Fundamental Concept of Algebra

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Transcript of Fundamental Concept of Algebra

PROJECT ON :-

FUNDAMENTAL CONCEPTS OF ALGEBRA.

MADE BY:-

SUNDEEP MALIK.

DEFINITION OF ALGEBRA:-

Algebra is a branch of mathematics that uses letters etc. to represent numbers and quantities.

ELEMENTARY TREATMENT:-1. CONSTANTS: In Arithmetic, we use digits

(0,1,2,3,4,5,6,7,8,9)each of which has a fixed value and so are called constants.

2. VARIABLES: We use letters of English alphabet which can be assigned any value according to the requirement. So the letters used in algebra are called variables.

3. TERM: A term is a number (constant),a variable or a combination (product or quotient) of number and variables. For e.g.:-

5, 3a, x, ax, -2xz, 4x/3y.etc. 4.ALGEBRIC EXPRESSION: An algebraic

expression is a collection of one or more terms, which are separated from each other by addition (+) or subtraction (-) sign(s). For e.g.:-5xy – 2 ,3x + 5z - 2xy , ax + by – cz + axy + dxyz.etc.

NAME

CONDITION EXAMPLES

i. Monomial

has only one term

X, 6xy, -4xy/9yz, etc.

ii. Binomial

has two terms 2a+x, 7z-15x, etc.

iii. Trinomial

has three terms ax2+bx+c, x2-xy-9y+x/2, etc.

iv.Multinomial

has more than three terms

4-a+ax+by, x2-2x+3xy-3y-b+ , etc.

v. Polynomial

has two or more than two terms.

Every binomial, trinomial & multinomial, etc.

5. TYPES OF ALGEBRIC EXPRESSION:-

6. PRODUCT : When two or more quantities (constants, variables or both) are multiplied together; the result is called their product. E.g.:- 4ay is the product of 4,a and y.

7.FACTOR: Each of the quantities (numbers or variables) multiplied together to form a term is called a factor of the term. E.g. :- 3,a and y are the factors of the term 3ay.

8.CO-EFFICIENT: In a monomial, any factor or a group of factors of a term is called the coefficient of the remaining part of the monomial. E.g. :- In 5xyz, 5 is the coefficient of xyz, x is the coefficient of 5yz, y is the coefficient of 5xz and so on.

9. TYPES OF CO-EFFICIENTS:-

i. NUMERICAL CO-EFFICIENTS: - If a factor is a numerical quantity , it is called numerical co-efficient. E.g. :- In 4abc, 4 is the numerical co-efficient.

ii. LITERAL CO-OEFFICIENT:- The factor involving letters is called the literal co-efficient. E.g. :- In 7xyz,

7, x , y , z , 7x , 7y , 7z , xy , yz , xz ,7xy ,7yz ,7xz ,xyz and 5xyz are the literal co-efficients.

10. DEGREE OF A MONOMIAL:- The degree of a monomial is the exponent of its variable or the sum of the exponents of its variable. E.g. :-

i. ii. iii.

11.DEGREE OF A POLYNOMIAL :- The degree of a polynomial is the degree of the highest degree term. E.g. :-

12.LIKE TERMS :- Terms having the same literal coefficients or alphabetic letters are called like terms. Like terms can be added, subtracted, multiplied and divided. E.g.:-

i. ii. iii.

13.UNLIKE TERMS:- Terms having different literal coefficients are called unlike terms. Unlike terms cannot be added, subtracted, multiplied or divided. E.g.:-

i. ii. iii.

ADDITION & SUBTRACTION( Like Terms )Method:- Add or subtract (as required) the

numerical coefficients of like terms. E.g. :-i. Addition of 8xy, 15xy and 3xy = 8xy + 15xy

+ 3xy = ( 8 + 15 +

3 ) xy = 26xy

( Ans )ii. Subtraction of 8xy from 15xy = 15xy – 8xy

= ( 15 – 8 )

xy = 7xy

( Ans )

Example 1 : Add: (i) 2ax + 3by + 4cy, 5by – 3cy – ax and

6cy + 4ax – 9by. Solution: 1st method : Re-write the given expression in

such a way that their like terms are one below the other, then operate (add or subtract, as the case may be) like terms column-wise. Thus:

2ax + 3by + 4cy - ax + 5by – 3cy 4ax – 9by + 6cy 5ax – by + 7cy (Ans.)

2nd method : Steps : i. Write each of the given polynomials

(expressions) in a bracket with plus sign(+) between the consecutive brackets.

ii. Remove the brackets without changing the sign of any term.

iii. Group the like terms and add. E.g.:- ( 2ab + 3yz + 4xz ) + ( 5yz – 3xz –ab ) +

( 6xz + 4ab – 9yz ) = 2ab + 3yz + 4xz + 5yz – 3xz –ab +

6xz + 4ab – 9yz = 2ab – ab + 4ab + 3yz + 5yz – 9yz +

4xz – 3xz + 6xz = 5ab – yz + 7xz

( Ans.)

Subtract: (i) 2a + 3b – c from 4a + 5b + 6c. Solution: 1st method:i. Write the given expressions in two rows in

such a way that the like terms are written one below the other, taking care that the expression to be subtracted is written in the second row.

ii. Change the sign of each term in the second row (lower row).

iii. With these new signs of the terms in lower row, add the like terms column wise.

Step 1: 4a + 5b + 6c 2a + 3b - c Step 2: - - + . Step 3: 2a + 2b + 7c . ( Ans.)

2nd method:i. Write both the expressions in a single row with the

expression to be subtracted in a bracket and put a minus sign (-) before(outside) this bracket (minus sign is for subtraction).

ii. Open the bracket by changing the sign of each term inside the bracket.

iii. Add the like terms. E.g. 4a + 5b + 6c – ( 2a + 3b – c )

[ Step 1 ] = 4a + 5b + 6c – 2a – 3b + c

[Step 2 ] = 4a - 2a + 5b – 3b + 6c + c = 2a + 2b + 7c ( Ans.)

[Step 3 ]

ADDITION & SUBTRACTION( Unlike Terms )

The two like terms can be added or subtracted to get a single term; but two unlike terms cannot be added or subtracted together to get a single term. All that can be done is to connect them by the sign as required.

E.g. :i. Addition of 5x and 7y = 5x + 7yii. Subtraction of 9y from 3xy = 3xy – 9y

MULTIPLICATIONMultiplication of monomials : Steps : 1. Multiply the numerical

coefficients together. 2. Multiply the literal coefficients

separately together. E.g. :i. 8x x 3y = ( 8 x 3 ) x ( x x y ) = 24 x xy = 24 xyii. 5a x 3b x 6c = ( 5 x 3 x 6 ) x ( a x b x c ) = 90 abc

Multiplication of a polynomial by a monomial :

Steps: 1. Write the given polynomial inside the brackets and the monomial outside it.

2. Multiply the monomial with each term of the polynomial and simplify.

E.g. :i. Multiplication of 2x + y – 8 and 4x = 4x ( 2x

+ y – 8 ) = 4x x

2x + 4x x y – 4x x 8 =

(Ans.)

Multiplication of a polynomial by a polynomial : Steps: 1. Multiply each term of one polynomial by

each term of other polynomial. 2. Combine (add or subtract) the like terms. E.g.: Multiplication of a + b and 2a + 3b, a + b 2a + 3b [Multiplying first

polynomial by 2a] ……………………………………… [Multiplying

first polynomial by 3b] ……………………………………… [Adding like

terms]

Alternative method: Multiplication of a + b and 2a + 3b =

(a + b) (2a + 3b)

= a(2a + 3b) + b(2a + 3b) =

a(2a) + a (3b) + b (2a) + (3b) =

= .

DIVISIONDividing a monomial by a monomial : Steps: 1.Write the dividend in numerator

and divisor in denominator. 2. Simplify the fraction, obtained in

Step (1). E.g.i. Division of 15xy by 5x = = 3y .

ii. .

iii. .

Dividing a polynomial by a monomial : Divide each term of the polynomial by

the monomial: E.g. :i. Division of

.

ii.

Dividing a polynomial by a polynomial: E.g.: Divide: by Step 1: Set the two expressions as: Step 2: Divide first term of the dividend by the first term

of divisor to get the first term of quotient. Here, ;

which is the first term of the quotient .

Step 3: Multiply quotient (2x) with each term of divisor and write as :

Step 4: Subtract the result of Step 3 and take the next term/terms of the dividend down :

Step 5 : Repeat the process from Step 2 to Step 4 taking remainder

15x + 10 as new dividend : Thus,

(Ans.)

The End