"Mathematics is a easy subject but maximum find them so hard to learn and remember Mathematics functions and rules. That all because we not give time to understand basics of Mathematics rule. So learn here the basic rules step by step"
I try to explain all topics in easy tips or steps so a learner can easily understand basic of mathematics. I try to give much knowledge in short notes. All explain with necessary examples, I sure if you read well my topic and try to solve all examples which I give here, then you can solve any problem without any teacher.
(A), POLYNOMIALS
Polynomials are algebraic expressions that consist of Variables , Coefficients , and Constant .
We can do easily addition , subtraction , multiplication , division and also exponential operations on polynomials but never be division by variables.
Standard form of polynomial is : P(X) = aX2+bX+c
Where P(X) = Given Polynomial,
a, b = Coefficients,
C = Constant Value,
X, Y = Variables,
In X2 , X ; 2 and 1 are exponents.
and + or - symbol called Operators.
Example : 2, 5, √2 , √5 , 21 , 35 , 5∕∕2 X , 12X + 5 XY + 6Z2 +0.5 are Polynomials.
(BUT) 1/X , (1/X) +3 , X−2 , √X , 3X−2, 2/ (X+5) are not Polynomials.
NOTE : All numbers are polynomials because we can arrenge numbers in polynomial form.
Example : for number 2 , we can arrenge as ; 0X2 + 0X + 2 etc.
1. CLASSIFICATION OF POLYNOMIAL :
There are 3 types of Polynomials which are ;-
1. Monomial
2. Binomial
3. Trinomial
1. Monomial :
That type polynomial consist only one term .
EXAMPLE : 2X , 5 , -2XY , 6Z2,6X2 , etc.
2. Binomial :
That type polynomial consist two terms .
EXAMPLE : 2X+3 , -5X+7 , 6X2+2X , 6X2Y+XY , etc.
3. Trinomial :
That type polynomial consist three terms .
EXAMPLE : 6X2- 2X+3 , 6X2-5X+7 , etc.
2. DEGREE OF POLYNOMIAL :
" Degree of Polynomial is a highest degree of a monomial in a polynomial. "
We can classify polynomial accordind to degree of polynomial , such as -
(a). ZERO POLYNOMIAL : Here degree of polynomial is zero (0).
Example : P(X) = 9
(b). LINEAR POLYNOMIAL : Here degree of polynomial is ONE (1).
Example : P(X) = 9X + 5
(c). QUADRATIC POLYNOMIAL : Here degree of polynomial is two (2).
Example : P(X) = 9X2 + 5X + 2
(d). CUBIC POLYNOMIAL : Here degree of polynomial is three (3).
Example : P(X) = 5X3 + 9X2 + 5X + 2
(e). QUARTIC POLYNOMIAL : Here degree of polynomial is four (4).
Example : P(X) = 5X4+ 5X3 + 9X2 + 5X + 2
3. TERMS OF POLYNOMIAL :
The parts of polynomial which separated by sign " + ' and " - " called terms of polynomial.
If P (X) = 9X2 + 5X - 2 then ,
Degree of polynomial = 2 ,
Terms of polynomial = 3 (As 9X2 , 5X , 2)
4. DIVISIBILITY RULE AND FACTOR THEOREM :
" If polynomial is P(X) , then P(X) is divisible by a binomial ( X - a) if and only P(a) = 0 "
If in polynomial P(X) ; P(a) = 0 then ( X - a) is a factor of given polynomial (OR) if P(-a) =0 ; then ( X + a) is a factor of given polynomial.
5. ZERO OF THE POLYNOMIAL P(X) :
If in polynomial P(X) = aX2+bX+c ;
If P(m) = 0 , then m is a Zero of Polynomial of the given polynomial.
EXAMPLE :
If P(X) = 5X2-5
Here, P(1) = 5( 1 )2-5 = 5 - 5 = 0
⇒ P(1) = 0
Hence 1 is a Zero of polynomial P(X).
NOTE : 1. Every linear polynomial has one and only one Zero.
2. A polynomial can have more than one Zero.
3. " 0 " may also be a Zero of a polynomial.
6. REMAINDER THEOREM :
" If a polynomial P(X) is divided by (X - a) ; then the remainder will be P(a) . "
If in polynomial P(X) if P(a) = 0 ; then a is a Zero of polynomial , but if P(a) ≠ 0 and
give some value then that value called remainder.
EXAMPLE 1 :
If P(X) = 3X2 + X - 1
Here , P( -1 ) = 3( -1 )2 + ( -1 ) - 1 = 3 - 1 -1 = 1
Here , 1 is remainder and ( 3X2 + X - 1 ) is ' dividend ' , ( X + 1 ) will be ' devisor '.
DIVIDEND = ( DIVISOR x QUOTIENT ) + REMAINDER
By divide method we can understand that problem well ; such as above equation :--
X + 1 ) 3X2 + X - 1 ( 3X - 2 ← ( QUOTIENT )
3X2 + 3X
- -
________________
- 2X - 1
- 2X - 2
+ +
_________________
1 ( REMAINDER )
EXAMPLE 2 :
Find the remainder if 2X4+ X3 - X2 + 3X + 1 is divided by X + 1 .
SOLUTION :
P(X) = 2X4+ X3 - X2 + 3X + 1
Here , Zero of ( X + 1 ) is ( - 1 )
Hence ,
P( -1 ) = 2( -1 )4+ ( -1 )3 - ( -1 )2 + 3( -1 ) + 1
= 2 - 1 -1 -3 + 1 = -2
Hence , -2 is the remainder when 2X4+ X3 - X2 + 3X + 1 is divided by X + 1 .
NOTE : If in polynomial P(X) ; if P(a) = 0 ; then (X - a) will be the factor of the given polynomial.
( Since X = a ; so that X - a = 0 )
7. FACTORISATION :
FACTOR :
" If an algebraic ex[pression can be written as the product of two or more expressions , then the each of these expressions are called Factor of the given expression. "
FACTORISATION
"To divide an expression in two or more small expressions called factorisation . Each small expression will be called the factor of given expression."
FACTOR THEOREM
" If in a polynomial P(x) , "a" is any real number,then if P(a) = 0 ; then (x-a) is a factor of given polynomial P(x). "
7(1). METHODS OF FACTORISATION :
(A). BY TAKE COMMON FACTOR :
EXAMPLE 1. :
5x2y + 125 xy2 - 25 x3y2
= 5xy ( x + 25 y - 5x2y )
EXAMPLE 2. :
28 ( 5x + 9y )2 + 21( 5x + 9y )
= 7 (5x + 9y ) ⟨ 4 (5x + 9y )+3(1)⟩
= 7 (5x + 9y ) ( 20x + 36y + 3 )
(B). FACTORISATION BY GROUPING :
If there possible to arrange given expression in suitable groups in which groups have a common factor , then we can solve expression by use common factor method after arrange in group.
EXAMPLE :
4ax2 + 6by2 + 2bx2 + 12ay2
⇒ 4ax2 + 2bx2 + 6by2 + 12ay2
⇒ 2x2 (2a+b) + 6y2(b+2a)
⇒ (2a+b) (2x2+6y2)
⇒ (2a+b) 2 (x2+3y2)
⇒ 2 (2a+b) (x2+3y2)
(C). BY USE SPECIAL IDENTITIES :
Following are main identities which can be use in factorisation of given expression :-
1. (a+b)(a-b) = a2-b2 2. (a+b)2 = a2+2ab+b2
3. (a-b)2 = a2-2ab+b2
EXAMPLE :
1. 25 - 16x2
= (5)2-(4x)2
By use a2-b2 = (a+b)(a-b) ; we get-
= (5+4x)(5-4x)
2. 75a2-48b2 = 3(25a2-16b2)
= 3⟨(5a)2-(4b)2⟩
By use a2-b2 = (a+b)(a-b) ; we get-
= 3(5a+4b)(5a-4b)
3. 16(a+b)2-25(a-2b)2
= ⟨4(a+b)⟩2-⟨5(a-2b)⟩2
By use a2-b2 = (a+b)(a-b) ; we get-
= ⟨4(a+b)+5(a-2b)⟩ ⟨4(a+b)-5(a-2b)⟩
= (4a+4b+5a-10b) (4a+4b-5a+10b)
= (9a-6b) (14b-a)
4. 16x2-4y2+20y-25
=16x2- (4y2-20y+25)
By use identity (x-y)2 = x2-2xy+y2 ; we get -
= (4x)2 - (2y-5)2
By use a2-b2 = (a+b)(a-b) ; we get-
= ⟨4x+(2y-5)⟩ ⟨4x-(2y-5)⟩
= (4x+2y-5)(4x-2y+5)
5. Evalute : (410)2-(390)2
(410)2-(390)2 = (400+10)2-(400-10)2
= ⟨(400+10)+(400-10)⟩ ⟨(400+10)-(400-10)⟩
= 800 x 20 = 16000
6. 25x2+10x+1
= (5x)2+2(5x)(1)+(1)2
By use identity (x+y)2 = x2+2xy+y2 ; we get -
= (5x+1)2
(D ) SOME SPECIAL EXAMPLES ON FACTORISATION OF POLYNIMIALS :
EXAMPLE 1. If x+1/x = 5 then find the value of -
(1). x2+1/x2 (2). x4+1/x4
SOLUTION :
(1). (x+1/x) = 5
⇒ (x+1/x)2 = (5)2
⇒ x2+1/x2+2(x)(1/x) = 25
⇒ x2+1/x2+2= 25
⇒ x2+1/x2= 23
(2). (X2+1/X2)2 = (23)2
⇒ X4+1/X4+2X2x1/X2 = 529
⇒ X4+1/X4 = 527
EXAMPLE 2. If x-1/x = 5 then find the value of -
(1). x2+1/x2 (2). x4+1/x4
SOLUTION :
(1). (X-1/X)2 =(5)2
⇒X2+1/X2-2Xx1/X = 25
⇒ X2+1/X2-2= 25
⇒ X2+1/X2 = 27
(2). (X2+1/X2)2 =(27)2
⇒X4+1/X4+2X2x1/X2 = 729
⇒ X4+1/X4= 727
EXAMPLE 3. (2X+3)(2X-3)(4X2+9)
SOLUTION : (2X+3)(2X-3)(4X2+9)
⇒ (4X2-9)(4X2+9) ( By use identity : (a+b)(a-b) = a2-b2 )
⇒ (4X2)2-(9)2 ( By use identity : (a+b)(a-b) = a2-b2 )
⇒ 16x4-81
EXAMPLE 4. If x+y = 15 , xy =20. Find the value of (x2+y2)
SOLUTION :
Here , x+y = 15
Hence, (x+y)2 =(15)2
⇒x2+y2+2xy = 225
⇒x2+y2+2(20) = 225
⇒x2+y2= 225 -40
⇒x2+y2= 185
EXAMPLE 5. If x-y = 15 , xy =20. Find the value of (x2+y2)
SOLUTION :
Here , x-y = 15 Hence, (x-y)2 =(15)2
⇒x2+y2-2xy = 225
⇒x2+y2-2(20) = 225
⇒x2+y2= 225 +40
⇒x2+y2= 265
EXAMPLE 6. Examine that X+1 is a factor of 3x2+x-2
SOLUTION :
Here zero of x+1 is -1.
Let, P(x) = 3x2+x-2 ⇒P(-1) = 3(-1)2+(-1)-2
⇒P(-1) = 3-1-2 = 0
So, by the factor theorem, (x+1) is a factor of given polynomial 3x2+x-2
EXAMPLE 7. Find the value of K if (x+1) is a factor of 5x4+2x3-4x+K
SOLUTION :
Here P(x) = 5x4+2x3-4x+K
Since (x+1) is a factor of P(x), hence
P(-1) = 0
⇒P(x) =5x4+2x3-4x+K
⇒P(-1) = 5(-1)4+2(-1)3-4(-1)+K = 0
⇒5-2+4+K = 0
⇒7+K = 0
⇒ K = -7
EXAMPLE 7. Find the value of x if equation is 5x-15
SOLUTION :
( Method: To solve polynomial equations we need to take "0" at right hand sides of equation; then use basic algebraic methods and factorisation concepts to solve equation.)
P(x) = 5x-15
⇒ 5x-15 = 0
⇒5x=15
⇒ x = 15/5 = 3
Hence,
x=3
EXAMPLE 8. Find the value of x if equation is 3x2-x+x3-3.
SOLUTION :
( Method: To solve polynomial equations we need to arrenge the equation in descending order of degree of coefficients then equal that equation as zero.)
3x2-x+x3-3
⇒ x3+3x2-x-3=0
⇒ x2(x+3)-1(x+3)=0
⇒ (x2-1)(x+3)=0
Hence,
If x2-1=0
⇒ x2 = 1
⇒ x = 1, -1 and,
If x+3 = 0
⇒ x = -3
Hence,
Solutions are = 1, -1, -3
EXAMPLE 9. Find the value of Polynomial as indicated the value of variable:-
(1). P(x) = 6x2-5x+5 ,at x=2
(2). P(m) = 4m4+5m3-2m2+2 ,at m=a
SOLUTION :
(1). P(x) = 6x2-5x+5
⇒ P(2) = 6(2)2-5(2)+5
= 24-10+5 = 19
(2). P(m) = 4m4+5m3-2m2+2
⇒ 4(a)4+5(a)3-2(a)2+2
⇒ 4a4+5a3-2a2+2
7(2). FACTORISATION OF QUADRATIC EXPRESSIONS :
Here the expression will be in the following form :-
ax2+bx+c
For factorisation we need to divide middle term " bx " in two parts such as which product (multiply) equals to (axc) and the sum should be eqyals to " b " .
EXAMPLE 1. Find the value of x for the expression 2x2+11x+15 .
SOLUTION : P(X) = 2x2+11x+15
Here the product of first and last term's coefficients = 2x15 =30
Now we need to divide 30 in two parts , which sum = (+11)
Here we get 6+5=11, 6x5=30
Hence,
2x2+11x+15 = 2x2+6x+5x+15
⇒ 2x(x+3)+5(x+3)
⇒ (x+3)(2x+5)
EXAMPLE 2. Find the value of x for the expression 2x2-x-15 .
SOLUTION : P(X) = 2x2-x-15
Here the product of first and last term's coefficients = 2x15 =30
Now we need to divide 30 in two parts , which sum = (-1)
Here we get -6+5=-11, -6x5=-30
Hence,
2x2-x-15 = 2x2-6x+5x-15
⇒ 2x(x-3)+5(x-3)
⇒ (x-3)(2x+5)
(B). LINEAR EQUATIONS IN ONE VARIABLE
1. EQUATION :
A statement of equality of two algebraic expressions which have or more variables is called equation.
2. LINEAR EQUATION :
If any equation having any linear polynomials is called linear equation.
In other words " A mathematical statement that has an equal to the "=" symbol is called an equation. Linear equation are of degree 1."
EXAMPLES :
2X+4 = 9X
2(X+2) = 5
1/3 X -2 = 2X+9
3. SOLUTION OF LINEAR EQUATION IN ONE VARIABLE :
We can understand the method of solve Linear equations in one variable by these examples :--
EXAMPLE 1. Find the value of x for the expression 2x+4=9x-7 .
SOLUTION :
2x+4=9x-7
We need to carry all x value in one sides and numbers in other side. By solve that result ,we can find unknown value of x .
(NOTE : When we change sides of a given term then sign will be change.)
⇒ 2x-9x= -7-4
⇒ -7x = -11
⇒ x = -11/(-7) = 11/7
Hence, x = 11/7
EXAMPLE 2. Find the value of x for the expression :-
(x+5)/4 + (x-3)/5 = (5x-8)/2
SOLUTION :
(x+5)/4 + (x-3)/5 = (5x-8)/2
Here LCM of 4,5,2 is 20. So, multiply equation by 20 both sides we get-
⇒ 20 ⟨ (x+5)/4 + (x-3)/5 ⟩ = 20 ⟨ (5x-8)/2 ⟩
⇒ 5(x+5)+4(x-3) = 10(5x-8)
⇒ 5x+25+4x-12 = 50x-80
⇒ 5x+4x-50x = -80-25+12
⇒ -41x = -93
⇒ x = -93/(-41)
Hence,
x = 93/41
EXAMPLE 3. Find the value of x for the expression :-
(5x+7)/(2x+4) = (15x+2)/(6x+5)
SOLUTION :
(5x+7)/(2x+4) = (15x+2)/(6x+5)
By cross multiply we get result-
⇒ (5x+7) (6x+5) = (2x+4) (15x+2)
⇒ 5x(6x+5)+7(6x+5) = 2x(15x+2)+4(15x+2)
⇒ 30x2+25x+42x+35 = 30x2+4x+60x+8
⇒ 30x2-30x2+25x+42x-4x-60x = 8-35
⇒ 3x = -27
⇒ x = -27/3 = -9
Hence,
x = -9
(C). LINEAR EQUATIONS IN TWO VARIABLES
1. STANDARD FORM OF LINEAR EQUATIONS IN TWO VARIABLE IS :-
Y=mx + b
Where, x,y = variables ,
m = Slop or gradient,
b = y- intercept
We can represent linear equations in two variables on straight line in graph. Above equation also called "Slop Intercept equation of a straight line".
For Example, if y=3x+5
then, Slop m=3 , Intercept b=5
2. POINT-SLOP FORM OF EQUATION :-
y-y1=m(x-x1)
EXAMPLE :
Y-5=2/3(X-2)
Here,
y1=5 , x1=2 , m=2/3
2. GENERAL FORM OF STRAIGHT LINE:-
Ax+By=C , or
Ax+By+c=0
Where, A,B ≠ 0
x,y are variables,
A,B are coefficients, and
c is constant
Here, A,B,C are real numbers.
EXAMPLE :
2x+3y-5=0
Where, A=2, B=3, C=-5
NOTE THAT :
1. These are not linear equations :-
y2-5=0 ,
y2/3=15 ,
5√x-y=8
In linear equation exponents such as x2, x3 etc. and square roots or cube roots not allowed.
2. These are not linear equations :-
y=3x+5 ,
y+2=3(x+5) ,
y-3x+5=0 ,
3x=6 ,
x/3 =5
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