# NCERT Solutions for Class 10 Math Chapter 2 - Polynomials

NCERT Solutions for Class 10 Math Chapter 2 Polynomials are provided here with simple step-by-step explanations. These solutions for Polynomials are extremely popular among class 10 students for Math Polynomials Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the NCERT Book of class 10 Math Chapter 2 are provided here for you for free. You will also love the ad-free experience on Meritnation’s NCERT Solutions. All NCERT solutions for class 10 Math are prepared by experts and are 100% accurate.

#### Page No 28:

#### Question 1:

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)

(v)

#### 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.

##### Video Solution for polynomials (Page: 28 , Q.No.: 1)

NCERT Solution for Class 10 math - polynomials 28 , Question 1

#### Page No 33:

#### Question 1:

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 4*s*^{2} − 4*s* + 1 is zero when 2*s* − 1 = 0, i.e.,

Therefore, the zeroes of 4*s*^{2} − 4*s* + 1 areand.

Sum of zeroes =

Product of zeroes

The value of 6*x*^{2} − 3 − 7*x* is zero when 3*x* + 1 = 0 or 2*x *− 3 = 0, i.e., or

Therefore, the zeroes of 6*x*^{2} − 3 − 7*x* are.

Sum of zeroes =

Product of zeroes =

The value of 4*u*^{2} + 8*u* is zero when 4*u* = 0 or *u* + 2 = 0, i.e., *u* = 0 or *u* = −2

Therefore, the zeroes of 4*u*^{2} + 8*u* are 0 and −2.

Sum of zeroes =

Product of zeroes =

The value of *t*^{2} − 15 is zero when or , i.e., when

Therefore, the zeroes of *t*^{2} − 15 are and.

Sum of zeroes =

Product of zeroes =

The value of 3*x*^{2} − *x* − 4 is zero when 3*x* − 4 = 0 or *x* + 1 = 0, i.e., when or *x* = −1

Therefore, the zeroes of 3*x*^{2} − *x* − 4 are and −1.

Sum of zeroes =

Product of zeroes

##### Video Solution for polynomials (Page: 33 , Q.No.: 1)

NCERT Solution for Class 10 math - polynomials 33 , Question 1

#### Page No 33:

#### Question 2:

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 4*x*^{2} − *x* − 4.

Let the polynomial be , and its zeroes be and .

Therefore, the quadratic polynomial is 3*x*^{2} − *x* + 1.

Let the polynomial be , and its zeroes be and .

Therefore, the quadratic polynomial is .

Let the polynomial be , and its zeroes be and .

Therefore, the quadratic polynomial is .

Let the polynomial be , and its zeroes be and .

Therefore, the quadratic polynomial is .

Let the polynomial be .

Therefore, the quadratic polynomial is.

##### Video Solution for polynomials (Page: 33 , Q.No.: 2)

NCERT Solution for Class 10 math - polynomials 33 , Question 2

#### Page No 36:

#### 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:

#### 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.

#### Page No 36:

#### 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.

#### 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 .

##### Video Solution for polynomials (Page: 36 , Q.No.: 2)

NCERT Solution for Class 10 math - polynomials 36 , Question 2

#### Page No 36:

#### Question 3:

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.

##### Video Solution for polynomials (Page: 36 , Q.No.: 3)

NCERT Solution for Class 10 math - polynomials 36 , Question 3

#### Page No 36:

#### Question 4:

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

#### Answer:

*g*(*x*) = ? (Divisor)

Quotient = (*x* − 2)

Remainder = (− 2*x* + 4)

Dividend = Divisor × Quotient + Remainder

*g*(*x*) is the quotient when we divide by

##### Video Solution for polynomials (Page: 36 , Q.No.: 4)

NCERT Solution for Class 10 math - polynomials 36 , Question 4

#### Page No 36:

#### Question 5:

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.

##### Video Solution for polynomials (Page: 36 , Q.No.: 5)

NCERT Solution for Class 10 math - polynomials 36 , Question 5

#### Page No 37:

#### Question 3:

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 .

##### Video Solution for polynomials (Page: 37 , Q.No.: 3)

NCERT Solution for Class 10 math - polynomials 37 , Question 3

#### Page No 37:

#### Question 4:

**]**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 − 4*x − *3

*= x*^{2} − 4*x + *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} − 4*x + *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.

##### Video Solution for polynomials (Page: 37 , Q.No.: 4)

NCERT Solution for Class 10 math - polynomials 37 , Question 4

#### Page No 37:

#### Question 5:

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

##### Video Solution for polynomials (Page: 37 , Q.No.: 5)

NCERT Solution for Class 10 math - polynomials 37 , Question 5

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