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.
In the given figure, two triangles ABC and PQR are shown such that \(\angle A = \angle P\) and \(\angle C = \angle R\). If \(AD \perp BC\) and \(PS \perp QR\), then prove that (i) \(\Delta ADB \sim \Delta PSQ\) (ii) \(AD \times QS = BD \times PS\).
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2.
Determine the ratio in which the line \(2x + y = 6\) divides the line segment joining the points (1, 3) and (2, 5).
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3.
Seema daily goes to a park to exercise on machines available there. When Seema spent 15 minutes on exercise bicycle and 30 minutes on double cross walker, she received a message of burning 435 calories on her fitness watch. When she spent 30 minutes on exercise bicycle and 40 minutes on double cross walker, she received a message of burning 690 calories. Based on above information, answer the following questions:

38(i) Represent the above situation in terms of a pair of linear equations in two variables.
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4.
In the given figure, PQ is a tangent to a circle with centre \(O(-5, 3)\). If coordinates of P and Q are \((3, 1)\) and \((0, 6)\) respectively, then using distance formula, show that \(PQ \perp OQ\).
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5.
If \( 2 \sin A = 1 \), then the value of \( \tan A + \cot A \) is :
- \( \sqrt{3} \)
- \( \frac{4}{\sqrt{3}} \)
- \( \frac{\sqrt{3}}{2} \)
- \( 1 \)
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6.
Find the missing frequencies p and q in the following frequency distribution, when sum of frequencies is 40 and mean is 19 :
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