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NCERT Class 12 maths solutions, CBSE Class 12 answers
NCERT Class 12 maths solutions, CBSE Class 12 answers
CBSE Class 12 answers of mathematics
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NCERT Class 12 maths solutions
NCERT Solutions for Class 12 Maths Chapter-wise PDF
Polynomials are as expressions which are composed of two algebraic terms. In this lesson, all the concepts of polynomials like its definition, terms and degree, types, functions, formulas and solution are covered along with solved example problems. Below is the list of topics covered in this lesson.
A polynomial is defined as an expression which contains two or more algebraic terms. It is made up of two terms namely Poly (meaning “many”) and Nominal (meaning “terms.”). Polynomials are composed of:
Constants such as 1, 2, 3, etc.
Variables such as g, h, x, y, etc.
Exponents such as 5 in x5 etc.
Degree of a Polynomial
The degree of a polynomials is defined as the highest degree of a monomial within a polynomial. Thus, a polynomial equation having one variable which has the largest exponent is called a degree of the polynomial.
Example: Find the degree of the polynomial 6s4+ 3x2+ 5x +19
Solution:
The degree of the polynomial is 4.
Terms of a Polynomial
The terms of polynomials are the parts of the equation which are generally separated by “+” or “-” signs. So, each part of a polynomial in an equation is a term. For example, in a polynomial, say, 2x2 + 5 +4, the number of terms will be 3. The classification of a polynomial is done based on the number of terms in it.
Types of Polynomials
Polynomials are of 3 different types and are classified based on the number of terms in it. The three types of polynomials are:
Monomial
Binomial
Trinomial
These polynomials can be combined using addition, subtraction, multiplication, and division but is never division by a variable. A few examples of Non Polynomials are: 1/x+2, x-3
Monomial
A monomial is an expression which contains only one term. For an expression to be a monomial, the single term should be a non-zero term. A few examples of monomials are:
5x
3
6a4
-3xy
Binomial
A binomial is a polynomial expression which contains exactly two terms. A binomial can be considered as a sum or difference between two or more monomials. A few examples of binomials are:
– 5x+3,
6a4 + 17x
xy2+xy
Trinomial
A trinomial is an expression which is composed of exactly three terms. A few examples of trinomial expressions are:
– 8a4+2x+7
4x2 + 9x + 7
Polynomial Properties and Theorems
Some of the important properties of polynomials along with some important polynomial theorems are as follows:
Property 1: Division Algorithm
If a polynomial P(x) is divided by a polynomial G(x) results in quotient Q(x) with remainder R(x), then,
P(x) = G(x) • Q(x) + R(x)
Property 2: Bezout’s Theorem
Polynomial P(x) is divisible by binomial (x – a) if and only if P(a) = 0.
Property 3: Remainder Theorem
If P(x) is divided by (x – a) with remainder r, then P(a) = r.
Property 4: Factor Theorem
A polynomial P(x) divided by Q(x) results in R(x) with zero remainders if and only if Q(x) is a factor of P(x).
Property 5: Intermediate Value Theorem
If P(x) is a polynomial, and P(x) ≠ P(y) for (x < y), then P(x) takes every value from P(x) to P(y) in the closed interval [x, y].
Property 6:
The addition, subtraction and multiplication of polynomials P and Q result in a polynomial where,
Degree(P ± Q) ≤ Degree(P or Q)
Degree(P × Q) = Degree(P) + Degree(Q)
Property 7:
If a polynomial P is divisible by a polynomial Q, then every zero of Q is also a zero of P.
Property 8:
If a polynomial P is divisible by two coprime polynomials Q and R, then it is divisible by (Q • R).
Property 9:
If P(x) = a0 + a1x + a2x2 + …… + anxn is a polynomial such that deg(P) = n ≥ 0 then, P has at most “n” distinct roots.
Property 10: Descartes’ Rule of Sign
The number of positive real zeroes in a polynomial function P(x) is the same or less than by an even number as the number of changes in the sign of the coefficients. So, if there are “K” sign changes, the number of roots will be “k” or “(k – a)”, where “a” is some even number.
Property 11: Fundamental Theorem of Algebra
Every non-constant single-variable polynomial with complex coefficients has at least one complex root.
Property 12:
If P(x) is a polynomial with real coefficients and has one complex zero (x = a – bi), then x = a + bi will also be a zero of P(x). Also, x2 – 2ax + a2 + b2 will be a factor of P(x).
Polynomial Equations
The polynomial equations are those expressions which are made up of multiple constants and variables. The standard form of writing a polynomial equation is to put the highest degree first then, at the last, the constant term. An example of a polynomial equation is:
b = a4 +3a3 -2a2 +a +1
Polynomial Functions
A polynomial function is an expression constructed with one or more terms of variables with constant exponents. If there are real numbers denoted by a, then function with one variable and of degree n can be written as:
Any polynomial can be easily solved using basic algebra and factorization concepts. To solve a polynomial equation, the first step is to set the right-hand side as 0. The explanation of a polynomial solution are explained for two types:
Solving Linear Polynomials
Solving Quadratic Polynomials
Solving Linear Polynomials
Getting the solution of linear polynomials is easy and simple. First, isolate the variable term and make the equation as equal to zero. Then solve as basic algebra operation. An example of finding the solution of a linear equation is given below:
Example: Solve 3x – 9
Solution:
First, make the equation as 0. So,
3x – 9 = 0
⇒ 3x = 9
⇒ x = 9/3
Or, x = 3.
Thus, the solution of 3x-9 is x = 3.
Solving Quadratic Polynomials
To solve a quadratic polynomial, first, rewrite the expression in the ascending order of degree. Then, equate the equation and perform polynomial factorization to get the solution of the equation. An example to find the solution of a quadratic polynomial is given below for better understanding.
Example: Solve 3x2 – 6x + x3 – 18
Solution:
First, arrange the polynomial in the ascending order of degree and equate to zero.
⇒ x3 + 3x2 -6x – 18 = 0
Now, take the common terms.
x2(x+3) – 6(x+3)
⇒ (x2-6)(x+3)
So, the solutions will be x =-3 and
x2 = 6
Or, x = √6
Polynomial Operations
There are four main polynomial operations which are:
Addition of Polynomials
Subtraction of Polynomials
Multiplication of Polynomials
Division of Polynomials
Each of the operations on polynomials are explained below using solved examples.
Addition of Polynomials
To add polynomials, always add the like terms i.e. the terms having the same variable and power. The addition of polynomials always results in a polynomial of the same degree. For example,
Example: Find the sum of two polynomials: 5x3+3x2y+4xy−6y2, 3x2+7x2y−2xy+4xy2−5
Solution:
First, combine the like terms while leaving the unlike terms as they are. Hence,
(5x3+3x2y+4xy−6y2)+(3x2+7x2y−2xy+4xy2−5)
= 5x3+3x2+(3+7)x2y+(4−2)xy+4xy2−6y2−5
= 5x3+3x2+10x2y+2xy+4xy2−6y2−5
Subtraction of Polynomials
Subtracting polynomials is similar to addition, the only difference being the type of operation. So, subtract the like terms to obtain the solution. It should be noted that subtraction of polynomials also result in a polynomial of the same degree.
Example: Find the difference of two polynomials: 5x3+3x2y+4xy−6y2, 3x2+7x2y−2xy+4xy2−5
Solution:
First, combine the like terms while leaving the unlike terms as they are. Hence,
(5x3+3x2y+4xy−6y2)-(3x2+7x2y−2xy+4xy2−5)
= 5x3-3x2+(3-7)x2y+(4+2)xy-4xy2−6y2+5
= 5x3-3x2-4x2y+6xy-4xy2−6y2+5
Multiplication of Polynomials
Two or more polynomial when multiplied always result in a polynomial of higher degree (unless one of them is a constant polynomial). An example of multiplying polynomials is given below:
Example: Solve (6x−3y)×(2x+5y)
Solution:
⇒ 6x ×(2x+5y)–3y × (2x+5y) ———- Using distributive law of multiplication
⇒ (12x2+30xy) – (6yx+15y2) ———- Using distributive law of multiplication
⇒12x2+30xy–6xy–15y2 —————– as xy = yx
Thus, (6x−3y)×(2x+5y)=12x2+24xy−15y2
Division of Polynomials
Division of two polynomial may or may not result in a polynomial. Let us study below the division of polynomials in details. To divide polynomials, follow the given steps:
Polynomial Division Steps:
If a polynomial has more than one term, we use long division method for the same. Following are the steps for it.
Write the polynomial in descending order.
Check the highest power and divide the terms by the same.
Use the answer in step 2 as the division symbol.
Now subtract it and carry down the next term.
Repeat step 2 to 4 until you have no more terms to carry down.
Note the final answer including remainder in the fraction form (last subtract term).
आयु सम्बन्धी प्रश्न और हल – Age Related Question of Maths in Hindiआयु सम्बन्धी प्रश्न और हल – Age Related Question of Maths in Hindi
Age Related Question of Maths in Hindi सभीप्रकारकीप्रतियोगितापरीक्षाओंमेंविभिन्नतरहकेगणितकेसवालपूछेजातेहैजिनमेसेआयुसंबंधितप्रश्न (Math
Age Trick) अवश्यपूछेजातेहै ! परीक्षाओंमेंजिसतरहकेप्र्श्नपूछेजातेहैउनप्रश्नोपरआधारितकुछप्रश्ननीचेदिएगएहैजिससे किआपकोयेपताचलसके किप्रतियोगितापरीक्षाओंमेंआयुसंबंधितप्रश्नोकाप्रकारक्याहोताहै , किसतरहकेसवालपूछेजातेहैउनकाकठिनाईस्तरक्याहोताहै ! नीचेदिएगएप्रश्नोकाअध्ययनकरनेसेआपकोआयुसंबंधितसवालोंकोपरीक्षामेंहलकरनेमेंआसानीहोगी ! उदाहरणकेरूपमेंनीचेकुछप्रश्नहलसहितदिएगएहै I
आयु सम्बन्धी प्रश्न और हल – Age Related Question of Maths in Hindi
Age Related Question of Maths in
Hindi
01 समीनातथासुहानाकीआयुकाअनुपात 7:3 है 6 वर्षबादइनकीआयुकाअनुपात 5:3 होगाइनकीआयुमेकितनाअंतरहै ?
हल – मानासमीनाकीवर्तमानआयु
= 7x वर्षतथासुहानाकीवर्तमानआयु
= 3x वर्ष
तब 7x+6 / 3x+6 = 5 / 3 ⇒ 3 (7x + 6) = 5 (3x =6)
⇒ 21x + 18 = 15x + 30 ⇒ 6x + 12 ⇒ x = 2
इनकीआयुमेअंतर = (7x – 3x) वर्ष = 4x वर्ष
= (4*2) वर्ष = 8 वर्ष
हल – मानारीनाकीआयु = x वर्षअतःउसकीमाताकीआयु = (49 – x) वर्ष
7 वर्षपूर्वरीनाकीआयु = (x – 7) वर्ष
7 वर्षपूर्वमाताकीआयु = (49 – x – 7) वर्ष = (42 -x) वर्ष
क्योकि 42 – x = 4 (x – 7) ⇒ 42 – x = 4x – 28
⇒ 5x = 70 ⇒ x =14
अतःरीनाकीमाताकीवर्तमानआयु = (49 – 14) वर्ष = 35 वर्ष
इस तरह के प्रश्न हमेशा प्रतियोगिता परीक्षा में पूछे जाते है अगर आपने एक बार अच्छे से इन सवलो का अध्ययन कर लिया तो आप परीक्षा में इस तरह के सवाल आसानी से हल कर पायेंगे ! इस तरह के प्रश्नो में अक्सर एक व्यक्ति की आयु दी जाती है एवं दूसरे व्यक्ति की आयु के बारे में पूछा जाता है , इन सवलो को बीजगणित की सहायता से आसानी से हल किया जा सकता है ! जिसकी आयु पूछी जाती है उसे x माना जाता है फिर प्रश्न के अनुसार हल करके उत्तर निकाल लिया जाता है.