NCERT Solutions Class 9 Maths Chapter 2 – Polynomials

NCERT Solutions Class 9 Maths Chapter 2 Polynomials are provided here. These NCERT Solutions are updated by Infinity Learn expert faculties to help students in the preparation for their second term exams. These renowned faculties solve and provide the NCERT Solutions for Class 9 so that it would assist students to solve the questions comfortably. They give a detailed and stepwise analysis of each answer to the problems given in the exercises in the NCERT textbook for Class 9.

In NCERT Solutions for Class 9, students are introduced to a lot of important topics which will be helpful for those who wish to pursue Mathematics as a subject in further classes. Based on these NCERT Solutions. These solutions help students to prepare for their upcoming Term II exams by covering the updated term wise CBSE syllabus for 2021-22 and its guidelines.

2. Polynomials (Final)

Also Check: Extra Questions for Class 9 Maths with Solutions Chapter Wise

Exercise 2.4 Class 9 – Question Answer

1. Determine which of the following polynomials has (x + 1) a factor:

(i) x³+x²+x+1

Solution:

Let p(x) = x³+x²+x+1

The zero of x+1 is -1. [x+1 = 0 means x = -1]

p(−1) = (−1)³+(−1)²+(−1)+1

= −1+1−1+1

= 0

∴By factor theorem, x+1 is a factor of x³+x²+x+1

(ii) x⁴+x³+x²+x+1

Solution:

Let p(x)= x⁴+x³+x²+x+1

The zero of x+1 is -1. . [x+1= 0 means x = -1]

p(−1) = (−1)⁴+(−1)³+(−1)²+(−1)+1

= 1−1+1−1+1

= 1 ≠ 0

∴By factor theorem, x+1 is not a factor of x⁴ + x³ + x² + x + 1

(iii) x⁴+3x³+3x²+x+1

Solution:

Let p(x)= x⁴+3x³+3x²+x+1

The zero of x+1 is -1.

p(−1)=(−1)⁴+3(−1)³+3(−1)²+(−1)+1

=1−3+3−1+1

=1 ≠ 0

∴By factor theorem, x+1 is not a factor of x⁴+3x³+3x²+x+1

(iv) x³ – x²– (2+√2)x +√2

Solution:

Let p(x) = x³–x²–(2+√2)x +√2

The zero of x+1 is -1.

p(−1) = (-1)³–(-1)²–(2+√2)(-1) + √2 = −1−1+2+√2+√2

= 2√2 ≠ 0

∴By factor theorem, x+1 is not a factor of x³–x²–(2+√2)x +√2

Also Check: Important Questions for CBSE Class 9 Mathematics Circles

2. Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases:

(i) p(x) = 2x³+x²–2x–1, g(x) = x+1

Solution:

p(x) = 2x³+x²–2x–1, g(x) = x+1

g(x) = 0

⇒ x+1 = 0

⇒ x = −1

∴Zero of g(x) is -1.

Now,

p(−1) = 2(−1)³+(−1)²–2(−1)–1

= −2+1+2−1

= 0

∴By factor theorem, g(x) is a factor of p(x).

(ii) p(x)=x³+3x²+3x+1, g(x) = x+2

Solution:

p(x) = x³+3x²+3x+1, g(x) = x+2

g(x) = 0

⇒ x+2 = 0

⇒ x = −2

∴ Zero of g(x) is -2.

Now,

p(−2) = (−2)³+3(−2)²+3(−2)+1

= −8+12−6+1

= −1 ≠ 0

∴By factor theorem, g(x) is not a factor of p(x).

(iii) p(x)=x³–4x²+x+6, g(x) = x–3

Solution:

p(x) = x³–4x²+x+6, g(x) = x -3

g(x) = 0

⇒ x−3 = 0

⇒ x = 3

∴ Zero of g(x) is 3.

Now,

p(3) = (3)³−4(3)²+(3)+6

= 27−36+3+6

= 0

∴By factor theorem, g(x) is a factor of p(x).

3. Find the value of k, if x–1 is a factor of p(x) in each of the following cases:

(i) p(x) = x²+x+k

Solution:

If x-1 is a factor of p(x), then p(1) = 0

By Factor Theorem

⇒ (1)²+(1)+k = 0

⇒ 1+1+k = 0

⇒ 2+k = 0

⇒ k = −2

(ii) p(x) = 2x²+kx+√2

Solution:

If x-1 is a factor of p(x), then p(1)=0

⇒ 2(1)²+k(1)+√2 = 0

⇒ 2+k+√2 = 0

⇒ k = −(2+√2)

(iii) p(x) = kx²–√2x+1

Solution:

If x-1 is a factor of p(x), then p(1)=0

By Factor Theorem

⇒ k(1)²-√2(1)+1=0

⇒ k = √2-1

(iv) p(x)=kx²–3x+k

Solution:

If x-1 is a factor of p(x), then p(1) = 0

By Factor Theorem

⇒ k(1)²–3(1)+k = 0

⇒ k−3+k = 0

⇒ 2k−3 = 0

⇒ k= 3/2

4. Factorize:

(i) 12x²–7x+1

Solution:

Using the splitting the middle term method,

We have to find a number whose sum = -7 and product =1×12 = 12

We get -3 and -4 as the numbers [-3+-4=-7 and -3×-4 = 12]

12x²–7x+1= 12x²-4x-3x+1

= 4x(3x-1)-1(3x-1)

= (4x-1)(3x-1)

(ii) 2x²+7x+3

Solution:

Using the splitting the middle term method,

We have to find a number whose sum = 7 and product = 2×3 = 6

We get 6 and 1 as the numbers [6+1 = 7 and 6×1 = 6]

2x²+7x+3 = 2x²+6x+1x+3

= 2x (x+3)+1(x+3)

= (2x+1)(x+3)

(iii) 6x²+5x-6

Solution:

Using the splitting the middle term method,

We have to find a number whose sum = 5 and product = 6×-6 = -36

We get -4 and 9 as the numbers [-4+9 = 5 and -4×9 = -36]

6x²+5x-6 = 6x²+9x–4x–6

= 3x(2x+3)–2(2x+3)

= (2x+3)(3x–2)

(iv) 3x²–x–4

Solution:

Using the splitting the middle term method,

We have to find a number whose sum = -1 and product = 3×-4 = -12

We get -4 and 3 as the numbers [-4+3 = -1 and -4×3 = -12]

3x²–x–4 = 3x²–4x+3x–4

= x(3x–4)+1(3x–4)

= (3x–4)(x+1)

5. Factorize:

(i) x³–2x²–x+2

Solution:

Let p(x) = x³–2x²–x+2

Factors of 2 are ±1 and ± 2

Now,

p(x) = x³–2x²–x+2

p(−1) = (−1)³–2(−1)²–(−1)+2

= −1−2+1+2

= 0

Therefore, (x+1) is the factor of p(x)

Now, Dividend = Divisor × Quotient + Remainder

(x+1)(x²–3x+2) = (x+1)(x²–x–2x+2)

= (x+1)(x(x−1)−2(x−1))

= (x+1)(x−1)(x-2)

(ii) x³–3x²–9x–5

Solution:

Let p(x) = x³–3x²–9x–5

Factors of 5 are ±1 and ±5

By trial method, we find that

p(5) = 0

So, (x-5) is factor of p(x)

Now,

p(x) = x³–3x²–9x–5

p(5) = (5)³–3(5)²–9(5)–5

= 125−75−45−5

= 0

Therefore, (x-5) is the factor of p(x)

Now, Dividend = Divisor × Quotient + Remainder

(x−5)(x²+2x+1) = (x−5)(x²+x+x+1)

= (x−5)(x(x+1)+1(x+1))

= (x−5)(x+1)(x+1)

(iii) x³+13x²+32x+20

Solution:

Let p(x) = x³+13x²+32x+20

Factors of 20 are ±1, ±2, ±4, ±5, ±10 and ±20

By trial method, we find that

p(-1) = 0

So, (x+1) is factor of p(x)

Now,

p(x)= x³+13x²+32x+20

p(-1) = (−1)³+13(−1)²+32(−1)+20

= −1+13−32+20

= 0

Therefore, (x+1) is the factor of p(x)

Now, Dividend = Divisor × Quotient +Remainder

(x+1)(x²+12x+20) = (x+1)(x²+2x+10x+20)

= (x+1)x(x+2)+10(x+2)

= (x+1)(x+2)(x+10)

(iv) 2y³+y²–2y–1

Solution:

Let p(y) = 2y³+y²–2y–1

Factors = 2×(−1)= -2 are ±1 and ±2

By trial method, we find that

p(1) = 0

So, (y-1) is factor of p(y)

Now,

p(y) = 2y³+y²–2y–1

p(1) = 2(1)³+(1)²–2(1)–1

= 2+1−2

= 0

Therefore, (y-1) is the factor of p(y)

Now, Dividend = Divisor × Quotient + Remainder

(y−1)(2y²+3y+1) = (y−1)(2y²+2y+y+1)

= (y−1)(2y(y+1)+1(y+1))

= (y−1)(2y+1)(y+1)

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Summary

As this is one of the important Chapters in Class 9 Maths, it comes under the unit – Algebra and has a weightage of 12 marks in Class 9 Maths CBSE Term II examination. This chapter talks about:

• Polynomials in One Variable
• Zeroes of a Polynomial

• Remainder Theorem
• Factorization of Polynomials
• Algebraic Identities

Students can refer to the NCERT Solutions for Class 9 Maths while solving exercise problems and preparing for their Class 9 Maths exams.

List of Exercises in Class 9 Maths Chapter 2:

Exercise 2.1 Solutions 5 Questions

Exercise 2.2 Solutions 4 Questions

Exercise 2.3 Solutions 3 Questions

Exercise 2.4 Solutions 5 Questions

Exercise 2.5 Solutions 16 Questions

NCERT Solutions for Class 9 Maths Chapter 2 – Polynomials

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials is the second chapter of Class 9 Maths. Polynomials are introduced and discussed in detail here. The chapter discusses Polynomials and their applications. The introduction of the chapter includes whole numbers, integers, and rational numbers.

The chapter starts with the introduction of Polynomials in section 2.1 followed by two very important topics in sections 2.2 and 2.3

• Polynomials in one Variable – Discussion of Linear, Quadratic and Cubic Polynomial.
• Zeroes of a Polynomial – A zero of a polynomial need not be zero and can have more than one zero.

• Real Numbers and their Decimal Expansions – Here you study the decimal expansions of real numbers and see whether they can help in distinguishing between rationals and irrationals.

Next, it discusses the following topics:

• Representing Real Numbers on the Number Line – In this the solutions for 2 problems in Exercise 2.4.
• Operations on Real Numbers – Here you explore some of the operations like addition, subtraction, multiplication, and division on irrational numbers.
• Laws of Exponents for Real Numbers – Use these laws of exponents to solve the questions.

Key Advantages of NCERT Solutions for Class 9 Maths Chapter 2 – Polynomials

• These NCERT Solutions for Class 9 Maths helps you solve and revise the updated term wise CBSE syllabus of Class 9 for 2021-22.
• After going through the stepwise solutions given by our subject expert teachers, you will be able to score more marks.
• It follows NCERT guidelines which help in preparing the students accordingly.
• It contains all the important questions from the examination point of view.

• It helps in scoring well in Class 10 CBSE Term II Maths exams.

In order to improve the problem solving skills among the Class 9 students, we at Infinity Learn have provided solutions for other textbooks of the CBSE Board. To learn more clearly about Polynomials, students can access the solutions link provided below.

• RD Sharma Solutions for Class 9 Maths Chapter 6 Factorization of Polynomials

Frequently Asked Questions on NCERT Solutions for Class 9 Maths Chapter 2

How many exercises are present in NCERT Solutions for Class 9 Maths Chapter 2?

NCERT Solutions for Class 9 Maths Chapter 2 has 5 exercises. The topics discussed in these exercises are polynomials in one Variable, zeros of a polynomial, real numbers and their decimal expansions, representing real numbers on the number line and operations on real numbers laws of exponents for real numbers. Practice is an essential task to learn and score well in Mathematics. Hence the solutions are designed by Infinity Learn experts to boost confidence among students in understanding the concepts covered in this chapter.

Why should I opt for NCERT Solutions for Class 9 Maths Chapter 2?

The concepts present in NCERT Solutions for Class 9 Maths Chapter 2 are explained in simple language, which makes it possible even for a student not proficient in Maths to understand the subject better. Solutions are prepared by a set of experts at Infinity Learn with the aim of helping students boost their CBSE Term II exam preparation.

Is NCERT Solutions for Class 9 Maths Chapter 2 difficult to learn?

No, if you practice regularly with NCERT Solutions for Class 9 Maths Chapter 2 you can achieve your goal by scoring high in CBSE Term II. These solutions are formulated by a set of Maths experts at Infinity Learn. Students can score good marks in the exams by solving all the questions and cross-checking the answers with the NCERT Solutions for Class 9 Maths Chapter 2.