# NCERT Solutions for Class 9 Maths Chapter 2

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials in Hindi & English Medium, PDF form as well as online digital contents. Download NCERT Apps and Solutions based on latest CBSE Syllabus for 2019-20.

 Class 9: Maths – गणित Chapter 2: Polynomials

## NCERT Solutions for Class 9 Maths Chapter 2

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### Class 9 Maths Chapter 2 Polynomials Solutions

#### Hindi Medium and English Medium Solutions

• Class 9 Maths Chapter 2 Exercise 2.1 Solutions
• Class 9 Maths Chapter 2 Exercise 2.2 Solutions
• Class 9 Maths Chapter 2 Exercise 2.3 Solutions
• Class 9 Maths Chapter 2 Exercise 2.4 Solutions
• Class 9 Maths Chapter 2 Exercise 2.5 Solutions

• A combination of constants and variables, connected by four fundamental arithmetical operations +, -, x and / is called an algebraic expression. e.g. 6x² – 5y² + 2xy
• An algebraic expression which have only whole numbers as the exponent of one variable, is called polynomial in one variable. e.g. 3x³ + 2x² – 7x + 5 etc.
• The part of a polynomial separated from each other by + or – sign is called a term and each term of a polynomial has a coefficient.
• Highest power of the variable in a polynomial, is known as degree of that polynomial.

• On the Basis of Number of Terms
1. A polynomial containing one non-zero term, is called a monomial.
2. A polynomial containing two non-zero terms, is called a binomial.
3. A polynomial containing three non-zero terms, is called a trinomial.
• On the Basis of Degree of Variables
1. A polynomial of degree 0, is called a constant polynomial.
2. A polynomial of degree 1, is called a linear polynomial.
3. A polynomial of degree 2, is called a quadratic polynomial.
4. A polynomial of degree 3, is called a cubic polynomial.
5. A polynomial of degree 4, is called a biquadratic polynomial.
• The value obtained on putting a particular value of the variable in polynomial is called value of the polynomial at the value of variable.
• Zero of a polynomial p(x) is a number alpha, such that p(alpha) = 0. It is also called root pf polynomial equation p(x) = 0.

• Let f(x) be any polynomial of degree n,(n ≥ 1) and a be any real number. If f(x) is divided by the linear polynomial (x-a), then the remainder is f(a).
• Let f(x) be a polynomial of degree n,(n ≥ 1) and a be any real number. Then,
1. If f(a) = 0, then (x – a) is a factor of f(x).
2. If (x – a) is a factor of f(x), then f(a) = 0.

#### उपयुक्त सर्वसमिकाओं को प्रयोग करके निम्न्लिखित गुणनफल ज्ञात कीजिए: (x+4)(x+10)

(x + 4)(x + 10)
=x^2 + (4 + 10)x + 4 × 10
[∵〖(x + a)(x + b) = x〗^2 + (a + b) x + ab] = x^2 + 14x + 40

#### Determine whether polynomials x^3+x^2+x+1 has x+1 a factor?

Let p(x) = x^3 + x^2 + x + 1
Putting x + 1 = 0,
we get, x = – 1
Using remainder theorem,
when p(x)=x^3+x^2+x+1 is divided by x + 1,
remainder is given by p(-1)
=〖(-1)^3 + (-1)〗^2 + (-1) + 1
= – 1 + 1 – 1 + 1
= 0
Since, remainder p(-1) = 0,
Hence x + 1 is a factor of x^3 + x^2 + x + 1.

#### Use the Factor Theorem to determine whether g(x) is a factor of p(x): p(x) = 2x^3 + x^2 – 2x – 1, g(x) = x + 1.

p(x) = 2x^3 + x^2 – 2x – 1 and
g(x) = x + 1

Putting x + 1 = 0,
we get, x = -1

Using remainder theorem,
when p(x) = 2x^3 + x^2 – 2x – 1 is divided by g(x) = x + 1, remainder is given by p(-1)
=〖(-1)^3 + (-1)〗^2 + (-1) + 1
= – 1 + 1 – 1 + 1
= 0
Since, remainder p(-1) = 0,
hence g(x) is a factor of p(x).

#### सीधे गुणा किए बिना निम्न्लिखित मान ज्ञात कीजिए: 103×107

103 × 107
= (100 + 3)(100 + 7)
= (100)^2 + (3 + 7)100 + 3 × 7
[∵〖(x + a)(x + b) = x〗^2 + (a + b)x + ab] = 10000 + 1000 + 21
= 11021

#### उपयुक्त सर्वसमिकाओं को प्रयोग करके निम्न्लिखित का गुणनखंड कीजिए: 9x^2 + 6xy + y^2

9x^2 + 6xy + y^2
=(3x)^2 + 2 × 3x × y + y^2
=(3x + y)^2
[∵a^2 + 2ab + b^2 = (a + b)^2]

#### उपयुक्त सर्वसमिकाओं को प्रयोग करके निम्न्लिखित के मान ज्ञात कीजिए: (99)^3

(99)^3 = (100 – 1)^3
= (100)^3 + (-1)^3 + 3(100)^2 (-1) + 3(100) (-1)^2
[〖∵(a + b)〗^3 = a^3 + b^3 + 3a^2 b + 3ab^2 ] = 1000000 – 1 -30000 + 300
= 970299

#### Find the value of k, if x – 1 is a factor of p(x) = x^2 + x + k.

p(x) = x^2 + x + k
Putting x – 1 = 0, we get, x = 1
Using remainder theorem,
When p(x) = x^2 + x + k is divided by x – 1, remainder is given by p(1)
= (1)^2 + (1) + k
= 2 + k
Since x – 1 is a factor of p(x), hence remainder p(1) = 0
⇒ 2 + k = 0
⇒ k = -2

#### 2+x^2 + x में x^2 का गुणांक लिखिए।

2 + x^2 + x में x^2 का गुणांक 1 है।

#### 〖x^3-ax〗^2+6x-a को x-a से भाग देने पर शेषफल ज्ञात कीजिए।

माना p(x)=〖x^3 – ax〗^2 + 6x – a
x – a = 0 रखने पर, x = a
शेषफल प्रमेय के अनुसार
p(x) =〖x^3 – ax〗^2 + 6x – a को x – a से भाग देने पर शेषफल
p(a) =〖(a)^3 – a(a)〗^2 + 6(a) – a
= a^3 – a^3 + 6a – a
= 5a

#### Find the zero of the polynomial p(x) = x + 5.

p(x) = x + 5
Putting p(x) = 0, we get
x + 5 = 0
⇒ x = – 5
Hence, x = – 5 is a zero of the polynomial p(x).