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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 𝑅. 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) [3𝑥(6𝑥)]]5 = 3𝑥[6𝑥(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
𝑎.1
𝑎= 1
Examples:
1) -2(−1
2)=1
2) 8(1
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)+0.45=8+(
9
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 𝛼𝑛 , α is the base and 𝘯 is the exponent. The expression 𝛼𝑛
means that the value α is multiplied 𝘯 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)
𝑎0 = 1
Examples:
1) (6𝑎0 + 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. α𝘯= αm+n
where m and n are integers.
Examples:
1) α5.α4=𝑎5+4 = 𝑎9
2) 4𝑥𝑦2(2𝑥2𝑦2) = 8𝑥 1+2𝑦2+2 = 8𝑥 3𝑦4
3) 𝑥 𝑎+3. 𝑥 𝑎+4 = 𝑥 2𝑎+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),
(𝑎
𝑏)𝑛 =
𝑎𝑛
𝑏𝑛
where n is any integer.
Examples:
1) (𝑎2
𝑏3 )2 =𝑎4
𝑏6
2) (3
4)3 =
33
43 =27
64
3) (𝑥
𝑦+2)2=
𝑥2
(𝑦+2)2 =𝑥2
𝑦2 +4𝑦+4
Theorem 6:
For any real numbers a(a≠0),
𝑎𝑚
𝑎𝑛= 𝑎𝑚−𝑛
where m and n are integers.
Examples:
1) 𝑎7
𝑎5 = 𝑎7−5 =𝑎2
2) 𝑥3𝑦4 𝑧5
𝑥𝑦𝑧= 𝑥 3−1𝑦4−1 𝑧5−1 = 𝑥 2𝑦3𝑧4
3) 𝑥4𝑦4
𝑥4𝑦4 = 𝑥 4−4𝑦4−4 = 𝑥0𝑦0 = 1(1) = 1
Theorem 7:
For any real numbers a(a≠0),
𝑎−𝑛 =1
𝑎𝑛
Where n is any positive integer.
Examples:
1) 3𝑥 3𝑦−2=3𝑥3
𝑦2
2) (4𝑥 2𝑦)−2 =1
(4𝑥2𝑦)2 =1
8𝑥4 𝑦2
3) (𝑥 2 + 𝑦)−2 =1
(𝑥2+𝑦)2 =1
𝑥4+𝑦2
Exercises 1.2.1
Simplify and express the following expressions with positive and negative
integrals only.
1. 50
2. 10𝑚4
30m
3. 16𝑏4 𝑐
−4𝑏𝑐3
4. 𝑦3. 𝑦4
5. (5𝑥𝑦)6
6. (𝑎𝑏)3
7. (𝑥 3𝑦2)3
8. [(−5)2]2
9. 𝑥5𝑦6
𝑥𝑦=
10. 𝑎7
𝑎3
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 𝑥1
2 is the same as √2 (read as square root of 2), and 𝑥2
3 is
the same as √𝑥23 (read as cube root of x squared). The expression √𝑎𝑚𝑛
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.
𝑎𝑚𝑛
= √𝑎𝑚𝑛
Laws of Radicals
Theorem 1:
For any real numbers a,
√𝑎𝑛 = 𝑎𝑛
Examples:
1) √42 = 4
2) √(𝑥 2𝑦)33 =𝑥 2𝑦
3) √333=3
Theorem 2:
For any real numbers a,and b.
√𝑎𝑛 . √𝑏𝑛= √𝑎𝑏𝑛
Examples:
1) √3.√3 = √3.3 = √9=3
2) √4.√3 = √4.3 = √12
3) √𝑎. √𝑏 = √𝑎. 𝑏
Theorem 3:
For any real numbers a,and b, (b≠0)
√𝑎𝑛
√𝑏𝑛= √
𝑎
𝑏
𝑛
Examples:
1) √𝑎3
√𝑏3 = √
𝑎
𝑏
3
2) √4
√5= √
4
5
3) √𝑥4
√𝑦4 = √
𝑥
𝑦
4
Theorem 4:
For any real numbers a ,
√𝑎𝑚𝑛 = √ √𝑎𝑛𝑚
= √ √𝑎𝑚𝑛
Examples:
1) √646
= √√643
= √83
= 2
2) √164 = √√162
= √42 =2
3) √1003 = √1003=√100 = 10
Theorem 5:
For any real numbers a
k √𝑎𝑘𝑚𝑛= √𝑎𝑚𝑛
Examples:
1) √246= √22.22.3
= √223= √43
2) √936= √93.13.2
= √92
=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) 𝑦√𝑥 3𝑦 − √𝑥 3𝑦3 + 𝑥√𝑥𝑦3 = 𝑦√𝑥 2.𝑥𝑦 − √𝑥 2. 𝑥. 𝑦2.𝑦 + 𝑥√𝑥. 𝑦2. 𝑦 =
𝑥𝑦√𝑥𝑦 − 𝑥𝑦√𝑥𝑦 + 𝑥𝑦√𝑥𝑦 = 𝑥𝑦√𝑥𝑦
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) √𝑥𝑦3 . √𝑥2𝑦3 . √𝑥𝑧3 = √𝑥𝑦.𝑥 2𝑦. 𝑥𝑧3 = √𝑥 4𝑦2𝑧3 = 𝑥 √𝑥𝑦2𝑧3
3) √163
÷ √−23
= √16 ÷ (−23 )= √−83
= −2
Exercise: 1.2.2
Simplify and solve.
1. (5√2)(3√6)
2. (3𝑎 √4𝑥 23)(4 √3𝑥𝑦3 )
3. 4√9
16
4. √2(3+√3)
5. 5√2+3√2
6. √18 − 2√27 + 3√3 − 6√8
7. √16𝑏 + √4𝑏
8. −12√24
3√2
9. √8 + √50
10. 4√𝑥 7𝑦10
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) 𝑥 𝑚.𝑥 𝑛 = 𝑥 𝑚+𝑛
2) 𝑥 −2.𝑥 2=𝑥 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.
𝑥 𝑚
𝑥𝑛= 𝑥 𝑚−𝑛 𝑎𝑛𝑑 𝑥 −𝑛 =
1
𝑥 𝑛
Examples:
1) 𝑥5
𝑥2 = 𝑥 3
2) 𝑥 −5=1
𝑥5
3) Divide 7𝑥 2 − 5𝑥 𝑏𝑦 𝑥
𝑥 is the divisor and 7𝑥 2 − 5𝑥 as the dividend, we have
7𝑥2 −5𝑥
𝑥=
7𝑥2
𝑥-
5𝑥
𝑥=7𝑥 − 5
Exercise: 1.3
Simplify.
1. (5𝑥 − 1) + (10𝑥 2 + 7𝑥)
2. (20𝑥2 + 2) + (15𝑥2 − 8) + (3𝑥 2 − 4)
3. (𝑥 2 + 𝑦2 + 8) + (4𝑥 2 − 2𝑦2 − 9)
4. (−3𝑥 2 + 5𝑦 − 4𝑥𝑦 + 𝑦2) 𝑓𝑟𝑜𝑚(2𝑥2 − 4𝑦 + 7𝑥𝑦 − 6𝑦2)
5. 2𝑥 2 + 6𝑥 + 5 𝑎𝑛𝑑 3𝑥2 − 2𝑥 − 1
6. (𝑥 + 2)(𝑥2 − 2𝑥 + 3)
7. 𝑎𝑏(2𝑎 + 1)
8. 𝑥2−3𝑥 −10
𝑥+2
9. 𝑥6+2𝑥4 +6𝑥−9
𝑥3+3
10. (3𝑥 3 − 11𝑥 2𝑦 + 11𝑥𝑦2 − 2𝑦3) ÷ (𝑥 − 2𝑦)
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) 8𝑎2𝑏 𝑎𝑛𝑑 18𝑎2𝑏2𝑐
8𝑎2𝑏=②. 2.2. 𝑎 ⃝ . 𝑎 ⃝ . 𝑏 ⃝
18𝑎2𝑏2 𝑐=②.3.3. 𝑎 ⃝ . 𝑎 ⃝ . 𝑏 ⃝. 𝑏. 𝑐
= 2𝑎2𝑏
The GCF of 8𝑎2 𝑏 𝑎𝑛𝑑 18𝑎2𝑏2𝑐 is 2𝑎2𝑏.
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 10𝑦2 + 15𝑦
10𝑦2=2.⑤. 𝑦 ⃝.𝑦
15𝑦 =3.⑤.𝑦
The GCF is 5y
10𝑦2 + 15𝑦=5y(2y)+5y(3)
=5y(2y+3) distributive property
2) Factor 21𝑎𝑏2 − 33𝑎2𝑏𝑐
21𝑎𝑏2=③.7. a ⃝ . 𝑏 ⃝. 𝑏
33𝑎2𝑏𝑐=③.11. a ⃝ . 𝑎. 𝑏 ⃝.c the GCF is 3ab
21𝑎𝑏2 − 33𝑎2𝑏𝑐= 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. 12𝑎𝑛2, 40𝑎4
7. 7𝑦2, 14𝑦2
8. 15, 10
9. 6𝑎2, 18𝑏2, 9𝑏3
10. 18𝑥 2𝑦2,6𝑦2, 42𝑥 2𝑦3
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 8𝑚2 𝑛 − 5𝑚 − 24𝑚𝑛 + 15
8𝑚2 𝑛 − 5𝑚 − 24𝑚𝑛 + 15=(8𝑚2𝑛 − 5𝑚) + (−24𝑚𝑛 + 15
= 𝑚(8𝑚𝑛 − 5) + (−3)(8𝑚𝑛 − 5)
=𝑚 − 3(8𝑚𝑛 − 5)
Check:
𝑚 − 3(8𝑚𝑛 − 5) = 𝑚(8𝑚𝑛) + 𝑚(−5) + (−3)(8𝑚𝑛) + (−3)(-5)
= 8𝑚2 𝑛 − 5𝑚 − 24𝑚𝑛 + 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 2𝑦2 + 7𝑦 + 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
2𝑦2 + (3 + 4)𝑦 + 6 Select the factors 3 and 4.
2𝑦2 + 3𝑦 + 4𝑦 + 6
(2𝑦2 + 3𝑦) + (4𝑦 + 6) Group terms that have a
𝑦(2𝑦 + 3) + 2(2𝑦 + 3) common monomials factor
(𝑦 + 2)(2𝑦 + 3) Factor (use the distirbutive property)
Therefore 2𝑦2 + 7𝑦 + 6= (𝑦 + 2)(2𝑦 + 3)
Example:
Factor 5𝑥 − 6 + 𝑥 2
The trinomials 5𝑥 − 6 + 𝑥 2 can be written as 𝑥 2 + 5𝑥 − 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, 𝑥 2 + 5𝑥 − 6 = (𝑥 − 1)(𝑥 + 6)
Exercise: 1.4.3-4
Factor each trinomials. If possible. If the trinomial cannot be factored using integers.
Write prime.
1. 6𝑚𝑥 − 4𝑚 + 3𝑥 − 2𝑟
2. 3𝑚𝑦 − 𝑎𝑏 + 𝑎𝑚 − 30𝑦
3. 𝑎2 − 2𝑎𝑏 + 𝑎 − 20
4. 3𝑚2 − 5𝑚2 𝑝 + 3𝑝2 − 5𝑝3
5. 4𝑎𝑥 − 14𝑏𝑥 + 35𝑏𝑦 − 10𝑎𝑦
6. 𝑦2 + 12𝑦 + 27
7. 𝑐2 + 2𝑐 − 3
8. 𝑥 2 − 5𝑥 − 24
9. 7𝑎2 + 22𝑎 + 3
10. 6𝑦2 − 11𝑦 + 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.
𝑎2 − 𝑏2 = (𝑎 − 𝑏)(𝑎 + 𝑏)
Examples:
1) 𝑓𝑎𝑐𝑡𝑜𝑟 𝑎2 − 64
𝑎2 − 64 = (𝑎)2 − (8)2
= (𝑎 − 8)(𝑎 + 8)
𝑎. 𝑎 = 𝑎2 𝑎𝑛𝑑 8.8 = 64 use the difference of square
2) 𝑓𝑎𝑐𝑡𝑜𝑟 𝑎𝑥 2 − 100𝑦2
𝑎𝑥 2 − 100𝑦2 = (3𝑥)2 − 10𝑦2
= (3𝑥 − 10𝑦)(3𝑥 + 10𝑦)
3𝑥. 3𝑥 = 9𝑥 2 𝑎𝑛𝑑 10𝑦. 10𝑦 = 100𝑦2
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 (𝑎 + 𝑏)2 𝑎𝑛𝑑 (𝑎 − 𝑏)2 are called perfect
squares and the expansions of these products are called perfect square trinomials.
(𝑎 + 𝑏)2 = 𝑎2 + 2𝑎𝑏 + 𝑏2
(𝑎 − 𝑏)2 = 𝑎2 − 2𝑎𝑏 + 𝑏2
Finding a Product Factoring
(𝑦 + 8)2 = 𝑦2 + 2(𝑦)(8) + 82 𝑦2 + 16𝑦 + 64 = (𝑦)2 + 2(𝑦)(8) +
(8)2
= 𝑦2 + 16𝑦 + 64 = (𝑦 + 8)2
Examples:
Determine whether 16𝑎2 + 81 − 72𝑎 is a perfect square trinomial.
1) 16𝑎2 + 81 − 72𝑎 = 16𝑎2 − 72𝑎 + 81
= (4𝑎)2 − 2(4𝑎)(𝑎) + (𝑎)2
= (4𝑎 − 9)2
2) 𝑥 2 + 22𝑥 + 121 = (𝑥)2 + 2(𝑥)(11) + (11)2
= (𝑥 + 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
𝑦 = −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)
𝐄𝐱𝐞𝐫𝐜𝐢𝐬𝐞 𝟏.𝟒.𝟓 − 𝟔
Factor.
1. 𝑥 2 − 49
2. 𝑥 2 − 36𝑦2
3. 16𝑎2 − 9𝑏2
4. 2𝑎2 − 25
5. 2𝑧2 − 98
6. 𝑛2 − 8𝑛 + 16
7. 4𝑘2 − 4𝑘 + 1
8. 𝑥 2 + 6𝑥 − 9
9. 1 − 10𝑧 + 25𝑧2
10. 50𝑥 240𝑥 + 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) 4𝑎2 𝑏𝑐3
6𝑎𝑏3 𝑐4 =2.2.𝑎.𝑎.𝑏.𝑐.𝑐.𝑐
2.3.𝑎.𝑏.𝑏.𝑏.𝑐.𝑐.𝑐 .𝑐=
2𝑎
3𝑏2𝑐
2) 𝑥2+2𝑥𝑦+𝑦2
𝑥2 −𝑦2 =(𝑥+𝑦)(𝑥+𝑦)
(𝑥+𝑦)−(𝑥−𝑦)=
𝑥+𝑦
𝑥−𝑦
3) 𝑥3+8𝑦3
4𝑥+8𝑦=
𝑥+2𝑦(𝑥2−2𝑥𝑦+4𝑦2
4(𝑥+2𝑦)=
𝑥2−2𝑥𝑦+4𝑦2
4
1.5.2 Multiplying and Dividing Rational Expressions
In multiplication if 𝑝
𝑞𝑎𝑛𝑑
𝑟
𝑠 are rational expressions and q and s are real numbers
not equal to 0, then 𝑝
𝑞.
𝑟
𝑠=
𝑝𝑟
𝑞𝑠.
Examples:
1) 4
3.
1
5=
4
15
2) 𝑐
𝑎2−𝑏2 . (𝑎 + 2𝑏)(𝑎 − 𝑏)
=𝑐
(𝑎 + 𝑏)(𝑎 − 𝑏). (𝑎 + 2𝑏)(𝑎 − 𝑏)
=𝑐(𝑎+2𝑏)
𝑎+𝑏
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
18𝑜𝑟
10
9
2) 8
7÷ 3 =
8
7.
1
2=
8
14𝑜𝑟
4
7
3) 𝑦2 −16
𝑦−5÷
2𝑦 −8
𝑥𝑦−5𝑥=
(𝑦−4)(𝑦+4)
𝑦−5.
𝑥(𝑦−5)
2(𝑦−4)=
𝑥𝑦+4𝑥
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) 3𝑥 − 2𝑦 +2𝑥2 −𝑦2
𝑥 +𝑦=
3𝑥(𝑥+𝑦)−2𝑦(𝑥+𝑦)+2𝑥2 −𝑦2
𝑥+𝑦=
3𝑥2 +3𝑥𝑦−2𝑥𝑦+2𝑦2 +2𝑥2 −𝑦2
𝑥 +𝑦=
5𝑥2 +𝑥𝑦−3𝑦2
𝑥+𝑦
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 𝑜𝑟 4
2) 3
2+1
3
=3
6+1
3
=37
3
= 3.3
7=
9
7
Exercise: 1.5
Solve and simplify.
1. 𝑎+1
𝑎3 −𝑎+2
𝑎2 +𝑎+3
𝑎
2. 5𝑥3
7𝑦4 .21𝑦2
10𝑥2
3. 9𝑥5
36𝑥2
4. 5−𝑎
𝑎2−25
5. 10𝑎2 −29𝑎+10
6𝑎2 −29𝑎 +10÷
10𝑎2 −19𝑎 +6
12𝑎 2 −28𝑎 +15
6.
1
𝑥+ℎ−
1
𝑥
ℎ
7. 𝑥6−7𝑥3 −8
4𝑥2−4𝑥−8 ÷ (2𝑥 2 + 4𝑥 + 8)
8.
𝑎
𝑏−
𝑏
𝑎𝑎
𝑏+
𝑏
𝑎
9. 𝑡2 −2𝑡−15
𝑡2 −9.
𝑡2−6𝑡+9
12−4𝑡
10. 𝑎−1+𝑏−1
𝑎−2−𝑏−2
1.6 Rational Exponents
We defined 𝑎𝑛 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.
That is, we wish to attach a meaning to 𝑎1
𝑛⁄ 𝑎𝑛𝑑 𝑎𝑚
𝑛⁄ , where the exponents are
fractions. Before discussing fractional exponents, we give the following
definition.
Definition
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 𝑛𝑡ℎ root of a real number
If n is a positive integer greater than
1 𝑎𝑛𝑑 𝑎 𝑎𝑛𝑑 𝑏 are real number such that
𝑏𝑛 = 𝑎, then b is an 𝑛𝑡ℎ root of a.
The principal 𝑛𝑡ℎ root of a real number. If n is a
positive integer greater than 1, a is a real number,
and √𝑎𝑛 denotes the princial 𝑛𝑡ℎ root of a, then
If a>0, √𝑎𝑛 is the positive 𝑛𝑡ℎ root of a.
.
The symbol √ is called a radical sign. The entire expression √𝑎𝑛 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) √814 = 3
3) √643 = 4
Definition
⁄
Examples 3:
1) 251
2⁄ = √25 = 5
2) −81
3=⁄ √−83
= −2
3) (1
81)1/4= √
1
81
4=
1
3
Definition
If n is a positive integer greater than 1, and a is
a real number, then if √𝑎𝑛 is a real number
𝑎1
𝑛⁄ = √𝑎𝑛
If m and n are positive integers that are
relatively prime, and a is a real number,
then if √𝑎𝑛 is a real number
𝑎𝑚
𝑛⁄ = ( √𝑎𝑛 )m ⇔ 𝑎𝑚
𝑛⁄ = (𝑎1
𝑛⁄ )m
Examples 4:
1) 93
2⁄ =(√9)3=33=27
2) 82
3⁄ = ( √83 )2=22 = 4
3) −274
3⁄ = ( √−273
)4=(-3)4=81
It can be shown that the commutative law holds for rational
exponents, and therefore
(𝑎𝑚)1/n=(𝑎1
𝑛⁄ )m
From which it follows that √𝑎𝑚𝑛 = ( √𝑎𝑛 )m
The next theorem follows from this equality and the definition of 𝑎𝑚
𝑛⁄
Theorem 1
Examples 5:
Theorem 1 is applied in the following:
1) 93
2⁄ =√93=729 =27
2) 82
3⁄ = √83 2= √643 = 4
3) −274
3⁄ = ( √−273)4= √5314413
=81
If m and n are positive integrers that are
relatively prime, and a is a real number,
then if √𝑎𝑛 is a real number
𝑎𝑚
𝑛⁄ = √𝑎𝑚𝑛 ⇔ 𝑎𝑚
𝑛⁄ = (𝑎𝑚)1/n
Observe that 𝑎𝑚
𝑛⁄ can be evaluated by finding either ( √𝑎𝑛 )m or √𝑎𝑚𝑛. Compare
example 4 and 5 and you will see the computation of ( √𝑎𝑛 )m in example 4 is simpler than
that for √𝑎𝑚𝑛 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 (𝑎𝑚)1/n=│a│m/n
A particular case of this equality occurs when m=n. We then have (𝑎𝑛)1/n=│a│
(if n is a positive even integer) or, equivalently, √𝑎𝑛𝑛= │a│ (if n is even)
If n is 2, we have √𝑎2 = │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
82
3⁄ =1
( √8)23 =1
22 =1
4
2) 8−2
3⁄ = (8−1
3⁄ )2=(1
81
3⁄ )2=(1
2)2=
1
4
3) 𝑥
13⁄
𝑥1
4⁄ =𝑥1
3⁄ .1
𝑥1
4⁄ =𝑥1
3⁄ .𝑥−1
4⁄ = 𝑥 (13⁄ )−1
4⁄ = 𝑥1
12⁄
If m and n positiv e integer that are
relatively prime and a is a real number and
a≠0, then if √𝑎𝑛 is a real number.
𝑎−𝑚
𝑛⁄ =1
𝑎𝑚
𝑛⁄
Exercise 1.6
Find the value.
1. 𝑎)811
2⁄ ; 𝑏)271
3⁄ ; 𝑐)6251
4⁄ ; 𝑑)321
5⁄
2. 𝑎)161
2⁄ ; 𝑏)1251
3⁄ ; 𝑐)161
4⁄ ; 𝑑)1000001
5⁄
3. 𝑎)𝑥−3
4⁄ .𝑥5
6⁄ .𝑥−1
3⁄ ; 𝑏)𝑦
−34⁄
𝑦3
2⁄ )-1/9
4. 𝑎)𝑦1
4⁄ .𝑦−3
2⁄ .𝑦−5
8⁄ ; 𝑏)𝑥
−35⁄
𝑥−7
10⁄ )-1/4
5. (𝑥1
3⁄ − 𝑥−2
3⁄ )(𝑥2
3⁄ − 𝑥−1
3⁄ )
6. (𝑎1
4⁄ − 𝑎1
2⁄ )(𝑎−1
4⁄ + 𝑎−1
2⁄ )
7. 𝑎)2𝑦3
2⁄ − 3𝑦5
2⁄ ; 𝑏)5𝑥−4
3⁄ + 4𝑥5
3⁄
8. 𝑎)6𝑡3
4⁄ + 𝑡7
4⁄ ; 𝑏)4𝑤4
5⁄ − 3𝑤−6
5⁄
9. a)(𝑎3)n/3(𝑎3𝑛 )3/n; b)( 𝑥𝑛
2⁄ )-1/2(𝑥−1
2⁄ )-n
10. a)(𝑦4)n/4(𝑦2𝑛)2/n; b)( 𝑡𝑛
3⁄ )-2/3(𝑡−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 6𝑎2 + (6𝑎 + 𝑎2 ) + 9𝑎
a. 6𝑎2 + (6𝑎 + 𝑎2 ) + 9𝑎 = 6𝑎2 + (𝑎2 + 6𝑎)+9a
b. = (6𝑎2 + 𝑎2 ) + (6𝑎+9a)
c. = (6𝑎2 + 1𝑎2) + (6𝑎+9a)
d. =6 + 1(𝑎2) + (6 + 9)𝑎
e. = 7𝑎2 + 15𝑎
Simplify and express the following.
1. 𝑎𝑚+5
𝑎𝑚−2
2. [(𝑥+𝑦)0+𝑎0+𝑏0
𝑎+𝑏+𝑐]-2
3. (𝑎−2 + 𝑦)-2
4. (37𝑥+5 )(34𝑥−4)
5. (9𝑥𝑦2)(4𝑥 3𝑦)
Rational Expression (simplify)
1. 9𝑛
63𝑛÷ 9𝑛
2. −15𝑚3 𝑛2 𝑝2
−35𝑚2 𝑛5 𝑝
3. 𝑥+𝑦
𝑥2−𝑦2
4. 3𝑚−1
9(𝑚−1)2−4
5. 4𝑚𝑛+6
10𝑚 +8𝑛
6. 2𝑥2 +3𝑥 −5
10𝑥 +25
7. 𝑥2−5𝑥 −24
4𝑥2 −27𝑥 −40
8. 25𝑎2 +70𝑎+49
25𝑎 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=
𝑟
24
24(5
12) = (
𝑟
24)24 multiply each side by 24
10 = 𝑟
Check: 5
12=
𝑟
24 replace r with 10
5
12=
10
24
5
12=
5
12
2) 24=-2a
24=-2a
−1
2(24) = −
1
2(2𝑎)
−12=a
𝑐ℎ𝑒𝑐𝑘:
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 𝑎
𝑐=
𝑏
𝑐.
Examples:
1. -6x=11 −6𝑥
−6=
11
−6 divide each side by -6
𝑥 = −11
6
Check: -6x=11
−6(−11
6)=11
11 = 11
2. 4x=24
4𝑥
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+ 𝑤 =
5
8
5. k+(-13)=21
6. 16-y=37
7. t-(-16)=9
8. (−41
2) 𝑥 = 36
9. −3
5𝑦 = −50
10. 3𝑥 = 42
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
2)-3
=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
𝑟
Father 24 r-12 24
𝑟−12
Equation:
28
𝑟=
24
𝑟 − 12
Solve the equation by first multiplying on both sides by the LCD:
𝑟(𝑟 − 12)28
𝑟= 𝑟(𝑟 − 12)
24
𝑟 − 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 11
2ℎ𝑟, what is each person’s average rate
of cycling?
