# Polynomials| NCERT Solutions| Class 10

NCERT Solutions of Chapter: Polynomials. NCERT Solutions along with worksheets and notes for Class 10.

Exercise 2.1

(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.

(i)

(ii)

(iii)

(iv)

(v)

(vi)

Ans-

(i) Number of zeroes for p(x) = 0

Curve does not intersect the x- axis.

(ii) Number of zeroes for p(x) = 1

Curve intersects the x- axis at one point only.

(iii) Number of zeroes for p(x) = 3

Curve intersects the x- axis at three points.

(iv) Number of zeroes for p(x) = 2

Curve intersects the x- axis at two points.

(v) Number of zeroes for p(x) = 4

Curve intersects the x- axis at four points.

(vi) Number of zeroes for p(x) = 3

Curve intersects the x- axis at one point and touches at two points.

Exercise 2.2

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

(i) x2 – 2x – 8                 (ii) 4s2 – 4s + 1                (iii) 6x2 – 3 – 7x

(iv) 4u2 + 8u                  (v) t2 – 15                         (vi) 3x2 – x – 4

Ans –

(i) x2 – 2x – 8

= x2 – 4x + 2x -8

=x(x – 4) +2(x – 4)

=(x – 4)(x + 2)

Therefore, the value of zeroes of polynomial x2 – 2x – 8 is

x – 4 = 0, ie, x = 4

x + 2 + 0, ie, x = -2

The zeroes of polynomial x2 – 2x – 8 are 4 and – 2.

Now,

Sum of the zeroes = 4 + (-2) = 2

Sum of the coefficient = Sum of the zeroes

Product of the zeroes = (4)(-2) = -8

Product of the coefficient = Product of the zeroes

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

(ii) 4s2 – 4s + 1

= 4s2 – 2s -2s -1

= 2s(2s – 1) +1(2s – 1)

=(2s – 1)(2s – 1)

Therefore, the value of zeroes of polynomial 4s2 – 4s + 1 is

Product of the coefficient = Product of the zeroes

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

(iii) 6x2 – 3 – 7x

6x2 – 7x – 3

= 6x2 – 9x + 2x -3

= 3x(2x – 3)+ 1(2x – 3)

= (2x – 3)(3x + 1)

Therefore, the value of zeroes of polynomial 6x2 – 7x – 3 is

Product of the coefficient = Product of the zeroes

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

(iv) 4u2 + 8u

= 4u (u + 2)

Therefore, the value of zeroes of polynomial 4u2 + 8u is

4u = 0, ie, u = 0

u + 2 = 0, ie, u = – 2

The zeroes of polynomial 4u2 + 8u are 0 and – 2.

Now,

Sum of the zeroes = 0 + -2 = -2

Sum of the coefficient = Sum of the zeroes

Product of the zeroes = (0)(-2) = 0

Product of the coefficient = Product of the zeroes

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

(v) t2 – 15

= t2 +225 – 30t

= t2 – 30t +225

= t2 – 15 t – 15t + 225

=t(t – 15) – 15(t – 15)

=(t – 15) (t – 15)

Therefore, the value of zeroes of polynomial t2 – 15 is

t – 15= 0, ie, t = 15

t – 15= 0, ie, t = 15

The zeroes of polynomial t2 – 30t +225 are 15 and 15.

Now,

Sum of the zeroes = 15 + 15 = 30

Sum of the coefficient = Sum of the zeroes

Product of the zeroes = (15)(15) = 225

Product of the coefficient = Product of the zeroes

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

(vi) 3x2 – x – 4

= 3x2 + 3x – 4x – 4

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

=(x + 1) (3x – 4)

Therefore, the value of zeroes of polynomial 3x2 – x – 4is

x + 1= 0, ie, x = -1

Product of the coefficient = Product of the zeroes

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

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

Ans-

Exercise 2.3

(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

Ans-

(i)

Dividend p(x) = x3 – 3x2 + 5x -3

Divisor g(x) = x2 – 2

Since degree (3) = 0 < 2 = degree (x2 – 2)

Therefore, quotient = x – 3, remainder = 7x – 9

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

Dividend p(x) = x4 –  3x2 + 4x + 5

Divisor g(x) = x2 + 1 –x

Since degree (4) = 0 < 3 = degree (x2 + 1 -x)

Therefore, quotient = x2 + x -3, remainder = 8

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

since degree (4) = 0 < 3 = degree (2 – x2)

Therefore, quotient = -x2 – 2, remainder = -5x + 10

(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

Ans

(i) t2 – 3, 2t4 + 3t3 – 2t2 – 9t – 12

Division of the second polynomial by the first polynomial is showing below-

Remainder is zero.

Hence, the first polynomial (t2 – 3) is a factor of the second polynomial.

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

Division of the second polynomial by the first polynomial is showing below-

Remainder is zero.

Hence, the first polynomial (x2 + 3x + 1) is a factor of the second polynomial 3x4 + 5x3 – 7x2 + 2x + 2.

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

Division of the second polynomial by the first polynomial is showing below-

Remainder is not zero,ie 2.

Hence, the first polynomial (x3 – 3x + 1) is not a factor of the second polynomial x5 – 4x3 + x2 + 3x + 1.

(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).

Ans-

Dividend = x3 – 3x2 + x + 2

Divisor = g(x)

Quotient = x – 2

Remainder = -2x + 4

Dividend = Divisor  Quotient + Remainder

x3 – 3x2 + x + 2 = g(x) × (x – 2) + (-2x + 4)

x3 – 3x2 + x + 2 – (-2x + 4) = g(x) × (x – 2)

x3 – 3x2 + x + 2 + 2x – 4 = g(x) × (x – 2)

x3 – 3x2 + 3x -2 = g(x) × (x – 2)

Hence, g(x) is x2 – x + 1.

(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

Ans-

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

Let 2x2 + 6x – 8 is divided by 2

p(x) = 2x2 + 6x – 8

g(x) = 2

q(x) = x2 + 3x – 4

r(x) = 0

deg p(x) = deg q(x) =2

Division algorithm

p(x) = g(x)  q(x) + r(x)

2x2 + 6x – 8 = 2  (x2 + 3x – 4) + 0

2x2 + 6x – 8 = 2x2 + 6x – 8

LHS = RHS

Hence, division algorithm verified.

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

Let x3 – 3x2 + 5x -3 is divided by x2 – 2

p(x) = x3 – 3x2 + 5x -3

g(x) = x2 – 2

q(x) = x – 3

r(x) = 7x – 9

deg q(x) = deg r(x) =1

Division algorithm

p(x) = g(x)  q(x) + r(x)

x3 – 3x2 + 5x -3 = (x2 – 2) (x – 3) + 7x – 9

x3 – 3x2 + 5x -3 = x3 – 3x2 – 2x + 6 + 7x – 9

x3 – 3x2 + 5x -3 = x3 – 3x2 + 5x -3

LHS = RHS

Hence, division algorithm verified.

(iii) deg r(x) =0

Let x2 + x + 2 is divided by x

p(x) = x2 + x + 2

g(x) = x

q(x) = x + 1

r(x) = 2

Therefore, deg r(x) =0

Division algorithm

p(x) = g(x)  q(x) + r(x)

x2 + x + 2 = (x) (x + 1) + 2

x2 + x + 2 = x2 + x + 2

LHS = RHS

Hence, division algorithm verified.

Exercise 2.4

(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:

(ii) x3 – 4x2 + 5x – 2; 2, 1, 1

The value of zeroes of polynomial x3 – 4x2 + 5x – 2 is

x – 2 = 0, ie, x =  2

x – 1 + 0, ie, x = 1

and

x – 1 = 1

Put value of zeroes in given polynomial

p(x) = x3 – 4x2 + 5x – 2

p(2) = 23 – 4(2)2 + 5(2) – 2

= 8 – 16 + 10 – 2

= 0

p(1) = x3 – 4x2 + 5x – 2

p(1) = (1)3 – 4(1)2 + 5(1) – 2

= 1 – 4 + 5 – 2 = 0

Hence, numbers are zeroes of the given polynomial.

Now, compare the given polynomial with ax3 + bx2 + cx + d, we get

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

The zeroes of polynomial x3 – 4x2 + 5x – 2 are 2 , 1 and 1.

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

(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.

Ans-

Given:

α + β + γ = 2

αβ + βγ + γα  = -7

αβγ = -14

But, we know

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

Ans

Comparing given polynomial to a1 x3 + b1 x2 + c1 x + d1

a1 = 1, b1 = -3 c1 = 1 and d1 = 1

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

And

x2 – 2x – 35

= x2 – 7x + 5x – 35

= x(x -7) +3(x – 7)

=(x – 7) (x + 3)

Therefore, zeroes are x = 7 and x = – 3

Hence, zeroes of the given polynomials are 2 +√3 , 2 -√3 , 7 and – 3.

(5) If the polynomial x4 – 6x3 + 16x2 – 25x + 10 is divided by another polynomial x2 – 2x + k, the remainder comes out to be x + a, find k and a.

Ans-

Divide polynomial x4 – 6x3 + 16x2 – 25x + 10 by another polynomial x2 – 2x + k

Comparing coefficient’s of it (2k – 9)x-(8-k)k + 10

with x + a

2k – 9 = 1

K = 5

And

-(8 – k)k + 10= a

Put value of k

-(8-5)5 + 10 = a

a = -15 + 10

= -5

Hence, value of k = 5 and value of a = -5.

Helping Topics

Polynomials

NCERT Solutions Class 10

Worksheet Class 10
The text passage is a set of NCERT solutions for Class 10 Polynomials, containing multiple questions and their corresponding answers, primarily focusing on finding the zeroes of polynomials and verifying relationships between the zeroes and coefficients.
multiple_questions_with_answers
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
# Polynomials| NCERT Solutions| Class 10

NCERT Solutions of Chapter: Polynomials. NCERT Solutions along with worksheets and notes for Class 10.

Exercise 2.1

(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.

(i)

(ii)

(iii)

(iv)

(v)

(vi)

Ans-

(i) Number of zeroes for p(x) = 0

Curve does not intersect the x- axis.

(ii) Number of zeroes for p(x) = 1

Curve intersects the x- axis at one point only.

(iii) Number of zeroes for p(x) = 3

Curve intersects the x- axis at three points.

(iv) Number of zeroes for p(x) = 2

Curve intersects the x- axis at two points.

(v) Number of zeroes for p(x) = 4

Curve intersects the x- axis at four points.

(vi) Number of zeroes for p(x) = 3

Curve intersects the x- axis at one point and touches at two points.

Exercise 2.2

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

(i) x2 – 2x – 8                 (ii) 4s2 – 4s + 1                (iii) 6x2 – 3 – 7x

(iv) 4u2 + 8u                  (v) t2 – 15                         (vi) 3x2 – x – 4

Ans –

(i) x2 – 2x – 8

= x2 – 4x + 2x -8

=x(x – 4) +2(x – 4)

=(x – 4)(x + 2)

Therefore, the value of zeroes of polynomial x2 – 2x – 8 is

x – 4 = 0, ie, x = 4

x + 2 + 0, ie, x = -2

The zeroes of polynomial x2 – 2x – 8 are 4 and – 2.

Now,

Sum of the zeroes = 4 + (-2) = 2

Sum of the coefficient = Sum of the zeroes

Product of the zeroes = (4)(-2) = -8

Product of the coefficient = Product of the zeroes

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

(ii) 4s2 – 4s + 1

= 4s2 – 2s -2s -1

= 2s(2s – 1) +1(2s – 1)

=(2s – 1)(2s – 1)

Therefore, the value of zeroes of polynomial 4s2 – 4s + 1 is

Product of the coefficient = Product of the zeroes

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

(iii) 6x2 – 3 – 7x

6x2 – 7x – 3

= 6x2 – 9x + 2x -3

= 3x(2x – 3)+ 1(2x – 3)

= (2x – 3)(3x + 1)

Therefore, the value of zeroes of polynomial 6x2 – 7x – 3 is

Product of the coefficient = Product of the zeroes

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

(iv) 4u2 + 8u

= 4u (u + 2)

Therefore, the value of zeroes of polynomial 4u2 + 8u is

4u = 0, ie, u = 0

u + 2 = 0, ie, u = – 2

The zeroes of polynomial 4u2 + 8u are 0 and – 2.

Now,

Sum of the zeroes = 0 + -2 = -2

Sum of the coefficient = Sum of the zeroes

Product of the zeroes = (0)(-2) = 0

Product of the coefficient = Product of the zeroes

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

(v) t2 – 15

= t2 +225 – 30t

= t2 – 30t +225

= t2 – 15 t – 15t + 225

=t(t – 15) – 15(t – 15)

=(t – 15) (t – 15)

Therefore, the value of zeroes of polynomial t2 – 15 is

t – 15= 0, ie, t = 15

t – 15= 0, ie, t = 15

The zeroes of polynomial t2 – 30t +225 are 15 and 15.

Now,

Sum of the zeroes = 15 + 15 = 30

Sum of the coefficient = Sum of the zeroes

Product of the zeroes = (15)(15) = 225

Product of the coefficient = Product of the zeroes

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

(vi) 3x2 – x – 4

= 3x2 + 3x – 4x – 4

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

=(x + 1) (3x – 4)

Therefore, the value of zeroes of polynomial 3x2 – x – 4is

x + 1= 0, ie, x = -1

Product of the coefficient = Product of the zeroes

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

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

Ans-

Exercise 2.3

(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

Ans-

(i)

Dividend p(x) = x3 – 3x2 + 5x -3

Divisor g(x) = x2 – 2

Since degree (3) = 0 < 2 = degree (x2 – 2)

Therefore, quotient = x – 3, remainder = 7x – 9

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

Dividend p(x) = x4 –  3x2 + 4x + 5

Divisor g(x) = x2 + 1 –x

Since degree (4) = 0 < 3 = degree (x2 + 1 -x)

Therefore, quotient = x2 + x -3, remainder = 8

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

since degree (4) = 0 < 3 = degree (2 – x2)

Therefore, quotient = -x2 – 2, remainder = -5x + 10

(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

Ans

(i) t2 – 3, 2t4 + 3t3 – 2t2 – 9t – 12

Division of the second polynomial by the first polynomial is showing below-

Remainder is zero.

Hence, the first polynomial (t2 – 3) is a factor of the second polynomial.

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

Division of the second polynomial by the first polynomial is showing below-

Remainder is zero.

Hence, the first polynomial (x2 + 3x + 1) is a factor of the second polynomial 3x4 + 5x3 – 7x2 + 2x + 2.

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

Division of the second polynomial by the first polynomial is showing below-

Remainder is not zero,ie 2.

Hence, the first polynomial (x3 – 3x + 1) is not a factor of the second polynomial x5 – 4x3 + x2 + 3x + 1.

(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).

Ans-

Dividend = x3 – 3x2 + x + 2

Divisor = g(x)

Quotient = x – 2

Remainder = -2x + 4

Dividend = Divisor  Quotient + Remainder

x3 – 3x2 + x + 2 = g(x) × (x – 2) + (-2x + 4)

x3 – 3x2 + x + 2 – (-2x + 4) = g(x) × (x – 2)

x3 – 3x2 + x + 2 + 2x – 4 = g(x) × (x – 2)

x3 – 3x2 + 3x -2 = g(x) × (x – 2)

Hence, g(x) is x2 – x + 1.

(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

Ans-

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

Let 2x2 + 6x – 8 is divided by 2

p(x) = 2x2 + 6x – 8

g(x) = 2

q(x) = x2 + 3x – 4

r(x) = 0

deg p(x) = deg q(x) =2

Division algorithm

p(x) = g(x)  q(x) + r(x)

2x2 + 6x – 8 = 2  (x2 + 3x – 4) + 0

2x2 + 6x – 8 = 2x2 + 6x – 8

LHS = RHS

Hence, division algorithm verified.

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

Let x3 – 3x2 + 5x -3 is divided by x2 – 2

p(x) = x3 – 3x2 + 5x -3

g(x) = x2 – 2

q(x) = x – 3

r(x) = 7x – 9

deg q(x) = deg r(x) =1

Division algorithm

p(x) = g(x)  q(x) + r(x)

x3 – 3x2 + 5x -3 = (x2 – 2) (x – 3) + 7x – 9

x3 – 3x2 + 5x -3 = x3 – 3x2 – 2x + 6 + 7x – 9

x3 – 3x2 + 5x -3 = x3 – 3x2 + 5x -3

LHS = RHS

Hence, division algorithm verified.

(iii) deg r(x) =0

Let x2 + x + 2 is divided by x

p(x) = x2 + x + 2

g(x) = x

q(x) = x + 1

r(x) = 2

Therefore, deg r(x) =0

Division algorithm

p(x) = g(x)  q(x) + r(x)

x2 + x + 2 = (x) (x + 1) + 2

x2 + x + 2 = x2 + x + 2

LHS = RHS

Hence, division algorithm verified.

Exercise 2.4

(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:

(ii) x3 – 4x2 + 5x – 2; 2, 1, 1

The value of zeroes of polynomial x3 – 4x2 + 5x – 2 is

x – 2 = 0, ie, x =  2

x – 1 + 0, ie, x = 1

and

x – 1 = 1

Put value of zeroes in given polynomial

p(x) = x3 – 4x2 + 5x – 2

p(2) = 23 – 4(2)2 + 5(2) – 2

= 8 – 16 + 10 – 2

= 0

p(1) = x3 – 4x2 + 5x – 2

p(1) = (1)3 – 4(1)2 + 5(1) – 2

= 1 – 4 + 5 – 2 = 0

Hence, numbers are zeroes of the given polynomial.

Now, compare the given polynomial with ax3 + bx2 + cx + d, we get

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

The zeroes of polynomial x3 – 4x2 + 5x – 2 are 2 , 1 and 1.

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

(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.

Ans-

Given:

α + β + γ = 2

αβ + βγ + γα  = -7

αβγ = -14

But, we know

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

Ans

Comparing given polynomial to a1 x3 + b1 x2 + c1 x + d1

a1 = 1, b1 = -3 c1 = 1 and d1 = 1

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

And

x2 – 2x – 35

= x2 – 7x + 5x – 35

= x(x -7) +3(x – 7)

=(x – 7) (x + 3)

Therefore, zeroes are x = 7 and x = – 3

Hence, zeroes of the given polynomials are 2 +√3 , 2 -√3 , 7 and – 3.

(5) If the polynomial x4 – 6x3 + 16x2 – 25x + 10 is divided by another polynomial x2 – 2x + k, the remainder comes out to be x + a, find k and a.

Ans-

Divide polynomial x4 – 6x3 + 16x2 – 25x + 10 by another polynomial x2 – 2x + k

Comparing coefficient’s of it (2k – 9)x-(8-k)k + 10

with x + a

2k – 9 = 1

K = 5

And

-(8 – k)k + 10= a

Put value of k

-(8-5)5 + 10 = a

a = -15 + 10

= -5

Hence, value of k = 5 and value of a = -5.

Helping Topics

Polynomials

NCERT Solutions Class 10

Worksheet Class 10
user
The text passage is a set of NCERT solutions for Class 10 Polynomials, containing multiple questions and their corresponding answers, primarily focusing on finding the zeroes of polynomials and verifying relationships between the zeroes and coefficients.

TYPE: MULTIPLE_QUESTIONS_WITH_ANSWERS