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Class 9 Mathematics Revision Notes for Polynomials of Chapter 2
The revision notes for Chapter 2 Mathematics Polynomials by Extramarks will help students get a clear idea of all the important topics covered in this chapter. The subjectmatter experts have curated Class 9 Mathematics Chapter 2 Notes as per the latest CBSE Syllabus for students who are preparing for their Class 9 exam.
Class 9 Mathematics Revision Notes for Polynomials of Chapter 2
Introduction
An algebraic expression of the form a0+a1x+a2x2+…+anxn where a0,a1,a2, ….a, where n is a positive integer and an are real numbers is referred to as a polynomial in x. Students must be aware of factors and products in the case of numbers. For instance, 8 is the product of 4 and 2. Thus, 8’s factors are 4 and 2.
In a similar manner, the algebraic expression a.b.c = abc is written as 1.a.b.c or 1.ab.c or 1,a,b,c,ab,bc,ac, abc are all factors of a.b.c and a.b.c is a product. The process of expressing a given expression or number as the product of its components is known as factorization.
(1) Polynomials in One Variable
Polynomials in one variable are formulas with only one variable. A mathematical statement made up of variables and coefficients involving operations like addition, subtraction, multiplication, and exponentiation is a polynomial.
Examples of Polynomials in One Variable:
a2+3a2
3b3+2b2b+1
c4–5c2+8c3
(2) Coefficient of Polynomials
A number or quantity that is associated with a variable is known as a coefficient. An integer multiplied by the variable immediately adjacent to it is generally the coefficient of polynomials.
For instance, in the expression 4x, 4 is the coefficient but in the expression x2+5, 1 is the coefficient of x2.
(3) Terms of Polynomial
Polynomial terms are the parts of the equation that are often denoted by plus (+) or minus () signs. In other words, every term in a polynomial equation is a component of the polynomial. The number of terms in a polynomial like 22 + 7x + 4 is 3.
(4) Types of Polynomials
Chapter 2 Mathematics Class 9 Notes discusses the following types of polynomials in detail:
Type  Meaning  Example 
Zero or constant polynomial  Polynomials having 0 degrees.  3 or 3x0 
Linear polynomial  Polynomials having a degree of 1 as the degree of a polynomial.  x+y4+5m+7n+2p 
Quadratic polynomial  Polynomials having a degree of 2 as the degree of a polynomial.  8x2 +7y+9m2+mn6 
Cubic polynomial  Polynomials having a degree of 3 as the degree of a polynomial.  3x3+ pq+7 
(5) Degree of Polynomial
A degree of the polynomial is the largest exponential power in a polynomial equation. When determining the degree of any polynomial, only variables are taken into account and coefficients are ignored. For example, 2a5+4a310. The polynomial degree would be 5.
(6) Zeros of Polynomials
The x values that fulfil the equation y = f(x) are polynomial zeros. The values of x for which the y value is equal to zero are the zeros of the polynomial, and f(x) is a function of x. The number of zeros in a polynomial is determined by the degree of the equation y= f(x).
Factorisation of Polynomials
Any polynomial of the form p(a) can also be written as p(a) = g(a) × h(a) + R(a)
Dividend = Quotient × Divisor + Remainder
p(a) =g(a) x h(a) if the remainder is zero, that is, the polynomial p(a) is a product of two other polynomials h(a) and g(a). For instance, 3a + 6a2 = 3a × (1+2a).
A polynomial may also be expressed as the product of two or more polynomials.
Now, let us take a look at the polynomial 3a + 6a2 = 3ax (1+2a).
It can also be factorised as 3a + 6a2 = 6a × (½ +a).
Methods of Factorising Polynomials
A polynomial can be factorised in the following three ways:
 By dividing the expression by the HCF of the words in the provided expression.
 By grouping the terms of the expression.
 By using identities.
HCF is the biggest monomial in a polynomial, which is a factor of each term in a polynomial. By determining the expression’s Highest Common Factor (HCF) and then dividing each term by its HCF, one can factorise a polynomial. HCF and the quotient are the factors of the above equation.
Steps for Factorisation
 You first need to find out the HCF of the supplied expression’s terms.
 Determine the quotient by dividing each term of the provided equation by the HCF.
 Multiply the HCF and quotient and write the expression.
Factorisation by Grouping the Expression’s Terms
We come across polynomials in several situations, and they may or may not have common factors among their components. In such cases, to make the common factors exist among the terms of the resulting groups, we arrange the expression’s terms.
Steps for Factorisation by Grouping
 Rearrange the terms if required.
 Assemble the phrase into groups, so that each group must have its own common component.
 Determine the HCF of each group.
 Determine the other component.
 Convert the expression as a product of the additional and common factors.
Factorisation Using Identities
To calculate the products, recall the following identities:
 (a+b)2 = a2 + 2ab + b2
 (a−b)2 = a2 − 2ab +b2
 (a+b) (a−b) = a2 − b2
 (a+b)3 = a3 + 3a2b + 3ab2 +b3
 (a−b)3 = a3 − 3a2b + 3ab2− b3
 (a+b+c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
Note that the LHS in the identities are all factors and the RHS are their products. Thus, the factors can be written as follows:
(a−b) and (a−b) are factors of a2 − 2ab + b2.
(a+b) and (a+b) are factors of a2 + 2ab + b2.
(a+b) and (a−b) are factors of a2 − b2.
(a+b), (a+b) and (a + b) are factors of a3 + 3a2b + 3ab2 + b3.
(a−b), (a−b) and (a−b) are factors of a3−3a2b + 3ab2 − b3.
(a+b+c) and (a+b+c) are factors of a2 + b2 + c2 + 2ab + 2bc + 2ca.
Steps for Factorisation Using Identities
 Determine the correct persona.
 Rewrite the provided statement in the form of the identity.
 Write the factors of the given equation using the identity.
a3 ± b3 ± 3ab(a±b)=(a±b)3
(a+b+c)2=a2+b2+c2+2ab+2bc+2ca
x3±y3=(x±y)(x2±xy+y2)
Factorisation of Trinomials
Expressions with three terms are trinomials. For instance, x2+14x+4 is a trinomial. It is not possible to factorise all trinomials using a single approach. We select the best approach for factorising the given binomial and investigate the pattern in trinomials.
Factorising a Trinomial by Splitting the Middle Term
The product is (x+a)×(x+b)=x2+x(a+b)+ab[a trinomial] of two binomials of the type (x+a) and (x+b).
Therefore, we need to first discover the two elements that meet the criteria mentioned before to factorise expressions of the type (x2+cx+d). That is, to make sure that the product of the components equals the last term, the middle word must be divided.
Factoring a Trinomial of the Type ax by Grouping
You need to find two numbers a and b such that their product is equal to the last term(constant) and their sum is equal to the coefficient of the middle term in order to factorise expressions of the type x2+bx+c.
Remainder Theorem
The remainder of the value f(x) at x=a i.e., Remainder =f(a), if f(x) is a polynomial in x and is divided by xa.
Proof:
Let the polynomial p(x) be divided by (x−a). Assume that the quotient is q(x) and R is the remainder. Dividend = ( Divisor x Quotient )+ Remainder by division algorithm i.e.
p(x) = q(x)⋅(x−a) + R
Now substitute x=a,
We will get
p(a) = q(a)(a−a) + R
p(a) = R(a − a = 0, 0 − q(a) = 0)
As a result, Remainder= p(a)
Steps for Factorisation Using Remainder Theorem
 Using the trialanderror method, identify the constant factor for which the provided expression equals zero.
 Subtract the first step’s factor from the expression.
 Factoring the quotient is necessary. Factorise the quotient if it is a trinomial.
 If the equation is of the fourth degree, first convert it to a binomial before factorising it further.
Factor Theorem
Statement:
(x−a) is a factor of p(x) if a polynomial p(x) in x is divided by x−a and the remainder =p (a) is zero.
Proof:
By remainder theorem p(x)=(x−a)⋅q(x)+p(a), R=p(a) when p(x) is divided by x−a.
Dividend = Divisor x quotient + Remainder Division Algorithm
But since it is given that p(a)=0, hence p(x)=(x−a)⋅q(x)
⇒(x−a) is a factor of p(x). This implies that if x−a is a factor of p(x), then p(a)=0. p(x)=(x−a)⋅q(x)+R
The remainder should be zero if (x−a) is a factor (x – a divides p(x) exactly) R=0
Thus, R=p(a) ⇒p(a)=0 by the remainder theorem.
Access Class 9 Mathematics Chapter 2 – Polynomials Notes (Summary, Examples, Important Points to Remember)
FAQs (Frequently Asked Questions)
1. What are the key concepts in Chapter 2 of Class 9 Mathematics?
Chapter 2 Polynomials is one of the most scoring topics in Class 9 Mathematics. Many important topics, formulas, theorems, and concepts related to polynomials are discussed in this chapter. These include polynomials in one variable, real numbers, decimal expansions of real numbers, zeros of a polynomial, exponents of real numbers, laws of real numbers and representing real numbers on the number line.
2. What is a polynomial?
An expression that contains terms, variables, constants, and exponents is known as a polynomial. For instance, 3x2 – 2x – 18 is a polynomial. Monomial, Binomial and Trinomial are mainly the three types of polynomials.
3. Define a zero polynomial.
When all of the variable coefficients in a polynomial are equal to zero, the polynomial is said to be zero. The polynomial function is equivalent to a constant function having a value of 0. In other words, it has a value of 0 and is a constant polynomial. Generally speaking, the zero polynomial is referred to as the polynomial additive group’s additive identity.
4. How can one solve linear polynomials?
One of the ways used to solve a polynomial is the linear method. One must isolate the variable terms in the first step of this method. After this, make sure that the equation is equal to zero. Once the equation is balanced, the equation can be solved by the use of a basic algebra operation.