Elementary Algebra

 

Elementary Algebra (Application of Algebraic Formula & SURDS)

 

ALGEBRAIC EXPRESSION

Algebraic expression consists of two types of symbols i.e., constants and variables with fundamental arithmetic operations\[(+,-,\div ,\times ).\]

(i) Constants Constants are fixed numerals

i.e., 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.

(ii) Variables Variables are the symbols representing any numeral.

e.g.,\[3{{x}^{2}}+4{{x}^{2}}+4xy-10x,\]\[2x+5y,\] etc. are algebraic expression.

 

POLYNOMIALS

Polynomials are the type of algebraic expression whose variable have only positive integer as power.

 

Degree of a Polynomial

For one variable: The highest power of the variable is called the degree of the polynomial.

e.g.,                  \[2x+3,\]degree = 1

\[3{{x}^{2}}+5,\] degree = 2 etc.

For Two or More Variables: The sum of the variables in each term is calculated and the highest sum so obtained is called degree of the polynomial.

e.g.,      \[3{{x}^{2}}y+5{{x}^{4}}-7{{x}^{2}}{{y}^{2}}+3x+2y-10;\]degree = 5

\[5{{x}^{3}}{{y}^{2}}z+3{{y}^{3}}{{z}^{2}}-6yz,\] degree = 6

 

Fundamental Operations on Algebraic Expression

I. Method of Addition

Step I   Collect different groups of like term.

Step II Find the sum of the numerical coefficients of like

              terms in each group.

Step III Write the required terms of the like terms.

e.g.,      \[4{{x}^{2}}+7{{x}^{2}}y+{{y}^{2}}+2{{x}^{2}}y+10{{x}^{2}}\]

\[-\,5{{y}^{2}}+16{{x}^{2}}-3{{y}^{2}}\]

Sol.      \[(4{{x}^{2}}+10{{x}^{2}}+16{{x}^{2}})+(7{{x}^{2}}y+2{{x}^{2}}y)\]

\[+\,({{y}^{2}}-5{{y}^{2}}-3{{y}^{2}})=30{{x}^{2}}+9{{x}^{2}}y-7{{y}^{2}}\]

 

II. Method of Subtraction

Step I Reverse the sign (from\['+'\]to \['-'\]and from\['-'\]to\['+'\])

 of all the terms of the expression which is to be subtracted.

Step II Follow the method used in addition.

e.g., Subtract\[(-\,2{{y}^{2}}+y-6)\]from\[(5{{y}^{2}}-4y+14)\]

Sol.      \[(5{{y}^{2}}-4y+14)-(-2{{y}^{2}}+y-6)\]

\[=5{{y}^{2}}-4y+14+2{{y}^{2}}-y+6\]

\[=(5{{y}^{2}}+2{{y}^{2}})+(-\,4y-y)+(14+6)\]

\[=7{{y}^{2}}-5y+20\]

 

III. Methods of Multiplication

Rule 1 Product of two factor with like sign is positive and

the product of two factor with unlike sign is negative.

i.e.,\[(+)\times (+)=+,\]\[(+)\times (-)=-,\]\[(-)\times (+)=-,\]                                              \[(-)\times (-)=+\]

Rule 2 If x is a variable and m, n are positive integer, then

\[{{x}^{m}}\times {{x}^{n}}={{x}^{m+n}}\]

e.g., Simplify \[(7xy)\times (5y).\]

Sol.      \[(7\times 5)(x{{y}^{1}}\times {{y}^{1}})=35\times x({{y}^{1+1}})=35x{{y}^{2}}\]  

e.g., Simplify \[(3x+5y)\times (7x-3x).\]

Sol.      \[3x\,(7x-3y)+5y\,(7x-3y)\]

\[=21{{x}^{2}}-9xy+35xy-15{{y}^{2}}=21{{x}^{2}}-15{{y}^{2}}+26xy\]

IV. Methods of Division

Step I Arrange the terms of both the polynomials in descending order.

Step II Divide the first term of the dividend by the first term of the divisor to obtain the first term of the quotient.

Step III Multiply all terms of the divisor by the first of the quotient and subtract the result from the dividend.

Step IV Consider the remainder (if any) as a new dividend and proceed as before.

Step V Repeat this process till we obtain a remainder which is either 0 or polynomial of degree less than that of the divisor.

e.g., Divide\[3{{x}^{4}}-3{{x}^{4}}-4{{x}^{2}}-4x\]by \[{{x}^{2}}-2x.\]

Sol.     

           

Hence, result\[=3{{x}^{2}}+3x+2\]

 

Factorization of Algebraic Expression

Factorization of an algebraic expression is the process of writing it in the form of the product of two or more algebraic expression.

Methods of Factorization of the Polynomial

\[\mathbf{a}{{\mathbf{x}}^{\mathbf{2}}}\mathbf{bx+c}\]

Splitting the Middle Term Coefficient of middle term is split as sum of two numbers such that the product of two members = Coefficient of \[{{x}^{2}}\times \] Constant term = \[a\times c\]

e.g., Factorize \[2{{x}^{2}}-x-15.\]

Sol.      According to the Equation,

\[ac=-15\times 2=-30\]

\[=-\,5\times 6=-\,6\times 5\]

\[=-10\times 3=-\,3\times 10\]

\[=-15\times 2\]

\[=30\times -1\]

Out of the possible factors, \[-\,6+5=-1\]is proper

\[\therefore \]     

Write as it is Middle term has been split as it is

\[=2x(x-3)+5(x-3)=(x-3)(2x+5)\]

e.g.,      Factorize\[{{x}^{2}}-y+xy+x.\]

Sol.      \[{{x}^{2}}-y+xy+x={{x}^{2}}+xy+x-y\]

\[=x(x+y)-1(x+y)=(x+y)(x-1)\]

e.g.,      Factorize\[144{{x}^{2}}-64{{y}^{2}}.\]

Sol.      \[144{{x}^{2}}-64{{y}^{2}}={{(12x)}^{2}}-{{(8y)}^{2}}\]\[=(12x+8y)(12x-8y)\]

[Using \[{{a}^{2}}-{{b}^{2}}=(a+b)(a-b)\]]

Application of Algebraic Formulae

Basic Algebraic Identities

Ø  \[{{a}^{2}}+2ab+{{b}^{2}}={{(a+b)}^{2}}\]  

[Square of the sum of two terms]

Ø  \[{{a}^{2}}-2ab+{{b}^{2}}={{(a-b)}^{2}}\]

[Square of difference of two terms]                   

Ø  \[{{a}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}+{{b}^{3}}={{(a-b)}^{3}}\] or \[{{a}^{3}}-{{b}^{3}}-3ab\,(a+b)\]

[Cube of the sum of two terms]

Ø  \[{{a}^{3}}-3{{a}^{2}}b+3a{{b}^{2}}-{{b}^{3}}={{(a-b)}^{3}}\] or \[{{a}^{3}}-{{b}^{3}}-3ab\,(a-b)\]

[Cube of the difference of two terms]

Ø  \[({{a}^{3}}-3{{a}^{3}}b+3a{{b}^{2}}-{{b}^{3}})={{(a-b)}^{3}}\] or \[{{(a+b)}^{3}}-3ab\,(a-b)\]

[Sum of cubes of two terms]

Ø  \[(a-b)({{a}^{2}}+ab+{{b}^{2}})={{a}^{3}}-{{b}^{3}}\]or \[{{(a-b)}^{3}}-3ab\,(a-b)\]

[Difference of cubes of two terms]

Ø  \[{{(a+b)}^{2}}-2ab={{a}^{2}}+{{b}^{2}}\]or \[{{(a-b)}^{3}}+3ab\,(a-b)\]

[Sum of squares of two terms]     

Ø  \[(a+b)(a-b)={{a}^{2}}-{{b}^{2}}\]        

[Difference of two squares]          

Ø  \[\frac{{{(a+b)}^{2}}}{4}-\frac{{{(a-b)}^{2}}}{4}=ab\]              

[Product of two terms]         

Ø  \[{{a}^{2}}+{{b}^{2}}+{{c}^{2}}2ab+2ac+2bc={{(a+b+c)}^{2}}\]

[Square of sum of three terms]    

Ø  \[{{(a-b)}^{2}}+4ab={{(a+b)}^{2}}\]

Ø  \[{{(a+b)}^{2}}-4ab=(a-b)\]

Ø  \[{{a}^{3}}+{{b}^{3}}{{c}^{3}}-3abc=(a+b+c)({{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ca)=\frac{1}{2}(a+b+c)[{{(a-b)}^{2}}+{{(b-c)}^{2}}+{{(c-a)}^{2}}]\]

Ø  If \[a+b+c=0,\]then \[{{a}^{3}}+{{b}^{3}}+{{c}^{3}}=3abc\]

Ø  \[{{a}^{3}}+{{b}^{3}}+{{c}^{3}}+(a+b)(b+c)(c+a)={{(a+b+c)}^{3}}\]

Ø  \[(x+a)(x+b)={{x}^{2}}+(a+b)x+ab\]

Ø  \[{{x}^{2}}+\frac{1}{{{x}^{2}}}={{\left( x+\frac{1}{x} \right)}^{2}}-2={{\left( x-\frac{1}{x} \right)}^{2}}+2\]

 

Application of surds 

Ø  \[{{(\sqrt{a}+\sqrt{b})}^{2}}=a+b+2\sqrt{ab}\]

Ø  \[{{(\sqrt{a}-\sqrt{b})}^{2}}=a+b-2\sqrt{ab}\]

Ø  \[(\sqrt{a}+\sqrt{b})(\sqrt{a}-\sqrt{b}=a-b)\]

Ø  \[\frac{1}{\sqrt{a}-\sqrt{b}}=\frac{\sqrt{a}-\sqrt{b}}{a-b}\]

Ø  \[\frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}=\frac{a+b+2\sqrt{ab}}{a-b}\]

 

Some Important Calculation Related to surds

(i) \[\sqrt{a}+n\sqrt{a}-p\sqrt{a}=(m+n-p)\sqrt{a}\]

(ii) \[\sqrt{a}\times \sqrt{b}\times \sqrt{c}\times ...=\sqrt{a\times b\times c\,x...}\]

(iii) \[\sqrt{a}\div \sqrt{b}=\sqrt{a+b}\]

(iv) \[\sqrt{a}\times \sqrt{a}=a\]