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.
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RD Sharma Solutions Class 9 Maths Chapter 6
Exercise-wise: RD Sharma Solutions Class 9 Maths Chapter 6
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.
RD Sharma Solutions Class 9 Maths Chapter 6 – Factorization of Polynomials
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