RD Sharma Solutions Class 9 Maths Chapter 6

RD Sharma Solutions Class 9 Maths Chapter 6 Factorization Of Polynomials: The best solution for all your factorization of polynomials practices which happens to be the most important topic in the CBSE Class 9 Algebra syllabus in Mathematics. The section teaches you the steps to factor a polynomial. You can start practicing the RD Sharma Solutions Class 9 Maths Chapter 6 and ace your Class 9 Maths exam.

Table of Contents

Download RD Sharma Solutions Class 9 Maths Chapter 6 PDF

RD Sharma Solutions Class 9 Maths Chapter 6

Exercise-wise: RD Sharma Solutions Class 9 Maths Chapter 6

RD Sharma Solutions Class 9 Chapter 6 Exercise 6.1

RD Sharma Solutions Class 9 Chapter 6 Exercise 6.2

RD Sharma Solutions Class 9 Chapter 6 Exercise 6.3

RD Sharma Solutions Class 9 Chapter 6 Exercise 6.4

RD Sharma Solutions Class 9 Chapter 6 Exercise 6.5

Access answers of RD Sharma Solutions Class 9 Maths Chapter 6 – Factorization of Polynomials

RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.1

Exercise 6.1 Page No: 6.2

Question 1: Which of the following expressions are polynomials in one variable, and which are not? State reasons for your answer:

(i) 3x² – 4x + 15

(ii) y² + 2√3

(iii) 3√x + √2x

(iv) x – 4/x

(v) x¹² + y³ + t⁵⁰

Solution:

(i) 3x² – 4x + 15

It is a polynomial of x.

(ii) y² + 2√3

It is a polynomial of y.

(iii) 3√x + √2x

It is not a polynomial since the exponent of 3√x is a rational term.

(iv) x – 4/x

It is not a polynomial since the exponent of – 4/x is not a positive term.

(v) x¹² + y³ + t⁵⁰

It is a three-variable polynomial, x, y and t.

Question 2: Write the coefficient of x² in each of the following:

(i) 17 – 2x + 7x²

(ii) 9 – 12x + x³

(iii) ∏/6 x² – 3x + 4

(iv) √3x – 7

Solution:

(i) 17 – 2x + 7x²

Coefficient of x² = 7

(ii) 9 – 12x + x³

Coefficient of x²=0

(iii) ∏/6 x² – 3x + 4

Coefficient of x²= ∏/6

(iv) √3x – 7

Coefficient of x²= 0

Question 3: Write the degrees of each of the following polynomials:

(i) 7x³ + 4x² – 3x + 12

(ii) 12 – x + 2x³

(iii) 5y – √2

(iv) 7

(v) 0

Solution:

As we know, degree is the highest power in the polynomial

(i) Degree of the polynomial 7x³ + 4x² – 3x + 12 is 3

(ii) Degree of the polynomial 12 – x + 2x³ is 3

(iii) Degree of the polynomial 5y – √2 is 1

(iv) Degree of the polynomial 7 is 0

(v) Degree of the polynomial 0 is undefined.

Question 4: Classify the following polynomials as linear, quadratic, cubic and biquadratic polynomials:

(i) x + x² + 4

(ii) 3x – 2

(iii) 2x + x²

(iv) 3y

(v) t² + 1

(vi) 7t⁴ + 4t³ + 3t – 2

Solution:

(i) x + x² + 4: It is a quadratic polynomial as its degree is 2.

(ii) 3x – 2 : It is a linear polynomial as its degree is 1.

(iii) 2x + x²: It is a quadratic polynomial as its degree is 2.

(iv) 3y: It is a linear polynomial as its degree is 1.

(v) t²+ 1: It is a quadratic polynomial as its degree is 2.

(vi) 7t⁴ + 4t³ + 3t – 2: It is a biquadratic polynomial as its degree is 4.

Exercise 6.2 Page No: 6.8

Question 1: If f(x) = 2x³ – 13x² + 17x + 12, find

(i) f (2)

(ii) f (-3)

(iii) f(0)

Solution:

f(x) = 2x³ – 13x² + 17x + 12

(i) f(2) = 2(2)³ – 13(2)² + 17(2) + 12

= 2 x 8 – 13 x 4 + 17 x 2 + 12

= 16 – 52 + 34 + 12

= 62 – 52

= 10

(ii) f(-3) = 2(-3)³ – 13(-3)² + 17 x (-3) + 12

= 2 x (-27) – 13 x 9 + 17 x (-3) + 12

= -54 – 117 -51 + 12

= -222 + 12

= -210

(iii) f(0) = 2 x (0)³ – 13(0)² + 17 x 0 + 12

= 0-0 + 0+ 12

= 12

Question 2: Verify whether the indicated numbers are zeros of the polynomials corresponding to them in the following cases:

(i) f(x) = 3x + 1, x = −1/3

(ii) f(x) = x² – 1, x = 1,−1

(iii) g(x) = 3x² – 2 , x = 2/√3 , −2/√3

(iv) p(x) = x³ – 6x² + 11x – 6 , x = 1, 2, 3

(v) f(x) = 5x – π, x = 4/5

(vi) f(x) = x² , x = 0

(vii) f(x) = lx + m, x = −m/l

(viii) f(x) = 2x + 1, x = 1/2

Solution:

(i) f(x) = 3x + 1, x = −1/3

f(x) = 3x + 1

Substitute x = −1/3 in f(x)

f( −1/3) = 3(−1/3) + 1

= -1 + 1

= 0

Since, the result is 0, so x = −1/3 is the root of 3x + 1

(ii) f(x) = x² – 1, x = 1,−1

f(x) = x² – 1

Given that x = (1 , -1)

Substitute x = 1 in f(x)

f(1) = 1² – 1

= 1 – 1

= 0

Now, substitute x = (-1) in f(x)

f(-1) = (−1)² – 1

= 1 – 1

= 0

Since , the results when x = 1 and x = -1 are 0, so (1 , -1) are the roots of the polynomial f(x) = x²– 1

(iii) g(x) = 3x² – 2 , x = 2/√3 , −2/√3

g(x) = 3x² – 2

Substitute x = 2/√3 in g(x)

g(2/√3) = 3(2/√3)² – 2

= 3(4/3) – 2

= 4 – 2

= 2 ≠ 0

Now, Substitute x = −2/√3 in g(x)

g(2/√3) = 3(-2/√3)² – 2

= 3(4/3) – 2

= 4 – 2

= 2 ≠ 0

The results when x = 2/√3 and x = −2/√3) are not 0. Therefore, (2/√3 , −2/√3 ) are not zeros of 3x²–2.

(iv) p(x) = x³ – 6x² + 11x – 6 , x = 1, 2, 3

p(1) = 1³ – 6(1)² + 11x 1 – 6 = 1 – 6 + 11 – 6 = 0

p(2) = 2³ – 6(2)² + 11×2 – 6 = 8 – 24 + 22 – 6 = 0

p(3) = 3³ – 6(3)² + 11×3 – 6 = 27 – 54 + 33 – 6 = 0

Therefore, x = 1, 2, 3 are zeros of p(x).

(v) f(x) = 5x – π, x = 4/5

f(4/5) = 5 x 4/5 – π = 4 – π ≠ 0

Therefore, x = 4/5 is not a zeros of f(x).

(vi) f(x) = x² , x = 0

f(0) = 0² = 0

Therefore, x = 0 is a zero of f(x).

(vii) f(x) = lx + m, x = −m/l

f(−m/l) = l x −m/l + m = -m + m = 0

Therefore, x = −m/l is a zero of f(x).

(viii) f(x) = 2x + 1, x = ½

f(1/2) = 2x 1/2 + 1 = 1 + 1 = 2 ≠ 0

Therefore, x = ½ is not a zero of f(x).

Important Topics RD Sharma CBSE Class 9 Chapter 6

With RD Sharma Solutions of Chapter 6 for Class 9, students will get to study the important concepts of Polynomials and Factorization that include:

Factorization of Polynomials Introduction
Terms and coefficients
Degree of a polynomial
Types of polynomials
Remainder Theorem
The factorization of polynomials by using the factor theorem

This is the complete blog on RD Sharma Solutions Class 9 Chapter 6. The solutions come at no cost and help you focus on your study for the CBSE class 9 Mathematics exam. To know more about the Class 9 Maths exams, ask in the comments.