NCERT Solutions for Class 10 Maths Chapter 2 Polynomials

NCERT Solutions for Class 10 Maths Chapter 2 Polynomials- This article is an important resource created by the GFG team of experts to help students solve problems in the NCERT Textbook in an easy manner. It contains free NCERT Solutions for Class 10 Maths Chapter 2 Polynomials, as per the latest CBSE syllabus 2023-2024 and guidelines.

Class 10 Maths NCERT Solutions Chapter 2 Polynomials Exercises
NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Exercise 2.1 – 1 Question (1 Short Answer)
NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Exercise 2.2 – 2 Questions (2 Short Answers)
NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Exercise 2.3 – 5 Questions (2 Short Answers, 3 Long Answers)
NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Exercise 2.4 – 5 Questions (2 Short Answers, 3 Long Answers)

Chapter 2 Polynomials of NCERT Class 10 Maths deal majorly in topics such as fundamental aspects of polynomials. Students learn about the basic definitions, terms, coefficients, and degrees of polynomials. They study several polynomial varieties and comprehend polynomial operations including addition, subtraction, multiplication, and division. Factorization methods for quadratic polynomials are introduced, along with concepts like the factor theorem and remainder theorem. The chapter also delves into finding zeros of polynomials.

All the solutions to NCERT Class 10 Maths Chapter 2 Polynomials exercises have been covered in NCERT Solutions for Class 10 Maths.

NCERT Solutions for Class 10 Maths Chapter 2 Polynomials: Exercise 2.1

Question 1. The graphs of y = p(x) are given in Fig. below, for some polynomials p(x). Find the number of zeroes of p(x), in each case.

👁 NCERT Solutions for Class 10 Maths Chapter 2 Polynomials

Solution:

So, we have equation y = p(x). When p(x) = 0, y = 0. 

That means we only need to find the number of times we get y = 0 in the graph.

In other words, the number of times the graph intersects the x-axis is the answer.

(i) 0. Because the graph doesn't intersect the x-axis.

(ii)1. Because the graph intersects the x-axis only at one point.

(iii) 3. Because the graph intersects the x-axis at three points.

(iv)2. Because the graph intersects the x-axis at two points.

(v)4. Because the graph intersects the x-axis at four points.

(vi) 3. Because the graph intersects the x-axis at three points.

NCERT Solutions for Class 10 Maths Chapter 2 Polynomials: Exercise 2.2

Question 1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.

(i) x² – 2x – 8

x² – 2x – 8 = x² – 4x + 2x – 8 

= x (x – 4) + 2(x – 4) 

= (x - 4) (x + 2)

Therefore, zeroes of equation x² – 2x – 8 are (4, -2)

Sum of zeroes is equal to [4 – 2]= 2 = -(-2)/1 

i.e. = -(Coefficient of x) / (Coefficient of x²)

Product of zeroes is equal to 4 × (-2) = -8 =-(8)/1  

i.e.= (Constant term) / (Coefficient of x²)

(ii) 4s2 – 4s + 1

4s² – 4s + 1 = 4s² – 2s – 2s +1 

= 2s(2s – 1) – 1(2s - 1) 

= (2s – 1) (2s – 1)

Therefore, zeroes of  equation 4s² – 4s +1 are (1/2, 1/2)

Sum of zeroes is equal to [(1/2) + (1/2)] = 1 = -4/4 

i.e.= -(Coefficient of s) / (Coefficient of s²)

Product of zeros is equal to [(1/2) × (1/2)] = 1/4 

i.e.= (Constant term) / (Coefficient of s² )

(iii) 6x² – 3 – 7x

6x² – 3 – 7x = 6x² – 7x – 3 

= 6x² – 9x + 2x – 3 

= 3x(2x – 3) + 1(2x – 3) 

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

Therefore, zeroes of equation 6x² – 3 – 7x are (-1/3, 3/2)

Sum of zeroes is equal to -(1/3) + (3/2) = (7/6) 

i.e.= -(Coefficient of x) / (Coefficient of x²)

Product of zeroes is equal to -(1/3) × (3/2) = -(3/6) 

i.e.= (Constant term) / (Coefficient of x² )

(iv) 4u² + 8u

4u² + 8u = 4u(u + 2)

Therefore, zeroes of equation 4u² + 8u are (0, -2).

Sum of zeroes is equal to [0 + (-2)] = -2 = -(8/4)

i.e. = -(Coefficient of u) / (Coefficient of u²)

Product of zeroes is equal to 0 × -2 = 0 = 0/4 

i.e. = (Constant term) / (Coefficient of u² )

(v) t² – 15

 t² – 15

⇒ t² = 15 or t = ±√15

Therefore, zeroes of equation t² – 15 are (√15, -√15)

Sum of zeroes is equal to [√15 + (-√15)] = 0 = -(0/1) 

i.e.= -(Coefficient of t) / (Coefficient of t2)

Product of zeroes is equal to √15 × (-√15) = -15 = -15/1 

i.e. = (Constant term) / (Coefficient of t2 )

(vi) 3x² – x – 4

3x² – x – 4 = 3x² – 4x + 3x – 4

= x(3x - 4) + 1(3x - 4) 

= (3x – 4) (x + 1)

Therefore, zeroes of equation 3x² – x – 4 are (4/3, -1)

Sum of zeroes is equal to (4/3) + (-1) = (1/3) = -(-1/3) 

i.e. = -(Coefficient of x) / (Coefficient of x²)

Product of zeroes is equal to (4/3) × (-1) = (-4/3) 

i.e. = (Constant term) / (Coefficient of x²)

Question 2. Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.

(i) 1/4, -1

Let two zeroes be α, β

∴ Sum of zeroes = α + β

∴ Product of zeroes = αβ

Given, Sum of zeroes = α + β = 1/4

Product of zeroes = α β = -1

∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as:-

x² – (α + β)x + αβ = 0

x² – (1/4)x +(-1) = 0

4x² – x - 4 = 0

∴ 4x² – x – 4 is the quadratic polynomial.

(ii) √2, 1/3

Let two zeroes be α, β

∴ Sum of zeroes = α + β

∴ Product of zeroes = αβ

Given Sum of zeroes = α + β =√2

Product of zeroes = αβ = 1/3

∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as:

x² – (α + β)x + αβ = 0

x² – (√2)x + (1/3) = 0

3x² - 3√2x + 1 = 0

∴ 3x² - 3√2x + 1 is the quadratic polynomial.

(iii) 0, √5

Let two zeroes be α, β

∴ Sum of zeroes = α + β

∴ Product of zeroes = αβ

Given, Sum of zeroes = α + β = 0

Product of zeroes = αβ = √5

∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as:-

x² – (α + β)x + αβ = 0

x² – (0)x + √5 = 0

∴ x² + √5 is the quadratic polynomial.

(iv) 1, 1

Let two zeroes be α, β

∴ Sum of zeroes = α + β

∴ Product of zeroes = αβ

Given, Sum of zeroes = α + β = 1

Product of zeroes = αβ = 1

∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as:-

x² – (α + β)x + αβ = 0

x² – x + 1 = 0

∴ x² – x + 1 is the quadratic polynomial.

(v) -1/4, 1/4

Let two zeroes be α, β

∴ Sum of zeroes = α + β

∴ Product of zeroes = αβ

Given, Sum of zeroes = α + β = -1/4

Product of zeroes = α β = 1/4

∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as:-

x² – (α + β)x + αβ = 0

x² – (-1/4)x + (1/4) = 0

4x² + x + 1 = 0

∴ 4x² + x + 1 is the quadratic polynomial.

(vi) 4, 1

Let two zeroes be α, β

∴ Sum of zeroes = α + β

∴ Product of zeroes = αβ

Given, Sum of zeroes = α + β = 4

Product of zeroes = αβ = 1

∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as:

x² – (α + β)x + αβ = 0

x² – 4x + 1 = 0

∴ x² – 4x + 1 is the quadratic polynomial.

NCERT Solutions for Class 10 Maths Chapter 2 Polynomials: Exercise 2.3

Question 1. Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following:

(i) p(x) = x³ – 3x² + 5x – 3, g(x) = x² – 2

(ii) p(x) = x⁴ – 3x² + 4x + 5, g(x) = x² + 1 – x

(iii) p(x) = x⁴– 5x + 6, g(x) = 2 – x²

Solution:

i)p(x) = x³ – 3x² + 5x – 3, g(x) = x² – 2

👁 NCERT Solutions for Class 10 Maths Chapter 2 Polynomials

R = 7x-9

Q = x-3

ii) p(x) = x⁴ – 3x² + 4x + 5, g(x) = x² + 1 – x

👁 NCERT Solutions for Class 10 Maths Chapter 2 Polynomials

R = 8

Q = x²+x-3

iii) p(x) = x⁴– 5x + 6, g(x) = 2 – x²

👁 NCERT Solutions for Class 10 Maths Chapter 2 Polynomials

Q = -x²-2

R = -5x+10

Question 2. Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial.

(i) t² – 3, 2t⁴ + 3t³ – 2t²– 9t – 12

(ii) x² + 3x + 1, 3x⁴ + 5x³ – 7x² + 2x + 2

(iii) x³ - 3x + 1, x⁵ – 4x³ + x² + 3x + 1

Solution:

i) t² – 3, 2t⁴ + 3t³ – 2t²– 9t – 12

👁 NCERT Solutions for Class 10 Maths Chapter 2 Polynomials

Q = 2t³+3t+4

R = 0

Yes 1^st polynomial is factor of 2^nd polynomial. 

ii) x² + 3x + 1, 3x⁴ + 5x³ – 7x² + 2x + 2

👁 NCERT Solutions for Class 10 Maths Chapter 2 Polynomials

R = 0

Q = 3x²-4x+2

iii) x³ - 3x + 1, x⁵ – 4x³ + x² + 3x + 1

👁 NCERT Solutions for Class 10 Maths Chapter 2 Polynomials

R = x²-1

Q = 2

Question 3. Obtain all other zeroes of 3x⁴ + 6x³ – 2x² – 10x – 5, if two of its zeroes are and √(5/3) and -√(5/3).

Solution:

👁 NCERT Solutions for Class 10 Maths Chapter 2 Polynomials

R = 0

Q = 3x²+6x+3

∴ we are factorizing

3x²+6x+3

x²+2x+1

(x+1)²

(x+1)  (x+1) = 0

∴ x = -1 and x = -1

Question 4. On dividing x³ – 3x² + x + 2 by a polynomial g(x), the quotient and remainder were x – 2 and -2x + 4 respectively. Find g(x).

Solution:

Dividend = Divisor * Quotient + Remainder

x³-3x²+3x-2/x-2

👁 NCERT Solutions for Class 10 Maths Chapter 2 Polynomials

R = 0

Q = x² -x +1

Answer: g(x)=x²-x+1

Question 5. Give examples of polynomials p(x), g(x), q(x) and r(x), which satisfy the division algorithm and:

(i) deg p(x) = deg q(x)

(ii) deg q(x) = deg r(x)

(iii) deg r(x) = 0

Solution:

i)deg p(x) = deg q(x)

p(x)=2x²-2x+14, g(x)=2

p(x)/g(x)=2x²-2x+14/2=(x²-x+7)

=x²-x+7=q(x)

=q(x)=x²-x+7

r(x)=0

ii) deg q(x)=deg r(x)

p(x)=4x²+4x+4, g(x)=x²+x+1

👁 NCERT Solutions for Class 10 Maths Chapter 2 Polynomials

q(x) = 4

r(x) = 0

∴Here deg q(x)=deg r(x)

iii) deg r(x)=0

p(x)=x³+2x²-x+2 ,g(x)=x²-1

👁 NCERT Solutions for Class 10 Maths Chapter 2 Polynomials

q(x) = x+2

r(x) = 4

deg of r(x) = 0 

NCERT Solutions for Class 10 Maths Chapter 2 Polynomials: Exercise 2.4

Question 1. Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also, verify the relationship between the zeroes and the coefficients in each case:

(i) 2x³ + x² - 5x + 2; 1/2, 1, -2

Solution:

p(x) = 2x³+x²-5x+2

p(1/2) = 2(1/2)³+(1/2)²-5(1/2)+2 

           = (1/4)+(1/4)-(5/2)+2 

           = 0

p(1) = 2(1)³+(1)²-5(1)+2 = 0

p(-2) = 2(-2)³+(-2)²-5(-2)+2 = 0

Therefore, 1/2, 1, -2 are the zeroes of 2x³+x²-5x+2.

Now, comparing the given polynomial with general expression

ax³+bx²+cx+d = 2x³+x²-5x+2

a=2, b=1, c= -5 and d = 2

α, β, γ are the zeroes of the cubic polynomial ax³+bx²+cx+d

α +β+γ = –b/a

αβ+βγ+γα = c/a

α βγ = – d/a.

α+β+γ = ½+1+(-2) 

            = -1/2 = –b/a

αβ+βγ+γα = (1/2×1)+(1 ×-2)+(-2×1/2) 

                  = -5/2 = c/a

α β γ = ½×1×(-2) 

         = -2/2 = -d/a

Hence, the relationship between the zeroes and the coefficients are satisfied.

(ii) x³ - 4x² + 5x - 2 ;2, 1, 1

Solution:

p(x) = x³-4x²+5x-2

Zeroes are 2,1,1.

p(2)= 2³-4(2)²+5(2)-2 

      = 0

p(1) = 1³-(4)(1² )+(5)(1)-2 = 0

Therefore, proved, 2, 1, 1 are the zeroes of x³-4x²+5x-2

On comparing the given polynomial with general expression

ax³+bx²+cx+d = x³-4x²+5x-2

a = 1, b = -4, c = 5 and d = -2

Therefore,

α + β + γ = –b/a

                = 2+1+1 

                = 4 

         –b/a = -(-4)/1

αβ + βγ + γα = c/a

                      = 2×1+1×1+1×2 

                      = 5 

                c/a = 5/1 

α β γ = – d/a.

          = 2×1×1 

          = 2 

 -d/a = -(-2)/1

Hence, the relationship between the zeroes and the coefficients is satisfied.

Question 2. Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time, and the product of its zeroes as 2, –7, –14 respectively.

Solution:

Let us consider the cubic polynomial as ax³+bx²+cx+d and zeroes of the polynomials be α, β, γ.

α+β+γ = -b/a = 2/1

αβ +βγ+γα = c/a = -7/1

α βγ = -d/a = -14/1

On comparing

a = 1, b = -2, c = -7, d = 14

Therefore, the cubic polynomial is x³-2x²-7x+14

Question 3. If the zeroes of the polynomial x³ - 3x² + x + 1are a – b, a, a + b, find a and b.

Solution:

p(x) = x³-3x²+x+1

Zeroes are given as a – b, a, a + b

px³+qx²+rx+s = x³-3x²+x+1

On comparing

p = 1, q = -3, r = 1 and s = 1

Sum of zeroes = a – b + a + a + b

-q/p = 3a

Putting the values q and p.

-(-3)/1 = 3a

a=1

Therefore, zeroes are 1-b, 1, 1+b.

Product of zeroes = 1(1-b)(1+b)

-s/p = 1-b²

-1/1 = 1-b²

b² = 1+1 = 2

b = √2

Therefore,1-√2, 1,1+√2 are the zeroes of x³-3x²+x+1.

Question 4. If two zeroes of the polynomial x⁴-6x³-26x²+138x-35are 2 ±√3,find other zeroes.

Solution:

Degree of polynomial is 4

Therefore, it has four roots

f(x) = x⁴-6x³-26x²+138x-35

As 2 +√3 and 2-√3 are zeroes of given polynomial f(x).

Therefore, [x−(2+√3)] [x−(2-√3)] = 0

(x−2−√3)(x−2+√3) = 0

Therefore, x²-4x+1 is a factor of polynomial f(x).

Let it be g(x) = x²-4x+1

By dividing f(x) by g(x) we get another factor of f(x)

👁 NCERT Solutions for Class 10 Maths Chapter 2 Polynomials

x⁴-6x³-26x²+138x-35 = (x²-4x+1)(x² –2x−35)

On factorizing (x²–2x−35) by splitting the middle term

x²–(7−5)x −35 = x²– 7x+5x-35

                         =x(x −7)+5(x−7)

(x+5)(x−7) = 0

x= −5 and x = 7.

Therefore, all four zeroes of given polynomial equation are: 2+√3, 2-√3, −5 and 7.

Key Features of NCERT Solutions for Class 10 Maths Chapter 2 Polynomials

• These NCERT solutions are developed by the GfG team, with a focus on students’ benefit.
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