topic 1 concepts of algebra

8/2/2019 Topic 1 Concepts of Algebra

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INTRODUCTIONIn this topic we shall begin with the set of real numbers. Then, we will learn aboutpolynomial and equations. We will also learn the operations of polynomials andapply these operations in solving problems in quadratic equations and partialfractions.

SETS OF REAL NUMBERSThe numbers that we use can be classified into various categories. We call themset of numbers. Let us begin with a set of natural numbers. Natural numbers arebasic counting numbers as follows:

N { 1, 2, 3, ... }

1.1

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11

Conceptsof Algebra

3. Factorise polynomials;

4. Identify proper and improper fractions; and

5. Express proper and improper fraction as partial fractions.

2. Solve mathematical operations involving polynomials;

1. Describe the sets of real numbers;By the end of this topic, you should be able to:

LEARNING OUTCOMES


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TOPIC 1 CONCEPTS OF ALGEBRA 3

As a simple exercise, can you give:(a) an integer which is not natural?

(b) a rational number which is not integer?

(c) a real number which is not rational?

SELF-CHECK 1.1

POLYNOMIALS1.2

An expression of the form:

axk

where a is a constant and x an unknown (or a variable) and k 0 is an integer iscalled a monomial . a is referred to as the coefficient of the monomial. We cancompute the sum or difference of any two monomials like axk and bxk and theresults of these operations are also monomials. These operations are performed byusing a distributive property as shown in Example 1.1:

Example 1.1

2 2 24 2 (4 2) 6 2 x x x x 2 and 3 52 2 2(3 5) 2 x x x x

0

0

0

Polynomial in a single variable is an algebraic expression of the following form:

11 1...

n nn na x a x a x a

where are constants, n 0, and x is a variab le. The constantsare coefficients of the polynomial and when a n 0, the

polynomial

1 1, , ..., ,n na a a a1 1 0, , ..., ,a a

n nn na x a x

n na a1

1 1... a x a is called a polynomial of degree n.Each of the monomials in a polynomial is called a term of the polynomial.

Hence, a polynomial is an algebraic sum of monomials in which no variablesappear in denominators and all variables that do appear are raised only to positive-integer powers.


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TOPIC 1 CONCEPTS OF ALGEBRA4

Often a polynomial is written in its standard form, with the highest degree non-zero term listed as the first, then followed by the rest of the terms in a descendingorder of their degrees. The highest power in the polynomial is known as the

degree of the polynomial.

The following is a number of polynomials and some of the important properties of each of the polynomials.

Table 1.1: Properties of Some Polynomials

Coefficient of Each of the MonomialsPolynomial

3 x 2 x 1 x 0 x Degree

3 3 2

2 3 8 2 0 3 8 x x x x x 2 0 3 8 32 3 32 11 0 ( 2) 0 1 x x x x 1 0 2 0 11 2

3 26 0 0 ( 6) x x x x 0 0 6 1

04 4.1 4 x 0 0 0 4 0

Note: Until now, we have only used x to represent variables in equations andpolynomials. In reality, we can use any other letter of an alphabet. Other lettersthat are often used are y and z. All the polynomials in Table 1.1 are in terms of x.

For example, 7 y3 3 y 4, is a polynomial of the 3rd-degree with y as the variable.

And, 9 z4 2 z3 10, is a polynomial of the 4th-degree with the variable z.

Remark: Polynomial of degree zero is called constant. Polynomial of degree 1 iscalled linear. A complete list of degree 2 to 100 is shown in Table 1.2.


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Table 1.2: Polynomials Classified by Degree

Degree Name Example

0 (non-zero) constant 1

1 linear x + 1

2 quadratic x2 + 1

3 cubic x3 + 1

4 quartic (or biquadratic) x4 + 1

5 quintic x5 + 1

6 sextic (or hexic) x6 + 1

7 septic (or heptic) x7 + 18 octic x8 + 1

9 nonic x9 + 1

10 decic x10 + 1

100 hectic x100 + 1

(a) Addition and Subtraction of PolynomialsThe addition and subtraction of polynomials are performed by combining

monomials of the same degree. That is, grouping monomials of the samedegree and combining them.

Example 1.2

2 3 2 3 2 2

3 2

3 2

(4 7 3) ( 6 5) (4 6 ) (7 ) ( 3 5)

(4 6) (7 1) ( 3 5)

2 8 2

x x x x x x x x x x

x x x

x x x

Example 1.3

In this example, we find the difference of two polynomials. The first step isto open up the brackets. When opening up the brackets, make sure the signof each of terms of the polynomial inside is changed accordingly due to thenegative sign before the bracket:
http://en.wikipedia.org/wiki/Constant_functionhttp://en.wikipedia.org/wiki/Linear_equationhttp://en.wikipedia.org/wiki/Quadratic_functionhttp://en.wikipedia.org/wiki/Cubic_functionhttp://en.wikipedia.org/wiki/Quartic_functionhttp://en.wikipedia.org/wiki/Quintic_equationhttp://en.wikipedia.org/wiki/Sextic_equationhttp://en.wikipedia.org/wiki/Septic_equationhttp://en.wikipedia.org/w/index.php?title=Octic_equation&action=edit&redlink=1http://en.wikipedia.org/w/index.php?title=Nonic_equation&action=edit&redlink=1http://en.wikipedia.org/w/index.php?title=Decic_equation&action=edit&redlink=1http://en.wikipedia.org/wiki/Hectic_equationhttp://en.wikipedia.org/wiki/Hectic_equationhttp://en.wikipedia.org/w/index.php?title=Decic_equation&action=edit&redlink=1http://en.wikipedia.org/w/index.php?title=Nonic_equation&action=edit&redlink=1http://en.wikipedia.org/w/index.php?title=Octic_equation&action=edit&redlink=1http://en.wikipedia.org/wiki/Septic_equationhttp://en.wikipedia.org/wiki/Sextic_equationhttp://en.wikipedia.org/wiki/Quintic_equationhttp://en.wikipedia.org/wiki/Quartic_functionhttp://en.wikipedia.org/wiki/Cubic_functionhttp://en.wikipedia.org/wiki/Quadratic_functionhttp://en.wikipedia.org/wiki/Linear_equationhttp://en.wikipedia.org/wiki/Constant_function


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4 3 2 4 3 2

4 3 2 4 3 2

4 3 2

4 3 2

(3 4 3 1) ( 3 5 9)

3 4 3 1 3 5 9

(3 1) (4 3) ( 3 1) (1 5) 1 92 2 4 10

x x x x x x x x

x x x x x x x x

x x x x x x x x

Signs changed

Grouping like terms

(b) Multiplication of PolynomialsMultiplication of polynomials is handled using the distributive propertiesand the rules of exponent repeatedly.

Example 1.42

2 2

2 2

3 2 2

3 2

(3 2)(4 7 3)

3 (4 7 3) 2(4 7 3)

3 4 3 7 3 3 2 4 2 7 2 3

12 21 9 8 14 6

12 29 5 6

x x x

x x x x x

x x x x x x x

x x x x x

x x x

distributive properties

ropertiesdistributive p

rules of exponent

combine like terms

(c) A Few Common ProductsThere are several products of polynomials that are often used in algebra.Among them are:

2 2

2 2 2

2 2 2

2

( )( )

( ) 2

( ) 2

( )( ) ( )

x a x a x a

x a x ax a

x a x ax a

x a x b x a b x ab

for any real number x, a, b, c and d.

(d)

Equations, Identities, Inequalities and FunctionsNow, let us consider the following algebraic expressions:

(i) 2( 1) 7 x x

(ii) 2 2( 1) 2 1 x x x

(iii) ( 2) 1 x

(iv) 2( 1) x


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It is obvious that there are dissimilarities among the four expressions. Let usdiscuss each of them in greater detail:

(i) 2( 1) 7 x x

When we replace x with 1 on the left hand side (LHS) of the equationand then on the right hand side (RHS) separately, we get

LHS (1 1)2 22 4

RHS 1 7 8

Therefore, LHS RHS.

Observe what happens if we replace x with 2:

LHS (2 1)2 32 9

RHS 2 7 9

Therefore, LHS RHS.

Notice that expression (i) can be re-arranged as follows:

x2 2 x 1 x 7

x2 x 6 0

Therefore, ( x 3)( x 2) 0

It is obvious that LHS RHS only when

x 3 0 i.e. x 3

or x 2 0 i.e. x 2

From this, ( x 1)2 x 7 only when x 3 or x 2 and this equationis not true for any other values for x.

Expression such as (i) is called an equation and an equation is trueonly for several values of an unknown.


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The process of obtaining the values for the unknown is called solvingthe equation.

(ii) ( x 1)2 x

2 2 x 1

When we replace x with 4, we see that

LHS (4 1)2 52 25

RHS 42 2(4) 1 25

Therefore, LHS RHS when x 4.

When we replace x with 2, we get

LHS ( 2 1)2 ( 1)2 1

RHS ( 2)2 (2)( 2) 1 1

Therefore, LHS RHS when x 2.

In fact, LHS RHS for all values of x. This is clearly shown in Figure 1.1which shows squares with sides equal to x 1.

2

Figure 1.1

Both of the rectangles shown are identical and hence their areas arealso identical. Therefore 2 2( 1) 2 1 x x x for all values of x and ( x

1)2 is said to be identical with x2 2 x 1.

Such relation is called an identity when both sides of the equationsare the same for any values of the unknown.


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FACTORING POLYNOMIALS1.3

In this section, we will learn how to factorise polynomials, in particularpolynomials of degree two, i.e., quadratic expression.

The factorisation of a quadratic expression is a process of finding two linearexpressions such that the product of these expressions produces the originalquadratic expression. We give several examples as follow.

Example 1.5

Factorise the following:

(a) x2 + 5 x

(b) 3 x2 + 9

(c) 4 x2 9

Solution

Factorise the common factor, i.e. x

(a) x2 + 5 x = x( x + 5)

Factorise the common factor, i.e. 3

(b) 3 x2 + 9 = 3( x2 + 3)

Change to square numbers.

(c) 4 x2 9 = (2 x)2 (3) 2

= (2 x 3) (2 x + 3)


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Example 1.6

(a) Factorise 2 4 5 x x

Solution

.)())((54 222 pq xq p x pqqx px xq x p x x x

Equate the coefficients of x and constant to obtain p + q = 4 and pq = 5.

pq p q p + q Check:

x 55 5 1 4 x 15 1 4

542 x xThere are two ways to obtain the

pq and we choose p + q = 4.

Then p = 5 and q = . So, .1 )1)(5())((542 x xq x p x x x (b) Factorise 232 x x

Solution

.)())((23 22 pq xq p xq x p x x x

Equate the coefficients of x and constant to obtain p + q = 2 and pq = 3.

pq p q p + q Check:

x 12 1 2 3 x 2

1 2 3232 x x

Then p = 1 and q = 2. So, ).2)(1())((232 x xq x p x x x There are two ways to obtain pq .


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Example 1.7

(a) Factorise 23 4 x x 4

Solution

.)())((443 22 pq xnqmpmnxqnx pmx x x

Equate the coefficients of x and constant to obtainmn = 3, mp + nq = 4 and pq = 4.

m n p q mp + nq Check

3 1 4 1 3 4 = 1 3 x 24 1 3 + 4 =12 2 6 + 2 = 4 x 2

2 2 6 2 = 4 3 x2+4 x 4

There are four possibilities to obtain pq = 4. Choose mp + nq = 4.Then m = 3, n =1, p = 2, q = 2. Then 3 ).2)(23(442 x x x x

(b) Factorise 3722

x x Solution

.)())((372 22 pq xnqmpmnxqnx pmx x x

Equate the coefficients of x and constant to obtainmn = 2, mp + nq = 7 and pq = 3.

m n p q mp + nq Check

2 1 3 1 2 + 3 = 5 3 x 21 3 6 + 1 = 7

x 23 x2+4 x 4


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There are two possibilities to obtain pq = 3. Choose mp + nq = 7.

Then m = 2, n =1, p = 1, q = 3. Then ).3)(12(372 2 x x x x

Example 1.8

(a) Factorise 23 12 1 x x 2

Solution

Step 1Factorise 3 since 3 is the common factor.

)

)

44(312123 22 x x x x

Step 2Factorise the RHS and Simplify.

2)(2(3 x x

(b) Factorise p 2 + 2 mp + 2 p + 4 m

Solution

p2 + 2mp + 2 p + 4m

p2 + 2mp + 2 p + 4m = p ( p + 2m) + 2 p + 4m

Step 2Factorise the common factor for the last two terms.

Step 1Factorise the common factor for the first two terms.

p2 + 2mp + 2 p + 4m = p ( p + 2m) + 2 ( p + 2m)


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Step 3Factorise the common factor ( p + 2m).

p2 + 2mp + 2 p + 4m = ( p + 2) ( p + 2m)

PARTIAL FRACTIONS1.4

The ratio of two polynomials like2

3

3,

2 5 x x

where both the numerator and

denominator are polynomials is called a proper fraction when the degree of the

numerator polynomial is smaller than the degree of the denominator polynomial.On the other hand, if the degree of the numerator polynomial is greater than orequal to the denominator polynomial, the resulting ratio function is called animproper fraction.

Try to remember that an improper fraction like43

can be written as3 1 1

1 .3 3

The same method can be used to change an improper fraction like2

2

41

x x

to the

form shown as follows:

2 2 2

2 2 2 2 2

4 1 3 1 3 31

1 1 1 1 x x x x x x x x

1

Consider a function like 23

( ) .2 1

x f x

x x

( ) f x can be written as a single fraction with a common denominator as follows:

2 2

2 2

3 3( 1) ( 2) 4 2( )

2 1 ( 2)( 1) ( 2)( 1 x x x x x x

f x x x x x x x2

3)


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Sometimes we need to reverse this operation, that is we need to write thepolynomial ratio as a summation of two or more polynomial ratios. The reverse

process of "taking the fraction

2

2

4 2 3( 2)( 1)

x x x x

apart" into the sum of simpler

fractions,

2

2 2

4 2 3 3,

( 2)( 1) 2 1 x x x

x x x x

is called "decomposing the fraction into the partial fractions".

1.4.1 The Cover-up RuleWhen the original fraction is a proper fraction, then the resulting partial fraction isalso a proper fraction.

In other words, a fraction such as2

( 3)( 2 x

x x ) can be written as

,3 2

A B x x

and 22

( 3)( 4) x

x x can be written as

2 ,3 ( 4) A Bx C

x x

where A, B and C are constants that need to be determined.

The method to find these constants depends on the factor of the denominator of the polynomials involved.

Example 1.9

Express2

( 3)( 2 x

x x ) in partial fractions.


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Example 1.10

Express 22

( 1)( 1 x x )as partial fractions.

SolutionObserve that the denominator in this example has a quadratic factor or apolynomial of the second degree. When such a factor exists, the resulting partialfractions can be a first degree polynomial (i.e. one degree lesser than the degree of the denominator polynomial). Hence we need to find the constants A, B, and C such that

2 2

2

( 1)( 1) ( 1) ( 1)

A Bx

x x x x

C

or

2

2 2

2 ( 1) ( )(( 1)( 1) ( 1)( 1)

A x Bx C x x x x x

1)

In other words,

22 ( 1) ( )( 1) ............(*) A x Bx C x Setting x 1 (to eliminate B and C ) gives us

2 A(12 1)

or

A 1.

We will not have any value of x that will eliminate A (because no real value of x that satisfies x2 1 0).

A simple choice that will eliminate B is x 0. Substituting this value, we get

2 A(1) C (1)


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Substituting A with 1 (the value obtained earlier) we have

2 1(1) C

or

C 1.

To find the value of the constant B, we can substitute any other value for x (best tochoose a small value for x in order to simplify the calculation). Let say we choose

x 1, we have

2 A((1) 2 1) ( B(1) C )(1 1)

or

2 2 A 2 B 2C .

But we have already gotten A 1 and C 1. Hence, B 1.

Therefore

2 2

2 1

( 1)( 1) ( 1) ( 1)

x

x x x x

1

1.4.2 The Combining Method

The method that we have used so far for determining the constants in a partialfraction is called the cover-up rule.

Another method is by expanding the right-hand side of equation (*) whichproduces

2 22 Ax A Bx Bx Cx C

or

22 ( ) ( ) ( ) A B x B C x A C


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This is an identity and hence the coefficients of x2, x and the constants in bothsides of the identity must be identical. Now, comparing the coefficients for each of them, we have

x2 : 0 A B

x : 0 B C

x0 (or 1) : 2 A C

The values for A, B and C can be found by solving those three equations.

Example 1.11

Express 2( 1)( 2) x

x x as partial fractions.

SolutionObserve that in this case the second factor of the denominator is a repeated factoras 2( 2) ( 2)( 2) x x x . Generally, any repeated factor of the form ( ax b)2 in the denominator will give

rise to two partial fractions of the form( )

A

ax band

2.

( )

B

ax b

Hence,

2 2( 1)( 2) ( 1) ( 2) ( 2) x A B C

x x x x x

.

In other words,

2( 2) ( 1)( 2) ( 1 x A x B x x C x ) And this is valid for all values of x.

Let us choose x 2 (to eliminate A and B). We get

2 A(0) 2 B(1)(0) C (2 1)

or

C 2


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Now let x 1 (to eliminate B and C ), we have

1 A(1 2) 2 B(0)(1 2) C (0)

or

A 1.

Finally, let us say we choose x 3. We get

3 A(3 2) 2 B(3 1)(3 2) C (3 1)

or

3 A 2 B 2C.

Substituting A 1 and C 2 obtained earlier gives

3 1 2 B 4

or B 1.

Therefore,

2 2

1 1 2( 1)( 2) ( 1) ( 2) ( 2)

x x x x x x

Note:Using similar technique, a repeated factor ( ax b)3 in the denominator will give

three partial fractions of the form ,( )

Aax b

2( ) B

ax band 3 .( )

C ax b

1.4.3 Improper Fractions

Example 1.12

Express3 3

( 1)( 1 x

x x )as partial fractions.


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1.Write the following expressions as partial fractions.

(a)1

( 2)( 2 x

x x )

(b)3

( 1) x x

x

(c)2

( 1)( 1 x

x x )

2. Determine the values of the constants A, B and C in the followingidentity:

3.

( 1)( 2)( 3) ( 1) ( 2) ( 3) x A B

x x x x x x C

(a) Polynomials

(i) Monomial is an algebraic expression of the form

axk

where a is a constant and x an unknown (or a variable), and k 0 is aninteger.

(ii) Polynomial in an algebraic expression of the form

11 1

n nn na x a x a x a 0

0

where are constants, n 0 and x is a variable.1 1, , , ,n na a a a

3. Assume that 23

( ) .( 4)( 2)

x f x

x x Express ( ) f x as partial

fractions .

EXERCISE 1.1


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The constants are known as the coefficients of the

polynomial and when a n 0, the polynomial1 1, , , ,n na a a a 0

0

11

n nn na x a x

1a x a is called a polynomial of degree n.

(b) Factorisation The process of writing an expression as a product of two or more factors iscalled a factorisation.

Important Formula:

a 2 + 2 ab + b2 = (a + b)(a + b) = ( a + b)2 a 2 2ab + b2 = (a b)(a b) = ( a b)2 a 2 b2 = (a + b)(a b)

Denominator

Equation

FactorisationFunctions

Identity

Inequalities

Monomial

Numerator

Partial fractionsPolynomial

Real numbers


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1. Write the quadratic equation with the following roots:

(a) 3 (repeating)

(b) j and k .

2. State expression

15

x x x

as partial fraction.

3. Obtain the partial fractions of

3.

2 x

x x

4. Solve 22 6 x x 0 without using the quadratic formula.

topic 1 concepts of algebra

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