**RD Sharma Class 9 Solutions Chapter 4 Exercise 4.1:** Practice questions from RD Sharma Solutions Class 9 Maths and ace your Maths exam this year. You don’t have to study too many books as RD Sharma Class 9 Solutions Chapter 4 Exercise 4.1 has everything you might need. To know more, you have to read the whole blog.

**Download RD Sharma Class 9 Solutions Chapter 4 Exercise 4.1 PDF**

RD Sharma Solutions for Chapter 4 Exercise 4.1 Class 9 PDF

**Access RD Sharma Class 9 Solutions Chapter 4 Exercise 4.1**

### 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) 3x ^{2} – 4x + 15**

**(ii) y ^{2} + 2√3**

**(iii) 3√x + √2x**

**(iv) x – 4/x**

**(v) x ^{12} + y^{3} + t^{50}**

**Solution:**

**(i)** 3x^{2} – 4x + 15

*It is a polynomial of x.*

**(ii)** y^{2} + 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)** x^{12} + y^{3} + t^{50}

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

**Question 2: Write the coefficient of x ^{2} in each of the following:**

**(i) 17 – 2x + 7x ^{2}**

**(ii) 9 – 12x + x ^{3}**

**(iii) ∏/6 x ^{2} – 3x + 4**

**(iv) √3x – 7**

**Solution:**

**(i)** 17 – 2x + 7x^{2}

*Coefficient of x ^{2} = 7*

**(ii)** 9 – 12x + x^{3}

*Coefficient of x ^{2 }=0*

**(iii)** ∏/6 x^{2} – 3x + 4

*Coefficient of x ^{2 }= ∏/6*

**(iv)** √3x – 7

*Coefficient of x ^{2 }= 0*

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

**(i) 7x ^{3} + 4x^{2} – 3x + 12**

**(ii) 12 – x + 2x ^{3}**

**(iii) 5y – √2**

**(iv) 7**

**(v) 0**

**Solution**:

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

**(i)** Degree of the polynomial 7x^{3} + 4x^{2} – 3x + 12 is *3*

**(ii)** Degree of the polynomial 12 – x + 2x^{3} 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 + x ^{2} + 4**

**(ii) 3x – 2**

**(iii) 2x + x ^{2}**

**(iv) 3y**

**(v) t ^{2} + 1**

**(vi) 7t ^{4} + 4t^{3} + 3t – 2**

**Solution:**

**(i)** x + x^{2} + 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 + x^{2}: It is a quadratic polynomial as its degree is *2*.

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

**(v)** t^{2}+ 1: It is a quadratic polynomial as its degree is *2*.

**(vi)** 7t^{4} + 4t^{3} + 3t – 2: It is a biquadratic polynomial as its degree is *4*.

### Exercise 6.2 Page No: 6.8

**Question 1: If f(x) = 2x ^{3} – 13x^{2} + 17x + 12, find**

**(i) f (2)**

**(ii) f (-3)**

**(iii) f(0)**

**Solution:**

f(x) = 2x^{3} – 13x^{2} + 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) = x ^{2} – 1, x = 1,−1**

**(iii) g(x) = 3x ^{2} – 2 , x = 2/√3 , −2/√3**

**(iv) p(x) = x ^{3} – 6x^{2} + 11x – 6 , x = 1, 2, 3**

**(v) f(x) = 5x – π, x = 4/5**

**(vi) f(x) = x ^{2} , 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) = x^{2} – 1, x = 1,−1

f(x) = x^{2} – 1

Given that x = (1 , -1)

Substitute x = 1 in f(x)

f(1) = 1^{2} – 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) = x ^{2 }– 1*

**(iii)** g(x) = 3x^{2} – 2 , x = 2/√3 , −2/√3

g(x) = 3x^{2} – 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 3x ^{2}–2.*

**(iv)** p(x) = x^{3} – 6x^{2} + 11x – 6 , x = 1, 2, 3

p(1) = 1^{3} – 6(1)^{2} + 11x 1 – 6 = 1 – 6 + 11 – 6 = 0

p(2) = 2^{3} – 6(2)^{2} + 11×2 – 6 = 8 – 24 + 22 – 6 = 0

p(3) = 3^{3} – 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) = x^{2} , x = 0

f(0) = 0^{2} = 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).*

This is the complete blog on RD Sharma Solutions Class 9 Maths Chapter 4 Exercise 4.1. To know more about the CBSE Class 9 Maths exam, ask in the comments.

**FAQs on RD Sharma Solutions Class 9 Maths Chapter 4 Exercise 4.1**

### How many questions are there in RD Sharma Solutions for Class 9 Maths Chapter 4 Exercise 4.1?

There are 14 questions in RD Sharma Solutions Class 9 Maths Chapter 4 Exercise 4.1.

### From where can I download the PDF of RD Sharma Solutions for Class 9 Maths Chapter 4 Exercise 4.1

You can find the download link from the above blog.

### How much does it cost to download the PDF of RD Sharma Solutions for Class 9 Maths Chapter 4 Exercise 4.1?

You can download it for free.