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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CBSE X Related Questions

1.
Assertion (A) : \((\sqrt{3} + \sqrt{5})\) is an irrational number.
Reason (R) : Sum of the any two irrational numbers is always irrational.
- Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
- Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
- Assertion (A) is true, but Reason (R) is false.
- Assertion (A) is false, but Reason (R) is true.
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2.
Prove that \(2 + 3\sqrt{5}\) is an irrational number given that \(\sqrt{5}\) is an irrational number.
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3.
For any natural number n, \( 5^n \) ends with the digit :
- 0
- 5
- 3
- 2
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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.
If \(PQ\) and \(PR\) are tangents to the circle with centre \(O\) and radius \(4 \text{ cm}\) such that \(\angle QPR = 90^{\circ}\), then the length \(OP\) is
- \(4 \text{ cm}\)
- \(4\sqrt{2} \text{ cm}\)
- \(8 \text{ cm}\)
- \(2\sqrt{2} \text{ cm}\)
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6.
A trader has three different types of oils of volume \(870 \text{ l}\), \(812 \text{ l}\) and \(638 \text{ l}\). Find the least number of containers of equal size required to store all the oil without getting mixed.
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