* You are going to go through Polynomials Chapter 2 NCERT Solutions For Class 10 CBSE Mathematics. This post presents to the students a clear conception of* how to move with the basics of Extra Questions and answers. The expert prepared The Extra Questions and And Answers. https://cbsencertanswers.com/is very much to make

*.*

*things way simpler and easier for the students. Especially those who are appearing for the board exams. We took every care to make sure that the effort serves the purpose. So, let us find out Polynomials Chapter 2 NCERT Solutions For Class 10 CBSE Mathematics*.*On this page, you can find Polynomials Chapter 2 NCERT Solutions For Class 10 CBSE Mathematics*## Ex 2.1

## 1. The graphs of y = p(x) are given in Fig. 2.10 below, for some polynomials p(x). Find the number of zeroes of p(x), in each case.

## Solutions:

**Graphical method to find zeroes:-**

Total number of zeroes in any polynomial equation = total number of times the curve intersects x-axis.

**(i)** In the given graph, the number of zeroes of p(x) is 0 because the graph is parallel to x-axis does not cut it at any point.

**(ii)** In the given graph, the number of zeroes of p(x) is 1 because the graph intersects the x-axis at only one point.

**(iii)** In the given graph, the number of zeroes of p(x) is 3 because the graph intersects the x-axis at any three points.

**(iv)** In the given graph, the number of zeroes of p(x) is 2 because the graph intersects the x-axis at two points.

**(v)** In the given graph, the number of zeroes of p(x) is 4 because the graph intersects the x-axis at four points.

**(vi)** In the given graph, the number of zeroes of p(x) is 3 because the graph intersects the x-axis at Three points.

### YOU ARE READING: Polynomials Chapter 2 NCERT Solutions For Class 10 CBSE Mathematics

## Ex 2.2

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

## Solutions:

(i) x^{2}–2x –8

⇒ x^{2} – 4x+2x–8 = x(x–4)+2(x–4) = (x-4)(x+2)

Therefore, zeroes of polynomial equation x^{2} –2x–8 are (4, -2)

Sum of zeroes = 4–2 = 2 = -(-2)/1 = -(Coefficient of x)/(Coefficient of x^{2} )

Product of zeroes = 4×(-2) = -8 =-(8)/1 = (Constant term)/(Coefficient of x^{2} )

(ii) 4s^{2} –4s+1

⇒4s^{2} –2s–2s+1 = 2s(2s–1)–1(2s-1) = (2s–1)(2s–1)

Therefore, zeroes of polynomial equation 4s^{2} –4s+1 are (1/2, 1/2)

Sum of zeroes = (½)+(1/2) = 1 = -4/4 = -(Coefficient of s)/(Coefficient of s^{2})

Product of zeros = (1/2)×(1/2) = 1/4 = (Constant term)/(Coefficient of s^{2} )

(iii) 6x^{2} –3–7x

⇒6x^{2} –7x–3 = 6x^{2} – 9x + 2x – 3 = 3x(2x – 3) +1(2x – 3) = (3x+1)(2x-3)

Therefore, zeroes of polynomial equation 6x^{2} –3–7x are (-1/3, 3/2)

Sum of zeroes = -(1/3)+(3/2) = (7/6) = -(Coefficient of x)/(Coefficient of x^{2} )

Product of zeroes = -(1/3)×(3/2) = -(3/6) = (Constant term) /(Coefficient of x^{2} )

### YOU ARE READING: Polynomials Chapter 2 NCERT Solutions For Class 10 CBSE Mathematics

(iv) 4u^{2}+8u

⇒ 4u(u+2)

Therefore, zeroes of polynomial equation 4u^{2} + 8u are (0, -2).

Sum of zeroes = 0+(-2) = -2 = -(8/4) = = -(Coefficient of u)/(Coefficient of u^{2} )

Product of zeroes = 0×-2 = 0 = 0/4 = (Constant term)/(Coefficient of u^{2} )

(v) t^{2}–15

⇒ t^{2} = 15 or t = ±√15

Therefore, zeroes of polynomial equation t^{2} –15 are (√15, -√15)

Sum of zeroes =√15+(-√15) = 0= -(0/1)= -(Coefficient of t) / (Coefficient of t^{2} )

Product of zeroes = √15×(-√15) = -15 = -15/1 = (Constant term) / (Coefficient of t^{2} )

(vi) 3x^{2}–x–4

⇒ 3x^{2} –4x+3x–4 = x(3x-4)+1(3x-4) = (3x – 4)(x + 1)

Therefore, zeroes of polynomial equation 3x^{2} – x – 4 are (4/3, -1)

Sum of zeroes = (4/3)+(-1) = (1/3)= -(-1/3) = -(Coefficient of x) / (Coefficient of x^{2} )

Product of zeroes=(4/3)×(-1) = (-4/3) = (Constant term) /(Coefficient of x^{2} )

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

## (i) 1/4 , -1

## Solution:

From the formulas of sum and product of zeroes, we know,

Sum of zeroes = α+β

Product of zeroes = α β

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:-

–(α+β)x +αβ = 0

–(1/4)x +(-1) = 0

4x^{2} –x-4 = 0

Thus,4x^{2} –x–4 is the quadratic polynomial.

### YOU ARE READING: Polynomials Chapter 2 NCERT Solutions For Class 10 CBSE Mathematics

## (ii)√2, 1/3

## Solution:

Sum of zeroes = α + β =√2

Product of zeroes = α β = 1/3

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

x^{2} –(α+β)x +αβ = 0

x^{2} –(√2)x + (1/3) = 0

3x^{2} -3√2x+1 = 0

Thus, 3x^{2} -3√2x+1 is the quadratic polynomial.

## (iii) 0, √5

## Solution:

Given,

Sum of zeroes = α+β = 0

Product of zeroes = α β = √5

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

x^{2}–(α+β)x +αβ = 0

x^{2} –(0)x +√5= 0

Thus, x

2+√5 is the quadratic polynomial.

## (iv) 1, 1

## Solution:

Given,

Sum of zeroes = α+β = 1

Product of zeroes = α β = 1

x^{2} –(α+β)x +αβ = 0

x^{2} –x+1 = 0

Thus , x^{2} –x+1is the quadratic polynomial.

## (v) -1/4, 1/4

## Solution:

Given,

Sum of zeroes = α+β = -1/4

Product of zeroes = α β = 1/4

x^{2} –(α+β)x +αβ = 0

x^{2} –(-1/4)x +(1/4) = 0

4x^{2} +x+1 = 0

Thus,4x^{2} +x+1 is the quadratic polynomial.

### YOU ARE READING: Polynomials Chapter 2 NCERT Solutions For Class 10 CBSE Mathematics

## (vi) 4, 1

## Solution:

Given,

Sum of zeroes = α+β = 4

Product of zeroes = αβ = 1

x^{2} –(α+β)x+αβ = 0

x^{2} –4x+1 = 0

Thus, x^{2} –4x+1 is the quadratic polynomial.

## Ex 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) = x^{3} -3x^{2} +5x–3 ,

## g(x) = x^{2} –2

## Solution:

Given,

Dividend = p(x) = x^{3} -3x^{2} +5x–3

Divisor = g(x) = x^{2} – 2

Therefore, upon division we get,

Quotient = x–3

Remainder = 7x–9

### YOU ARE READING: Polynomials Chapter 2 NCERT Solutions For Class 10 CBSE Mathematics

## (ii) p(x) =x^{4} – 3x^{2} +4x+5 ,

## g(x) = x^{2} +1-x

## Solution:

Given,

Dividend = p(x) = x^{4} – 3x^{2} +4x+5

Divisor = g(x) = x^{2} +1-x

Therefore, upon division we get,

Quotient = x^{2} + x–3

Remainder = 8

## (iii) p(x) = x^{4} –5x+6,

## g(x) = 2– x^{2}

## Solution:

Given,

Dividend = p(x) = x^{4} –5x+6 = x^{4} +0x^{2}–5x+6

Divisor = g(x) = 2– x^{2} = –x^{2} +2

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

## (i) t^{2}-3, 2t^{4} +3t^{3} -2t^{2} -9t-12

## Solutions:

Given,

First polynomial = t^{2} -3

Second polynomial = 2t^{4} +3t^{3} -2t^{2} -9t-12

As we can see, the remainder is left as 0. Therefore, we say that,

t^{2} -3 is a factor of 2t^{2} +3t +4.

### YOU ARE READING: Polynomials Chapter 2 NCERT Solutions For Class 10 CBSE Mathematics

## (ii) x^{2}+3x+1 , 3x^{4}+5x^{3}-7x^{2}+2x+2

## Solutions:

Given,

First polynomial = x^{2}+3x+1

Second polynomial = 3x^{4}+5x^{3}-7x^{2}+2x+2

As we can see, the remainder is left as 0. Therefore, we say that,

x^{2}+3x+1 is a factor of 3x^{4}+5x^{3}-7x^{2}+2x+2.

## (iii) x^{2} -3x+1, x^{5}-4x^{3} + x^{2} +3x+1

## Solutions:

Given,

First polynomial = x^{2} -3x+1

Second polynomial = x^{5}-4x^{3} + x^{2} +3x+1

As we can see, the remainder is not equal to 0. Therefore, we say that, x^{3}-3x+1

is not a factor of x^{5}-4x^{3} + x^{2} +3x+1.

√(5/3),- √(5/3) , −1 and −1.

Hence, is the answer

### YOU ARE READING: Polynomials Chapter 2 NCERT Solutions For Class 10 CBSE Mathematics

## 4. On dividing x^{3}-3x^{2}+x+2 by a polynomial g(x), the quotient and remainder were x–2 and –2x+4, respectively Find g(x).

## Solutions:

Given,

Dividend, p(x) = x^{3}-3x^{2}+x+2

Quotient = x-2

Remainder = –2x+4

We have to find the value of Divisor, g(x) =?

As we know,

Dividend = Divisor × Quotient + Remainder

## 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

## Solutions:

According to the division algorithm, dividend p(x) and divisor g(x) are two polynomials, where g(x)≠0. Then we

can find the value of quotient q(x) and remainder r(x), with the help of below given formula;

Dividend = Divisor × Quotient + Remainder

∴ p(x) = g(x)×q(x)+r(x)

Where r(x) = 0 or degree of r(x)< degree of g(x).

Now let us proof the three given cases as per division algorithm by taking examples for each.

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

Degree of dividend is equal to degree of quotient, only when the divisor is a constant term.

Let us take an example, 3x^{2}+3x+3 is a polynomial to be divided by 3.

So, (3x^{2} +3x+3)/3 = x^{2} +x+1 = q(x)

Thus, you can see, the degree of quotient is equal to the degree of dividend.

Hence, division algorithm is satisfied here.

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

Let us take an example , p(x)= x^{2} +x is a polynomial to be divided by g(x)=x.

So, ( x^{2} +x)/x = x+1 = q(x)

Also, remainder, r(x) = 0

Thus, you can see, the degree of quotient is equal to the degree of remainder.

Hence, division algorithm is satisfied here.

### YOU ARE READING: Polynomials Chapter 2 NCERT Solutions For Class 10 CBSE Mathematics

**(iii) deg r(x) = 0**

The degree of remainder is 0 only when the remainder left after division algorithm is constant.

Let us take an example, p(x) = x^{2} + 1 is a polynomial to be divided by g(x)=x.

So,( x^{2} +1)/x= x=q(x)

And r(x)=1

Clearly, the degree of remainder here is 0.

Hence, division algorithm is satisfied here

### YOU ARE READING: Polynomials Chapter 2 NCERT Solutions For Class 10 CBSE Mathematics