Module 4 - Lecture notes Chapter 4 notes PDF

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Module 4: ALGEBRA 4.1 EXPONENTS AND RADICALS Exponents are used to indicate repeated multiplication.

x n = x  x  ...  x n times

where x is the base that is multiplied by itself and base need to be multiplied.

n is the exponent that shows how many times the

The following rules are satisfied when simplifying exponential expressions: -

x 0 =1

-

x x = x m

n

m+n

-

xm = x m− n , x  0 xn n (x m ) = x mn

-

x -n =

-

x n = n xm

-

1 n x

m

For example: -

(2 x )(3 x ) = ( 2 3) ( x ) = 6 x

-

10 a 2bc 2a = 5 ab 2c b

-

2

5

2+5

(3 x ) = (9 x ) = 9 x 2 3

2 3

6

3

3

3

=

7

9 x 6 −3 9 x 3 = 5 5

5x 5x 5x 2 2 −3 a ab2 ab ab 2 = = = 3 b −2 2 3  3 8  3 24 2  −3x 5 y 3z −2  9x 10 y 6z −4 9x 10x 4 y 6z 12 9x 14z 8 = = =   −2 5 −6 − 4 10 −12 16 y 10z 4 16 y 4  4 x y z  16 x y z 2

-

 34 −21 a b  −3  c 4 

3 2 1 2 1 1 1 1 1  −  − 3 4 3 2 3 2 3 2 2 2 a c ( ) a b a b a c ac  = = = 1 = 1 = 3 3 2 1  −  − b  c 43 c 2 b3 b3 

Example 1: Simplify and express the answer in simplest form with no negative exponents:

…1

a)

(2 x y )(3x y )

b)

− 3x 2 y 5 6x5 y 3

2

3

5

4

2x 3y 2 c)

( 4x y ) 5

3 2

d)

(3x y )(7 x

e)

2x − 3y 2z − 4 3 x −2 y −4 z −6

−4

3

−3

y3

)

5 x −2 y −3 f)

( 2x

4

y−2 z−3

)

2

2x 3 y 2 g)

(3

−1

x 2 y −4 z −3

)

h)

 3 x−2 y −3 z −5   −4 2 2   x y z 

i)

 x −3 y 2 z 5   2 −3   x y z 

j)

 x 2 y −3   −3 −5 −2  x y z 

−2

2

−2

−3

4.2 ALGEBRAIC EXPRESSIONS An algebraic expression describes numbers and variables combined with different arithmetic operations (addition, subtraction, multiplication, division, exponents and radicals). …2

The parts in an algebraic expression separated by a plus or minus are referred as terms. A term is a number, variable or a product of a number and variable(s). For example −5 , x , 7x2 or 3ab are terms. The number part of the term is called a coefficient, and the variables including their exponents are making the variable part of the term. For example, in 4a2 b , the coefficient is 4 and the variable part is a2 b A constant term is a term that contains only a number. In other words, there is no variable in a constant term. For example, 1, 2013, or -5.6 are constants. Like terms are terms that have the exact same variables raised to the exact same exponents. For example, 5x and −7x are like terms; or 3a2 b , −4a 2b and a2 b are like terms. Simplifying algebraic expressions means combining the terms that are like terms. We will count up how many variables we have the same and write the number in front of the common variable part. To simplify variable expressions that contain parentheses we may use the following properties:

-

Commutative Property of Multiplication: a  b = b  a

-

Associative Property of Multiplication: a ( b  c) = ( a  b)  c

-

Distributive Property of Multiplication: a ( b + c) = ab + ac or a( b − c) = ab − ac

For example, to simplify the following algebraic expressions: -

3 x + 5x = 8x

-

2a − 5a + 7b = −3 a + 7b

-

−2 ( 2 x 2 +5 x −7 ) = −4x 2 −10 x +14

-

3 ( x + 4 ) = 3 x + 12

-

2 (5x − 4 ) + 3 (3 x −1 ) = 10x − 8 + 9x − 3 =19x −11

-

3 ( 2 x + 3) − 4( 2− 5 x) = 6 x + 9 − 8 + 20 x = 26 x + 1

Example 1: Simplify the following variable expressions:

…3

a)

2 x ( 3 x 2 + 5 x − 7)

b)

3( x 2 − 5x ) − 2x ( 4 − x )

c) 4 − x (3 + x ) − 2 (x − 1)

d) 2 − 3 x (2 x − 1) + (2 − 3 x )(4 − x )

e) 2 ( x − 3 )( x + 5 )

f)

2 xy ( 3 x − 5 y ) − xy ( x − 3 )( y − 5 )

g) 3 ( xy −1)(3 − x )(5 − y )

4.3 EQUATIONS An equation is a statement that two expressions are equal. In this module we will be solving linear equations with one variable. A linear equation is one which can be written with the variable appearing only in the numerator and raised to the first power. A solution of an equation is a number that when substituted back in the equation instead of the variable, we receive a correct statement. For example, for the equation x + 7 = 10 , a solution is x = 3 , because 3 + 7 = 10 …4

10 = 10 A strategy for solving equations is to: - eliminate fractions and grouping symbols. - isolate the variable terms. - simplify by combining like terms. - solve for the variable. The following examples demonstrate this strategy. Sol Solvi vi ving ng aan n equat equatio io ion n of tthe he for form m x+a =b In order to solve this type of an equation we need to review the following property If a = b , then a + x = b + x . In other words we can add a number (or expression) on both sides of the equation without changing the validity of the equation. For example, in order to solve the equation: x − 6 = 4 , we would add 6 on both sides so we can get the variable x by itself on one side of the equation and have:

x− 6 + 6= 4 + 6 x − 6 +6 = 4 + 6 x = 10 For example, in order to solve the equation: x + 2 = −5 , we would add (−2 ) on both sides so we can get the variable x by itself on one side of the equation and have:

x + 2 − 2 = −5 − 2 x = −7 Example 1: Solve the following equations:

a) −3 + x = 9

1 3 b) x + 4 = − 5

c) 9 − x = −

4 3

Sol Solvi vi ving ng aan n equat equatio io ion n of tthe he for form m ax = b In order to solve this type of equation we will review the following property: If a = b , then a  x = b x and

a b = . x x

In other words we can multiply both sides of the equation with a number (or expression) without changing the validity of the equation. For example, in order to solve the equation: 5 x = 15 , we would divide with 5 both sides so we can get the variable x by itself on one side of the equation and have:

…5

5x 15 = 5 5 x=3 And when we have a fraction that multiplies the variable we need to divide with the same fraction both sides. Note that dividing with a fraction means multiplying with its reciprocal. For example, in solving the equation

2 3 both sides and we have: x = 4 we would multiply by 2 3

 3  2 x = 3  4    2  3 2  x= 6 Example 2: Solve the following equations:

a)

−3 x = 15

b)

3 4 x=− 4 5

c) −8 = −

4 x 3

Sol Solvi vi ving ng aan n equat equatio io ion n of tthe he for form m ax + b = c In order to simplify and solve this type of equation we need to use both, the addition and the multiplication properties. For example, solve the following equations:

 5 x + 4 = 29

2  3 x+ 2 = 8

…6

5 x + 4 −4 = 29 − 4

2 x + 2 − 2= 8 − 2 3 2 x =6 3 3 63  3 2 2 3 x = 1  2     1

5x 25 = 5 5 x =5

x=9

Example 3: Solve the following equations: a) 3 x + 5 = 5 x + 9

c)

b) 5 − 2 (9 − 6 x ) = 3 x − 4 (2 − 7 x

x−7 = 2x + 3 5

d)

)

3 x 1  − 7 = − ( x + 2)  5 2  3

4.4 FACTORING To factor a polynomial means to write the polynomial as a product of two or more polynomials. Through examples this section presents techniques for a variety of common factoring situations. For example, to factor a monomial from a polynomial

(

)

5 x3 − 35 x2 + 1 0 x = 5 x x2 − 7 x + 2

…7

For example, to factor by grouping,

6 x2 − 9x − 4 xy + 6 y = 3x ( 2 x − 3) − 2 y ( 2x − 3) = (3 x − 2 y )(2 x −3 ) For example, to factor a difference of squares, we use A − B = ( A − B)( A + B) 2

2

. The

difference of two squares factors to the product of two binomials which are the sum and difference of the two quantities that are squared.

4 x2 − 9 y 2 = ( 2 x) −( 3 y) 2

2

= (2 x − 3 y )(2 x + 3 y ) For example, to factor a trinomial from the form x + bx + c we use trial and error, 2

x 2 + 7x + 6 = ( x −

)( x − ) = ( x + 6)( x + 1)

For example, to factor a trinomial from the form ax + bx + c we use factor by grouping after expanding the middle term with the help of the factors of the product ac 2

2 x2 + 5x − 12 = 2x2 + 8x − 3x − 12

a  c = 2  ( − 12 ) = −24

8 − 24    − 3

= 2 x ( x +4 ) −3 ( x +4 ) = ( 2x − 3)( x + 4) Example 1: Factor completely the following expressions: a)

4 x3 y4 z 2 − 32x4 y3 z + 16x 2 yz 4

b)

x4 − y2

c) 18 − 8 x2 d) 16 x 4 − 81 2 e) x + 10 x + 16

f)

x2 + x − 6

…8

g) x 2 + 5x − 14

h) x 2 − 5x − 6

i)

x 2 − 3x − 180

j)

3xy2 + 6 xy − 45x

k) 2x − x − 3 2

l)

8x 2 + 6x − 9

m) 4x 2 + 33x + 8

…9

4.5 ALGEBRAIC FRACTIONS A fraction that has polynomials for a numerator and/or denominator is called algebraic fraction. We say that an algebraic fraction is in its simplest form if the numerator and denominator have no common factors. 4.5.1 SIMPLIFICATION OF ALGEBRAIC FRACTIONS To simplify a given algebraic fraction we factor both, the numerator and the denominator and we cancel the common factors. For example, to simplify the following algebraic fraction:

( x − 3) ( x + 3) x − 3 x2 − 9 = = x2 − x −12 ( x + 3) ( x − 4 ) x − 4 Example 1: Simplify the following algebraic fractions

32 x3 y2 a) 24 xy5

…10

b)

x2 + 4 x xy + 4y

c)

x 2 − 3x 6 − 2x

d)

16 x3 − x 36 x − 9

e)

x 2 + 3x − 10 x 2 + 2x − 8

f)

x 2 + 3x − 4 x 2 + 7x + 12

g)

x 2 + 4x − 12 x 2 − 3x + 2

h)

x 2 − 5x + 4 x2 + 2 x − 3

i)

2 x 2 − 13 x + 15 2 x2 + x − 6

j)

3 x 2 − 2 x −1 2 3 x + 10 x + 3

…11

k)

2 x 2 + 5 x − 12 3 x 2 + 10 x − 8

l)

4 x  + 11x − 3 2x 2 +9x +9

4.5.2 MULTIPLICATION AND DIVISION OF ALGEBRAIC FRACTIONS To multiply algebraic fractions, we multiply the numerators and we multiply the denominators. During the process we can divide common factors and express the answer in its simplest form. For example,

( x +1) ( x + 3 ) y ( x −1) ( x +1) ( x + 3)( x −1) x2 + 4 x + 3 x2 y − y =  =  x 2 + 2 x +1 x2y x2 y x2 ( x + 1) ( x + 1) Example 1: Multiply the following algebraic fractions and simplify:

a)

x −2 5 xy3  2 x − 4 25x 4y 2

b)

10 x 2 −15 x 3x − 2  12 x − 8 20 x − 25

…12

c)

3 x 2 + 2 x 2 xy 3 − 3 y 3  2 xy −3 y 3 x 3 + 2 x 2

d)

x5 y3 x2 + 2 x − 3  x 2 + 13x + 30 x 7y 2

e)

8 x3 + 4 x2 x2 − 4  x 2 − 3x + 2 16x 2 + 8x

f)

x 2 − 49 2 x 2 − 13x − 7  x 2 + 10x + 21 x 2 − 14x − 49

…13

To divide algebraic fractions, we multiply the dividend (the first fraction) with the reciprocal of the divisor (second fraction). During the process of multiplication we can also divide common factors and express the answer in its simplest form. For example,

2 x2 + 9 x + 9 2 x2 + 9 x + 9 x2 − x − 6 (2 x + 3 ) ( x + 3 ) ( x + 2 ) ( x − 3) ( x + 2) 6x + 9  =  =  = x2 − 9 x2 − x − 6 x2 − 9 6x + 9 3 x + 3 x − 3 3 ( 2 x + 3) ( )( ) Example 2: Divide the following algebraic fractions and simplify:

a)

xy 2 − 3x 2y 6x 2 − 2xy  z3 z3

b)

3 x2 + 2 x 3 x3 + 2 x 2  2 xy −3 y 2 xy 3 −3 y 3

c)

x 2 − 49 x 2 − 14x + 49  x4 y 3 x4 y3

d)

x 2 − 5x + 6 x 2 − 6x + 8  x 2 − 9x + 18 x 2 − 9x + 20

…14

e)

x2 − 9 x +1 x −1   2 ( x + 1) 2 x − 6 4( x 2 − 1)

4.5.3 ADDITION AND SUBTRACTION OF ALGEBRAIC FRACTIONS To add and/or subtract algebraic fractions that have a same denominator we add and/or subtract the numerators only while copying the same denominator. We reduce the resulting algebraic fraction to the simplest form if possible. For example,

(

)

2 x 2 + 4 x + 4 x 2 − 2 x 2 + 4 x + 4 + x 2 − 2 2 x2 + 4 x + 2 2 x + 2 x + 1 x 2 + 2 x + 1 = + 2 = = = 2 x2 y 2x 2 y 2x 2 y x2 y 2x y 2 x2y

Example 1: Perform the given operations and simplify: a)

5x + 4 x 2 − 2 + 4x 4x

b)

4x + 9 2x − 5 − 6x2 6x2

To add and/or subtract algebraic fractions that have different denominator we first determine a common denominator and then expand each algebraic fraction to an equivalent fraction with the same common denominator. We then continue with adding and/or subtracting the numerators and reduce the answer to its simplest form if possible.

…15

For example,

4x + 5 2x 2 y

2y

x −2y + 4xy2

x

=

2 y( 4 x + 5) x( x − 2 y) 8 xy +10 y x2 − 2 xy + = + 4 x2 y2 4 x2 y2 4 x2 y2 4 x2 y2

=

8xy + 10y + x2 − 2xy x2 + 6xy + 10y = 4 x2 y 2 4 x2y 2

Example 2: Perform the given operations and simplify: a)

x +1 x − 3 + 3x 2 6x

b)

2x 1 − 2 x − 3 x +1

c)

7 x − 3 3 x +1 − 8 x −8 4 x − 4

d)

2x 3 − x − 4 x+ 2

e)

4 −x x − 2 − 2x 4 x − 2x

f)

2x − 3 5 1 + − 2 3x − x − 2 3x + 2 x −1

2

…16

g)

2 2 x + x − 6 x + 5x + 4 2  2 − 2 x + 2x − 8 x + 2x − 3 x − 1

…17...