2.2.3.7 the least prime factor of 21 is 3 All of the factors in tha last row are prime. Thus, the prime factorization of 84 is 2.2.3.7 or 22.3.7. 2) Factor 20a2b 20a2b=2.10.a.a.b =2.2.5.a.a.b The greatest common factor of two or more integer is the product of the prime factors common to the integers. Examples: 1) Find the GCF of 54, 63, and 180. 54=2.③.③.3 factor each number 63③.③7 180=2.2.③.③.5 then circle the common factors The GCF of 54, 63, and 180 is 3.3 or 9. 2) 8푎2푏 푎푛푑 18푎2푏2푐 8푎2푏=②. 2.2. 푎 ⃝ . 푎 ⃝ . 푏 ⃝ 18푎2푏2 푐=②.3.3. 푎 ⃝ . 푎 ⃝ . 푏 ⃝. 푏. 푐 = 2푎2푏 The GCF of 8푎2 푏 푎푛푑 18푎2푏2푐 is 2푎2푏. 1.4.2 Factoring Using the Distributive Property To multiplied a polynomial by a monomial by using the distributive property. Multiplying Polynomials Factoring Polynomials 3(a+b)=3a+3b 3a+3b=3(a+b) x(y-z)=xy-xz xy-xz= x(y-z) 3y(4x+2)=3y(4x)+3y(2) 12xy+6y=3y(4x)+3y(2) =12xy+6y =3y(4x+2) Examples: 1) Use the distributive property to factor 10푦2 + 15푦
  • 15. 10푦2=2.⑤. 푦 ⃝. 푦 15푦 =3.⑤.푦 The GCF is 5y 10푦2 + 15푦=5y(2y)+5y(3) =5y(2y+3) distributive property 2) Factor 21푎푏2 − 33푎2푏푐 21푎푏2=③.7. a ⃝ . 푏 ⃝. 푏 33푎2푏푐=③.11. a ⃝ . 푎. 푏 .⃝c the GCF is 3ab 21푎푏2 − 33푎2푏푐= 3ab(7b)-3ab(11ac) =3ab(7b-11ac) distributive property Exercise: 1.4.1-2 Find the factor and GCF. 1. 21 2. 4, 12 3. 63 4. 304 5. 18, 35 6. 12푎푛2, 40푎4 7. 7푦2, 14푦2 8. 15, 10 9. 6푎2, 18푏2, 9푏3 10. 18푥 2푦2, 6푦2, 42푥 2푦3 1.4.3 Factoring by Grouping Polynomial with four or more terms, like 3xy-21y+5x-35, can sometimes be factored by grouping terms of the polynomials. One method is to group the terms into binomials that can each be factored using the distributive property. Then use the the distributive property again with a binomial as the common factor. Examples: 1) Factor 3xy-21y+5x-35 3xy-21y+5x-35= (3xy-21y)+(5x-35) =3y(x-7)+5(x-7) =3y+5(x-7) Check by using FOIL ; (3y+5)(x-7)=3y(x)+3y(-7)+5(x)-5(7) =3xy-21y+5x-35
  • 16. 2) Factor 8푚2 푛 − 5푚 − 24푚푛 + 15 8푚2 푛 − 5푚 − 24푚푛 + 15=(8푚2푛 − 5푚) + (−24푚푛 + 15 = 푚(8푚푛 − 5) + (−3)(8푚푛 − 5) =푚 − 3(8푚푛 − 5) Check: 푚 − 3(8푚푛 − 5) = 푚(8푚푛) + 푚(−5) + (−3)(8푚푛) + (−3)(-5) = 8푚2 푛 − 5푚 − 24푚푛 + 15 1.4.4 Factoring Trinomials When two numbers are multiplied each number is a factor of the product. Similarly if two binomials are multiplied, each binomials is factor of the product. Consider the binomials 5x+2 and 3x+7. You can use the FOIL method to find their product. (5x+2)( 3x+7)=(5x)(3x)+(5x)(7)+(2)(3x)+(2)(7) =15x2+35x+6x+14 =15x2+(35+6)x+14 =15x2+41x+14 You can be use this pattern to factor quadratic trinomials, such as 2푦2 + 7푦 + 6 Factors of 12 Sum of Factors 1.12 1+12=13 no 2.6 2+6=8 no 3.4 3+4=7 yes 2푦2 + (3 + 4)푦 + 6 Select the factors 3 and 4. 2푦2 + 3푦 + 4푦 + 6 (2푦2 + 3푦) + (4푦 + 6) Group terms that have a 푦(2푦 + 3) + 2(2푦 + 3) common monomials factor (푦 + 2)(2푦 + 3) Factor (use the distirbutive property) Therefore 2푦2 + 7푦 + 6= (푦 + 2)(2푦 + 3) Example: Factor 5푥 − 6 + 푥 2 The trinomials 5푥 − 6 + 푥 2 can be written as 푥 2 + 5푥 − 6. For this trinomials, the constant terms is -6 and the coefficient of x is 5. Thus, we need ti find two factors two factors of -6 whose sum is 5. Factors of -6 Sum of factors 1, -6 1+(-6)=-5 no
  • 17. -1, 6 -1+6=5 yes Select the factors -1 and 6 Therefore, 푥 2 + 5푥 − 6 = (푥 − 1)(푥 + 6) Exercise: 1.4.3-4 Factor each trinomials. If possible. If the trinomial cannot be factored using integers. Write prime. 1. 6푚푥 − 4푚 + 3푥 − 2푟 2. 3푚푦 − 푎푏 + 푎푚 − 30푦 3. 푎2 − 2푎푏 + 푎 − 20 4. 3푚2 − 5푚2 푝 + 3푝2 − 5푝3 5. 4푎푥 − 14푏푥 + 35푏푦 − 10푎푦 6. 푦2 + 12푦 + 27 7. 푐2 + 2푐 − 3 8. 푥 2 − 5푥 − 24 9. 7푎2 + 22푎 + 3 10. 6푦2 − 11푦 + 4 1.4.5 Factoring Differences of Square The product of the sum and ifference of two expressions is called the differences of squares. The process for finding this product can be reversed in order to factor the differenceof squres. Factoring the difference of square can also be modeled geometrically. 푎2 − 푏2 = (푎 − 푏)(푎 + 푏) Examples: 1) 푓푎푐푡표푟 푎2 − 64 푎2 − 64 = (푎)2 − (8)2 = (푎 − 8)(푎 + 8) 푎. 푎 = 푎2 푎푛푑 8.8 = 64 use the difference of square 2) 푓푎푐푡표푟 푎푥 2 − 100푦2 푎푥 2 − 100푦2 = (3푥)2 − 10푦2 = (3푥 − 10푦)(3푥 + 10푦) 3푥. 3푥 = 9푥 2 푎푛푑 10푦. 10푦 = 100푦2
  • 18. 1.4.6 Perfect Square and Factoring Numbers such as 1,4,9 and 16 are called perfect squares. Since they can be expressed as the square of an integer. Products of the form (푎 + 푏)2 푎푛푑 (푎 − 푏)2 are called perfect squares and the expansions of these products are called perfect square trinomials. (푎 + 푏)2 = 푎2 + 2푎푏 + 푏2 (푎 − 푏)2 = 푎2 − 2푎푏 + 푏2 Finding a Product Factoring (푦 + 8)2 = 푦2 + 2(푦)(8) + 82 푦2 + 16푦 + 64 = (푦)2 + 2(푦)(8) + (8)2 = 푦2 + 16푦 + 64 = (푦 + 8)2 Examples: Determine whether 16푎2 + 81 − 72푎 is a perfect square trinomial. 1) 16푎2 + 81 − 72푎 = 16푎2 − 72푎 + 81 = (4푎)2 − 2(4푎)(푎) + (푎)2 = (4푎 − 9)2 2) 푥 2 + 22푥 + 121 = (푥)2 + 2(푥)(11) + (11)2 = (푥 + 11)2 1.4.7 Solving Equations by Factoring Zero Product Property For all numbers a and b, if ab=0, then a=0, b=0 or both a and b equal 0. Example: 1) Solve 16t(9-t)=0 16t(9-t)=0, then 16t=0 or 9-t zero product property 16t=0 or 9-t=0 solve each equation t=0 9=t check: Substitute 0and 9 for t in the original.
  • 19. 16t(9-t)=0 16(0)(9-0)=0 or 16(9)(9-9)=0 0(9)=0 144(0)=0 0=0 0=0 SOLUTION SET: (0,9) 2) (y+2)(3y+5)=0 If (y+2)(3y+5)=0, then y+2=0 or 3y+5=0 y+2=0 or 3y+5=0 y=-2 3y=-5 푦 = − 5 3 3) Check: (y+2)(3y+5)=0 (-2+2)[(3)(-2)+5]=0 or (− 5 3 + 2) [(3) (− 5 3 ) + 5] = 0 0(-1)=0 1 3 (0) = 0 0=0 0=0 SOLUTION SET: (-2, − 5 3 ) 퐄퐱퐞퐫퐜퐢퐬퐞 ퟏ. ퟒ. ퟓ − ퟔ Factor. 1. 푥 2 − 49 2. 푥 2 − 36푦2 3. 16푎2 − 9푏2 4. 2푎2 − 25 5. 2푧2 − 98 6. 푛2 − 8푛 + 16 7. 4푘2 − 4푘 + 1 8. 푥 2 + 6푥 − 9 9. 1 − 10푧 + 25푧2 10. 50푥 240푥 + 8
  • 20. 1.5 Rational Expressions A fraction where the numerator and denominator are polynomials, and is defined for all values of the variable that do not make the denominator zero. 1.5.1 Reducing Rational Expression to Lowest Terms We need to lowest term the fraction, if the numerator and denominator have no common factor. Examples: 1) 4푎2 푏푐3 6푎푏3 푐4 = 2.2.푎.푎 .푏.푐.푐.푐 2.3.푎.푏.푏.푏.푐.푐.푐 .푐 = 2푎 3푏2푐 2) 푥2+2푥푦 +푦2 푥2 −푦2 = (푥+푦)(푥+푦 ) = (푥+푦)−(푥−푦 ) 푥+ 푦 푥− 푦 3) 푥3+8푦 3 4푥+8푦 = 푥+2푦 (푥2−2푥푦 +4푦2 4(푥+2푦 ) 푥2−2푥푦+4푦2 = 4 1.5.2 Multiplying and Dividing Rational Expressions In multiplication if 푝 푞 푎푛푑 푟 푠 are rational expressions and q and s are real numbers not equal to 0, then 푝 푞 . 푟 푠 = 푝푟 푞푠 . Examples: 1) 4 3 . 1 5 = 4 15
  • 21. 2) 푐 . (푎 + 2푏)(푎 − 푏) 푎2−푏2 푐 = (푎 + 푏)(푎 − 푏) . (푎 + 2푏)(푎 − 푏) 푐(푎+2푏) = 푎+푏 In dividing algebraic fractions, multiply the dividend by the reciprocal of the divisor. The reciprocal of a fraction is its multiplicative inverse. Examples: 1) 4 3 ÷ 6 5 = 4 3 . 5 6 = 20 18 표푟 10 9 2) 8 7 ÷ 3 = 8 7 . 1 2 = 8 14 표푟 4 7 3) 푦2 −16 푦 −5 ÷ 2푦 −8 푥푦−5푥 (푦−4)(푦+4) = 푦 −5 . 푥(푦−5) 2(푦−4) = 푥푦 +4푥 2 1.5.3 Adding and Subtracting Rational Expressions. To add and subtract rational expressions, it is the important that the least common denominator is accurately determined. Examples: 1) 5 6 − 2 3 + 1 8 = 20−16+3 24 = 7 24 2) 4 5 + 3 5 + 2 5 = 4+3+2 5 = 9 5 3) 3푥 − 2푦 + 2푥2 −푦2 푥 +푦 3푥(푥+푦)−2푦 (푥+푦)+2푥2 −푦2 = 푥+푦 = 3푥2 +3푥푦 −2푥푦 +2푦2 +2푥2 −푦2 푥 +푦 = 5푥2 +푥푦− 3푦2 푥+푦 1.5.4 Simplifying Complex Rational Expressions A factor which contains one or more fractions either in the numerator or denominator or in both. Examples:
  • 22. 1) 4 31 3 = 4 3 . 3 1 = 12 3 표푟 4 2) 3 2+ 1 3 = 3 6+1 3 = 3 7 3 = 3. 3 7 = 9 7 Exercise: 1.5 Solve and simplify. 1. 푎+1 푎3 − 푎 +2 푎2 + 푎 +3 푎 2. 5푥3 7푦4 . 21푦 2 10푥2 3. 9푥5 36푥2 4. 5−푎 푎2− 25 5. 10푎2 −29푎 +10 6푎2 −29푎 +10 ÷ 10푎2 −19푎 +6 12푎 2 −28푎 +15 6. 1 푥+ℎ − 1 푥 ℎ 7. 푥6−7푥3 − 8 4푥2−4푥−8 ÷ (2푥 2 + 4푥 + 8) 8. 푎 푏 − 푏 푎 푎 푏 + 푏 푎 9. 푡2 −2푡 −15 푡2 −9 . 푡2− 6푡 +9 12−4푡 10. 푎−1+푏−1 푎−2−푏−2
  • 23. 1.6 Rational Exponents We defined 푎푛 if n is any integer (positive, negative or zero). To define a power of a where the exponent is any rational number, not specifically an integer. 1 That is, we wish to attach a meaning to 푎 ⁄푛 푎푛푑 푎 푚 ⁄푛, where the exponents are fractions. Before discussing fractional exponents, we give the following definition. Definition The 푛푡ℎ root of a real number If n is a positive integer greater than 1 푎푛푑 푎 푎푛푑 푏 are real number such that 푏푛 = 푎, then b is an 푛푡ℎ root of a. Examples 1: 1) 2 is a square root of 4 because 22 = 4 2) 3 is a fourth root of 81 because 34 = 81 3) 4 is a cube root of 64 because 43 = 64 Definition The principal 푛푡ℎ root of a real number. If n is a positive integer greater than 1, a is a real number, and √푎 푛 denotes the princial 푛푡ℎ root of a, then If a>0, √푎 푛 is the positive 푛푡ℎ root of a.
  • 24. . The symbol √ is called a radical sign. The entire expression √푎 푛 is called a radical, where the number a is the radicand and the number n is the index that indicates the order of the radical. Examples 2: 1) √4 = 2 2) √81 4 = 3 3) √64 3 = 4 Definition If n is a positive integer greater than 1, and a is a real number, then if √푎 푛 is a real number ⁄ Examples 3: 1) 25 1 ⁄2 = √25 = 5 2) −8 1 ⁄3= √−8 3 = −2 3) ( 1 81 )1/4=√ 1 81 4 = 1 3 Definition 푎 1 ⁄푛 = √푎 푛 If m and n are positive integers that are relatively prime, and a is a real number, then if √푎 푛 is a real number 푎 푚 ⁄푛 = (√푎 푛 )m ⇔ 푎 푚 ⁄푛 = (푎 1 ⁄푛)m
  • 25. Examples 4: 1) 9 3 ⁄2=(√9)3=33=27 2) 8 2 ⁄3 = (√8 3 )2=22 = 4 3) −27 4 ⁄3 = (√−27 3 )4=(-3)4=81 It can be shown that the commutative law holds for rational exponents, and therefore (푎푚)1/n=(푎 1 ⁄푛)m From which it follows that 푛√푎푚 = ( √푎 푛 )m The next theorem follows from this equality and the definition of 푎 푚 ⁄푛 Theorem 1 If m and n are positive integrers that are relatively prime, and a is a real number, then if √푎 푛 is a real number Examples 5: Theorem 1 is applied in the following: 1) 9 3 ⁄2=√93=729 =27 2) 8 2 ⁄3 = √8 3 2=√64 3 = 4 3) −27 4 ⁄3 = (√−27 3 )4=√531441 3 =81 푎 푚 ⁄푛 = 푛√푎푚 ⇔ 푎 푚 ⁄푛 = (푎푚)1/n
  • 26. Observe that 푎 푚 ⁄푛 can be evaluated by finding either ( √푎 푛 )m or 푛√푎푚. Compare example 4 and 5 and you will see the computation of ( √푎 푛 )m in example 4 is simpler than that for 푛√푎푚 in example 5. The laws of positive-integer exponents are satisfied by positive-rational exponents with one exception: For certain values of p and q, (ap)q≠apq for a<0. This situation arises in the following example. Examples 6: 1) [(-9)2]1/2=811/2=9 and (-9)2(1/2)=(-9)1=-9 Therefore [(-9)2]1/2≠(-9)2(1/2). 2) [(-9)2]1/4=811/4=3 and (-9)2(1/4)=(-9)1/2 (not a real number) Therefore [(-9)2]1/4≠(-9)2(1/4). The problems that arise in example 6 are avoided by adopting the following rule: If m and n are positive even integers and a is a real number, then (푎푚)1/n=│a│m/n A particular case of this equality occurs when m=n. We then have (푎푛)1/n=│a│ (if n is a positive even integer) or, equivalently, 푛√푎푛 = │a│ (if n is even) If n is 2, we have √푎2 = │a│ Examples 7: 1) [(-9)2]1/2=│-9│=9 2) [(-9)2]1/4=│-9│2/4=91/2=3 Definition Example:8 1) 8 −2 ⁄3 = 1 8 2 ⁄3= 1 ( 3√8)2 = 1 22= 1 4 2) 8 −2 ⁄3 = (8 −1 ⁄3)2=( 1 1 ⁄3)2=(1 8 2 1 4 )2= 3) 1 푥 ⁄3 푥 1 ⁄4 =푥 1 ⁄3. 1 푥 1 ⁄4=푥 1 ⁄3.푥 −1 ⁄4 = 푥(1 ⁄3)−1 ⁄4 = 푥 1 ⁄12 If m and n positiv e integer that are relatively prime and a is a real number and a≠0, then if √푎 푛 is a real number. 푎 −푚 ⁄푛 = 1 푎 푚 ⁄푛
  • 27. Exercise 1.6 Find the value. 1. 푎)81 1 ⁄2;푏)27 1 ⁄3;푐)625 1 ⁄4;푑)32 1 ⁄5 2. 푎)16 1 ⁄2;푏)125 1 ⁄3;푐)16 1 ⁄4;푑)100000 1 ⁄5 3. 푎)푥 −3 ⁄4.푥 5 ⁄6.푥 −1 ⁄3;푏) 푦 −3 ⁄4 푦 3 ⁄2 )-1/9 4. 푎)푦 1 ⁄4.푦 −3 ⁄2.푦 −5 ⁄8;푏) 푥 −3 ⁄5 푥 −7 ⁄10)-1/4 5. (푥 1 ⁄3 − 푥 −2 ⁄3)(푥 2 ⁄3 − 푥 −1 ⁄3) 6. (푎 1 ⁄4 − 푎 1 ⁄2)(푎 −1 ⁄4+ 푎 −1 ⁄2) 7. 푎)2푦 3 ⁄2− 3푦 5 ⁄2;푏)5푥 −4 ⁄3+ 4푥 5 ⁄3 8. 푎)6푡 3 ⁄4 + 푡 7 ⁄4; 푏)4푤 4 ⁄5 − 3푤 −6 ⁄5 9. a)(푎3)n/3(푎3푛 )3/n; b)( 푥 푛 ⁄2)-1/2(푥 −1 ⁄2)-n 10. a)(푦4)n/4(푦2푛)2/n; b)( 푡 푛 ⁄3)-2/3(푡 −1 ⁄3)-n
  • 28. Review exercise Name the property that justifies each step. 1. Simple 6a+(8b+2a) a. 6a+(8b+2a)=6a+(2a+8b) b. =(6a+2a)+8b c. =(6+2)a+8b d. =8a+8b 2. Simplify 6푎2 + (6푎 + 푎2 ) + 9푎 a. 6푎2 + (6푎 + 푎2 ) + 9푎 = 6푎2 + (푎2 + 6푎)+9a b. = (6푎2 + 푎2 ) + (6푎+9a) c. = (6푎2 + 1푎2) + (6푎+9a) d. =6 + 1(푎2) + (6 + 9)푎 e. = 7푎2 + 15푎 Simplify and express the following. 1. 푎푚+5 푎푚−2 2. [ (푥+푦)0+푎0+ 푏0 푎+푏+푐 ]-2 3. (푎−2 + 푦)-2 4. (37푥+5 )(34푥−4) 5. (9푥푦2)(4푥 3푦) Rational Expression (simplify) 1. 9푛 63푛 ÷ 9푛
  • 29. 2. −15푚3 푛2 푝2 −35푚2 푛5 푝 3. 푥+푦 푥2−푦2 4. 3푚− 1 9(푚−1)2−4 5. 4푚푛+6 10푚 +8푛 6. 2푥2 +3푥 −5 10푥 +25 7. 푥2−5푥 −24 4푥2 −27푥 −40 8. 25푎2 +70푎 +49 25푎 2 −49 Factor each polynomial into two binomials 1. a2+ 12a+ 27 2. y2+ 21y+ 110 3. n2-4n+ 4 4.x2-12x + 20 5. x2+ 11x -12 Answer the following word problemsand multiple choice questions 1. The area of a rectangle is (x2–12x + 35). If the length is (x-5),find the width.(hint:“x5” times“something” will give you “x2–12x + 35.”). 2. The area of a rectangle is 3a2+ 5a–28. If the length is (a+ 4), find the width. 3. A rectangle has an area of 3x2+ 5x –12. What factors are the length and width of the rectangle? a. (3x + 4)(x –3) b. (3x –4)(x + 3) c. (3x + 3)(x –4) d. (3x –3)(x + 4) 4. The area of a certain rectangle is 5n2–6n–27. Which factors are the width and length of the rectangle? a. (5n + 3)(n –9) b. (5n –3)(n + 9) c. (5n + 9)(n –3) d. (5n –9)(n + 3) 5. If the area of a certain rectangle is 6m2–2a –28, and the length is (2m + 4), what is the width? a. (3m + 7) b. (4m –7) c. (4m + 7) d. (3m –7)
  • 30. Chapter II Equations and Inequalities
  • 31. 2.1 Equations 2.1.1 Equations by Using Addition To sove an equation means to isolate the variable having a coefficient of 1 on one side of the equation. By using Addition Property of Equality. Examples: 1) solve r+16=-7 r+16=-7 r+16+(-16)=-7+(-16) add -16 to each side r=-23 the sum of -16 and 16 is 0 check: r+16=-7 -23+16=-7 -7=-7 2) x+(3.28)=-17.56 x+(3.28)=-17.56 x+(3.28)+(3.28)=-17.56+3.28 x=-14.28 check: x+(3.28)=-17.56 -14.28+(-3.28)=-17.56 -17.56=-17.56 3) y+21=-7 y+21+(-21)=-7+(-21) y=-28
  • 32. check: y+21=-7 -28+21=-7 -7=-7 2.1.2 Equations by Using Subtraction The property that used to subtract the same number from each side of an equation is called the subtraction property of equality. Examples: 1) x+15=-6 x+15-15=-6-15 x=-21 check: x+15=-6 -21+15=-6 -6=-6 2) b-(-8)=23 b+8=23 b+8-8=23-8 b=15 check: b-(-8)=23 15-(-8)=23 23=23 2.1.3. Equations by Using Multiplication and Division To solve the equation by using multiplication, we use the multiplication property of equality. For any numbers a,b, and c, if a=b, then ac=bc Eamples: 1) 5 12 = 푟 24 24( 5 12 ) = ( 푟 24 )24 multiply each side by 24 10 = 푟 Check: 5 = 푟 12 24 replace r with 10
  • 33. 5 12 = 10 24 5 12 = 5 12 2) 24=-2a 24=-2a − 1 2 (24) = − 1 2 (2푎) −12=a 푐ℎ푒푐푘: 24 =−2a 24 =−2a(-12) 24 = 24 To solve the equation by using division, we use the division property of equality. For any numbers a,b,c with c≠ 0, If a=b, then 푎 푐 = 푏 푐 . Examples: 1. -6x=11 −6푥 −6 = 11 −6 divide each side by -6 푥 = − 11 6 Check: -6x=11 −6(− 11 6 )=11 11 = 11 2. 4x=24 4푥 4 = 24 4 X=6 Check: 4x=24 4(6)=24 24=24 Exercise: 2.1 Solve and check each equation. 1. a-15=-32 2. b+(-14)=6 3. -1.43+w=0.89
  • 34. 4. − 5 8 + 푤 = 5 8 5. k+(-13)=21 6. 16-y=37 7. t-(-16)=9 8. (−4 1 2 ) 푥 = 36 9. − 3 5 푦 = −50 10. 3푥 = 4 2 3 2.2 Appplication of Linear Equations In many applications of algebra, the problems are stated in words. They are called word problems, and they give relatiomships between known numbers and unknown numbers to be determined. In this section we solve word problems by using linear equations. There is no specific method to use. However, here are some steps that give a possible procedurefor you to follow. As you read through the examples, refer to these steps to see how they are applied. 1. Read the problem carefully so that you understand it. To gain understanding, it is often helpful to make a specific axample that involves a similar situation in which all the quatities are known. 2. Determine the quantities that are known and those that are unknown. Use a variable to represent one of the unknown quantities inthe equation you will obtain. When employing only one equation, as we are in this section, any other unknown quantities should be expressed in terms of this one variable. Because the variable is a number, its definition should indicate this fact. For instance, if the unknown quantity is a length and lengths are mesured in feet, then if x is a variable, x should be defined as the number of feet in the length or, equivalently, x feet is the length. If the unknown quuantity is time, and time is measured in seconds, then if t is the variable, t should be defined as the number of seconds in the time or, equivalently, t seconds is the time. 3. Write down any numerical facts known about the variable.
  • 35. 4. From the information in step 3, determined two algebraic expressions for the same number and form an equation from them. The use of a table as suggested in step 3 will help you to discover equal algebraic expressions. 5. Solve the equation you obtained in step 4. From the solution set, write a conclusion that answers the questions of the problem. 6. It is important to keep in mind that the variable represents a number and the equation involves numbers. The units of measurement do not appear in the equation or its solution set. 7. Check your results by determining whether the condition of the word problem are satisfied. This check is to verify the accuracy of the equation obtained in step 4 as well as the accuracy of its solution set. Example 1 If a rectangle has a length that is 3cm less than four times its width and its perimeter is 9cm, what arethe dimension? Solution w: the number of centimeters in the width of the rectangle 4w-3: the number of centimeters in the length of the rectangle (4w-3)cm w cm w cm (4w-3)cm w+(4w-3)+ w+(4w-3)=19 10w-6=19 10w=25 w= 5 2 4w-3=4(5 )-3 2 =7 Example 2 A man invested part of $15,000 at 12 percent and the remainder at 8 percent. If his annual income from the two investments is $1456, how much does he have invested at each rate? Solution x: the number of dollars invested at 12 percent 15,000-x: the number of dollars invested at 8 percent
  • 36. Number of Dollar × Rate = Number of Dollars invested in Interest 12 percent investment x 0.12 0.12x 8 percent investment 15,000-x 0.08 0.08(15,000-x) 0.12x+0.08(15,000-x)=1456 0.12x+1200-0.08x=1456 0.04x=256 x=6400 15,000-x=15,000-6400 =8600 Thus the man has $6400 invested at 12 percent and $8600 at 8 percent. Example 3. A father and daughter leave home at the same time in separate automobiles. The father drives to his office, a distance of 24 km, and the daughter drives to school, a distance of 28 km. They arrive at their destinations at the same time. What are their average rates, if the father’s average rate is 12km/hr less than his daughter’s? Solution: r: the number of kilometers per hour in the daughter’s average rate r-12: the number of kilometers per hour in the father’s average rate Number of Kilometers ÷ Number of Kilometers = number of hours In Distance per hour in rate in time Daughter 28 r 28 푟 Father 24 r-12 24 푟−12
  • 37. Equation: 28 = 푟 24 푟 − 12 Solve the equation by first multiplying on both sides by the LCD: 푟(푟 − 12) 28 푟 = 푟(푟 − 12) 24 푟 − 12 (r-12)28=r(24) (r-12)7=r(6) 7r-84=6r 7r-6r=84 r=84 r-12=84-12 =72 Therefore, the daughter’s average rate is 84km/hr and the father’s average rate is 72km/hr. Exercise: 2.2 1. The sum of two numbers is 9 and their difference is 6. What are the numbers? 2. Find two numbers whose sum is 7, given that one is 3 times the other. 3. A woman invested $25,000 in two business ventures. Last year she made a profit of 15 percent from the first venture but lost 5 percent from the second venture. If last year’s income from the two investments was equivalent to a return of 8 percent on the entire amount invested, how much had she invested in each venture? 4. An investor wishes to realize a return of 12 percent on a total of two invesments.if he has $10,000 invested at 10 percent, how much additional money should be invested at 16 percent? 5. Two friends, living 39 mi apart, leave their homes at the same time on bicycles and travel toward each other. If one person acerages 2mi/hr more than the other, and they meet in 1 1 2 ℎ푟, what is each person’s average rate of cycling?