# CBSE Class 10 NCERT Solutions for Maths Chapter 2, Polynomials

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*सीबीएसई कक्षा 10 एनसीईआरटी समाधान)***ou can download the CBSE Class 10 NCERT Solutions for Maths Chapter 2, Polynomials PDF is also available in this article.**

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**CBSE Class 10 Maths NCERT Problems & Solution**s

#### Question 1:

The graphs of *y* = *p*(*x*) are given in the following the 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.

#### 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

#### 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.

#### 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 .

#### 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.

#### 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

#### 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.

#### 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 .

#### 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.

#### 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

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