NCERT Solutions for Class 9 Maths Chapter 2: Polynomials

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The NCERT Solutions for Class 9 Maths Chapter 2 Polynomials are provided in this article. A polynomial is an expression composed of variables and coefficients that contain fundamental arithmetic operations such as addition, subtraction, and multiplication, as well as the exponential negative exponential of variables.

Chapter 2 Polynomials belongs to Unit 2 Algebra which has a weightage of 20 marks in the CBSE Class 9 Maths Examination. NCERT Solutions for Class 9 Maths for Chapter 2 cover the following important concepts:

Download: NCERT Solutions for Class 9 Mathematics Chapter 2 pdf

NCERT Solutions for Class 9 Maths Chapter 2

Class 9 Chapter 2 NCERT Solutions are given below:

Important Topics in Class 9 Maths Chapter 2 Polynomials

Important Topics in Class 9 Maths Chapter 2 Polynomials are elaborated below:

Remainder Theorem

Remainder Theorem is an Euclidean approach of division of polynomials. It says that if we divide a polynomial P(x) by a factor ( x – a); which is not necessarily an element of the polynomial; then we can find a smaller polynomial along with a remainder.

Example: Assume that f(a) = a³-12a²-42 is divided by (a-3). The quotient will be a²-9a-27 and the remainder is -123. Determine whether it satifies the Reaminder Theorem?

Solution: First let’s put a-3 = 0
Then, a = 3
Therefore, f(a) = f(3) = -123
Hence, it satisfies the remainder theorem.

Degree of Polynomial

Degree of a polynomial is known to be the greatest exponent of a variable in the polynomial.

Example: Determine the degree of polynomial: 3x⁸+ 4x³ + 9x + 1.

Solution: As pe the question, the degree of the polynomial, 3x⁸+ 4x³ + 9x + 1 is 8.

Algebraic Identities

Algebraic identities are equations that are valid for every value of variables in them. Algebraic identities are also widely used for the factorization of polynomials.

A few examples of Algebraic Identities:

(x + y)² = x² + 2xy + y²
(x – y)² = x² – 2xy + y²
x² – y² = (x + y) (x – y)
(x + a) (x + b) = x² + (a + b)x + ab.
(x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx
(x + y)³ = x³ + y³ + 3xy(x + y)

Polynomials in One Variable

Polynomials in one variable are simply algebraic expressions. These can be found in axn, where n is a non-negative integer (i.e. positive or zero) and a is a real number, also known as the coefficient of the term.

Example:

P(x) = 4x – 3
G(y) = y⁴ – y²+ 2y + 9

Factorisation of Polynomials

Polynomials can also be represented as the product of its factors with a degree less than or equal to the original polynomial. In other words, the method of factoring is called factorization of polynomials.

Example: Factorise the Polynomial: x⁴ – 16.

Solution: Let’s consider the following
x⁴ – 16 = (x² + 4) (x² – 4)
Now, we can factorise (x²-4). Hence, the factorization will be,
x⁴ – 16 = (x² + 4) (x + 2) (x – 2)

NCERT Solutions for Class 9 Maths Chapter 2 Exercises:

The detailed solutions for all the NCERT Solutions for Real Numbers under different exercises are:

Read More:

Unit Related Topics
Polynomials in One Variable	Factorisation of Polynomials	Algebraic Identities
Remainder Theorem	Synthetic Division	Remainder

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Maths Important Formulas	Maths MCQs	Comparison Articles in Maths
Difference Between Topics in Maths	Math Study Material	Algebra
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CBSE X Related Questions

1.
In a class test, the sum of Anamika's marks obtained in Maths and Science is 30. Had she got 2 marks more in Maths and 3 marks less in Science, the product of the marks would have been 210. Find the marks she got in the two subjects.
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2.
There are two sections A and B of Grade X. There are 28 students in Section A and 30 students in Section B. What is the minimum number of books you will acquire for the class library so that they can be distributed equally among students of Section A or Section B ?
- 144
- 2
- 420
- 272
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3.
The HCF of 960 and 432 is :
- 48
- 54
- 72
- 36
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4.
The graph of \(y = f(x)\) is given. The number of zeroes of \(f(x)\) is :
- 0
- 1
- 3
- 2
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5.
\(ABCD\) is a parallelogram such that \(AF = 7 \text{ cm}\), \(FB = 3 \text{ cm}\) and \(EF = 4 \text{ cm}\), length \(FD\) equals
- \(\frac{21}{4} \text{ cm}\)
- \(\frac{28}{3} \text{ cm}\)
- \(\frac{12}{7} \text{ cm}\)
- \(5.5 \text{ cm}\)
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6.
PQ and PR are two tangents to a circle with centre O and radius 5 cm. AB is another tangent to the circle at C which lies on OP. If \(OP = 13\) cm, then find the length AB and PA.
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