Step by Step NCERT Solutions for Class 10 Maths Chapter Chapter 2 Polynomials. This solutions covers all questions of Chapter 2 Polynomials Class 10 as per CBSE Board guidelines from the latest NCERT book for class 10 maths. Following The topics and sub-topics in Chapter 2 Polynomials are covered.

2.1 Introduction,

2.2 Geometrical Meaning of the Zeroes of a Polynomial,

2.3 Relationship between Zeroes and Coefficients of a Polynomial,

2.4 Division Algorithm for Polynomials and

2.5 Summary

## Polynomials NCERT Solutions – Class 10 Maths

### NCERT Solutions for Class 10 Maths Chapter 2 – Free PDF Download

**Exercise 2.1 : **Solutions of Questions on Page Number **: 28**

**Q1 :**

**The graphs of y = p(x) are given in following figure, for some polynomials p(x). Find the number of zeroes of p(x), in each case.**

**(i)**

**(ii)**

**(iii)**

**(iv)**

**(v)**

**(vi)**

**Answer :**

(i) The number of zeroes is 0 as the graph does not cut the x-axis at any point.

(ii) The number of zeroes is 1 as the graph intersects the x-axis at only 1 point.

(iii) The number of zeroes is 3 as the graph intersects the x-axis at 3 points.

(iv) The number of zeroes is 2 as the graph intersects the x-axis at 2 points.

(v) The number of zeroes is 4 as the graph intersects the x-axis at 4 points.

(vi) The number of zeroes is 3 as the graph intersects the x-axis at 3 points.

**Exercise 2.2 :**Solutions of Questions on Page Number

**: 33**

**Q1 : Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.**

**Answer :**

The value of is zero when x – 4 = 0 or x + 2 = 0, i.e., when x = 4 or x = – 2

Therefore, the zeroes of are 4 and – 2.

Sum of zeroes =

Product of zeroes

The value of 4s² – 4s + 1 is zero when 2s – 1 = 0, i.e.,

Therefore, the zeroes of 4s² – 4s + 1 areand.

Sum of zeroes =

Product of zeroes

The value of 6x

^{2}– 3 – 7x is zero when 3x + 1 = 0 or 2x – 3 = 0, i.e., or

Therefore, the zeroes of 6x

^{2}– 3 – 7x are.

Sum of zeroes =

Product of zeroes =

The value of 4u

^{2}+ 8u is zero when 4u = 0 or u + 2 = 0, i.e., u = 0 or u = – 2

Therefore, the zeroes of 4u

^{2}+ 8u are 0 and – 2.

Sum of zeroes =

Product of zeroes =

The value of t

^{2}– 15 is zero when or , i.e., when

**Q2 : Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.**

**Answer :**

Let the polynomial be , and its zeroes be and .

Therefore, the quadratic polynomial is 4x

^{2}– x – 4.

Let the polynomial be , and its zeroes be and .

Therefore, the quadratic polynomial is 3x

^{2}– x+ 1.

Let the polynomial be , and its zeroes be and .

Therefore, the quadratic polynomial is .

Let the polynomial be, and its zeroes beand.

Therefore, the quadratic polynomial is x

^{2}– 1+ 1.

**Exercise 2.3 :**Solutions of Questions on Page Number

**: 36**

**Q1 : Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following:**

**(i)**

**(ii)**

**(iii)**

**Answer :**

Quotient = x – 3

Remainder = 7x – 9

Quotient = x

^{2}+ x – 3

Remainder = 8

Quotient = – x

^{2}– 2

Remainder = – 5x +10

**Q2 : Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial:**

**Answer :**

=

Since the remainder is 0,

Hence, is a factor of .

Since the remainder is 0,

Hence, is a factor of .

Since the remainder ,

Hence, is not a factor of

**Q3 : Obtain all other zeroes of , if two of its zeroes are .**

**Answer :**

Since the two zeroes are ,

is a factor of .

Therefore, we divide the given polynomial by .

We factorize

Therefore, its zero is given by x + 1 = 0

x = – 1

As it has the term , therefore, there will be 2 zeroes at x = – 1.

Hence, the zeroes of the given polynomial are, – 1 and – 1.

**Q4 : On dividing by a polynomial g(x), the quotient and remainder were x – 2 and – 2x + 4, respectively. Find g(x).**

**Answer :**

g(x) = ? (Divisor)

Quotient = (x – 2)

Remainder = ( – 2x + 4)

Dividend = Divisor × Quotient + Remainder

g(x) is the quotient when we divide by

**Q5 : Give examples of polynomial 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**

**Answer :**

According to the division algorithm, if p(x) and g(x) are two polynomials with

g(x) ≠ 0, then we can find polynomials q(x) and r(x) such that

p(x) = g(x) × q(x) + r(x),

where r(x) = 0 or degree of r(x) < degree of g(x)

Degree of a polynomial is the highest power of the variable in the polynomial.

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

Degree of quotient will be equal to degree of dividend when divisor is constant ( i.e., when any polynomial is divided by a constant).

Let us assume the division of by 2.

Here, p(x) =

g(x) = 2

q(x) = and r(x) = 0

Degree of p(x) and q(x) is the same i.e., 2.

Checking for division algorithm,

p(x) = g(x) × q(x) + r(x)

= 2()

=Thus, the division algorithm is satisfied.

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

Let us assume the division of x

^{3}+ x by x

^{2},

Here, p(x) = x

^{3}+ x

g(x) = x

^{2}

q(x) = x and r(x) = x

Clearly, the degree of q(x) and r(x) is the same i.e., 1.

Checking for division algorithm,

p(x) = g(x) × q(x) + r(x)

x

^{3}+ x = (x

^{2}) × x + x

x

^{3}+ x = x

^{3}+ x

Thus, the division algorithm is satisfied.

(iii)deg r(x) = 0

Degree of remainder will be 0 when remainder comes to a constant.

Let us assume the division of x

^{3}+ 1by x

^{2}.

Here, p(x) = x

^{3}+ 1

g(x) = x

^{2}

q(x) = x and r(x) = 1

Clearly, the degree of r(x) is 0.

Checking for division algorithm,

p(x) = g(x) × q(x) + r(x)

x

^{3}+ 1 = (x

^{2}) × x + 1

x

^{3}+ 1 = x

^{3}+ 1

Thus, the division algorithm is satisfied.

**Exercise 2.4 :**Solutions of Questions on Page Number

**: 37**

**Q 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:**

**Answer :**

(i)

Therefore,, 1, and – 2 are the zeroes of the given polynomial.

Comparing the given polynomial with, we obtain a = 2, b = 1, c = – 5, d = 2

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

(ii)

Therefore, 2, 1, 1 are the zeroes of the given polynomial.

Comparing the given polynomial with, we obtain a = 1, b = – 4, c = 5, d = – 2.

Verification of the relationship between zeroes and coefficient of the given polynomial

Multiplication of zeroes taking two at a time = (2)(1) + (1)(1) + (2)(1) =2 + 1 + 2 = 5

Multiplication of zeroes = 2 × 1 × 1 = 2

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

**Q2 : 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.**

**Answer :**

Let the polynomial be and the zeroes be .

It is given that

If a = 1, then b = – 2, c = – 7, d = 14

Hence, the polynomial is .

**Q3 : If the zeroes of polynomial are, find a and b.**

**Answer :**

Zeroes are a – b, a + a + b

Comparing the given polynomial with , we obtain

p = 1, q = – 3, r = 1, t = 1

The zeroes are .

Hence, a = 1 and b = or- .

**Q4 : It two zeroes of the polynomial are, find other zeroes.**

**Answer :**

Given that 2 + and 2-are zeroes of the given polynomial.

Therefore, = x

^{2}+ 4 – 4x – 3

= x

^{2}– 4x + 1 is a factor of the given polynomial

For finding the remaining zeroes of the given polynomial, we will find the quotient by dividing by x

^{2}– 4x + 1.

Clearly, =

It can be observed that is also a factor of the given polynomial.

And =

Therefore, the value of the polynomial is also zero when or

Or x = 7 or – 5

Hence, 7 and – 5 are also zeroes of this polynomial.

**Q5 : If the polynomial is divided by another polynomial, the remainder comes out to be x + a, find k and a.**

**Answer :**

By division algorithm,

Dividend = Divisor × Quotient + Remainder

Dividend – Remainder = Divisor × Quotient

will be perfectly divisible by .

Let us divide by

It can be observed that will be 0.

Therefore, = 0 and = 0

For = 0,

2 k =10

And thus, k = 5

For = 0

10 – a – 8 × 5 + 25 = 0

10 – a – 40 + 25 = 0

– 5 – a = 0

Therefore, a = – 5

Hence, k = 5 and a = – 5