NCERT Solutions for Class 10 Maths Chapter 2 Polynomials

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Exercise 2.1 Page: 28

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.

Exercise 2.2 Page: 33

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

Solutions:

(i) x²–2x –8

⇒x²– 4x+2x–8 = x(x–4)+2(x–4) = (x-4)(x+2)

Therefore, zeroes of polynomial equation x²–2x–8 are (4, -2)

Sum of zeroes = 4–2 = 2 = -(-2)/1 = -(Coefficient of x)/(Coefficient of x²)

Product of zeroes = 4×(-2) = -8 =-(8)/1 = (Constant term)/(Coefficient of x²)

(ii) 4s²–4s+1

⇒4s²–2s–2s+1 = 2s(2s–1)–1(2s-1) = (2s–1)(2s–1)

Therefore, zeroes of polynomial equation 4s²–4s+1 are (1/2, 1/2)

Sum of zeroes = (½)+(1/2) = 1 = -(-4)/4 = -(Coefficient of s)/(Coefficient of s²)

Product of zeros = (1/2)×(1/2) = 1/4 = (Constant term)/(Coefficient of s² )

(iii) 6x²–3–7x

⇒6x²–7x–3 = 6x² – 9x + 2x – 3 = 3x(2x – 3) +1(2x – 3) = (3x+1)(2x-3)

Therefore, zeroes of polynomial equation 6x²–3–7x are (-1/3, 3/2)

Sum of zeroes = -(1/3)+(3/2) = (7/6) = -(Coefficient of x)/(Coefficient of x²)

Product of zeroes = -(1/3)×(3/2) = -(3/6) = (Constant term) /(Coefficient of x² )

(iv) 4u²+8u

⇒ 4u(u+2)

Therefore, zeroes of polynomial equation 4u² + 8u are (0, -2).

Sum of zeroes = 0+(-2) = -2 = -(8/4) = = -(Coefficient of u)/(Coefficient of u²)

Product of zeroes = 0×-2 = 0 = 0/4 = (Constant term)/(Coefficient of u² )

(v) t²–15

⇒ t² = 15 or t = ±√15

Therefore, zeroes of polynomial equation t² –15 are (√15, -√15)

Sum of zeroes =√15+(-√15) = 0= -(0/1)= -(Coefficient of t) / (Coefficient of t²)

Product of zeroes = √15×(-√15) = -15 = -15/1 = (Constant term) / (Coefficient of t² )

(vi) 3x²–x–4

⇒ 3x²–4x+3x–4 = x(3x-4)+1(3x-4) = (3x – 4)(x + 1)

Therefore, zeroes of polynomial equation3x² – x – 4 are (4/3, -1)

Sum of zeroes = (4/3)+(-1) = (1/3)= -(-1/3) = -(Coefficient of x) / (Coefficient of x²)

Product of zeroes=(4/3)×(-1) = (-4/3) = (Constant term) /(Coefficient of x² )

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:-

x²–(α+β)x +αβ = 0

x²–(1/4)x +(-1) = 0

4x²–x-4 = 0

Thus, 4x²–x–4 is the quadratic polynomial.

(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:-

x²–(α+β)x +αβ = 0

x² –(√2)x + (1/3) = 0

3x²-3√2x+1 = 0

Thus, 3x²-3√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:-

x²–(α+β)x +αβ = 0

x²–(0)x +√5= 0

Thus, x²+√5 is the quadratic polynomial.

(iv) 1, 1

Solution:

Given,

Sum of zeroes = α+β = 1

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:-

x²–(α+β)x +αβ = 0

x²–x+1 = 0

Thus, x²–x+1 is the quadratic polynomial.

(v) -1/4, 1/4

Solution:

Given,

Sum of zeroes = α+β = -1/4

Product of zeroes = α β = 1/4

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

x²–(α+β)x +αβ = 0

x²–(-1/4)x +(1/4) = 0

4x²+x+1 = 0

Thus, 4x²+x+1 is the quadratic polynomial.

(vi) 4, 1

Solution:

Given,

Sum of zeroes = α+β =4

Product of zeroes = αβ = 1

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

x²–(α+β)x+αβ = 0

x²–4x+1 = 0

Thus, x²–4x+1 is the quadratic polynomial.

Exercise 2.3 Page: 36

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

Solution:

Given,

Dividend = p(x) = x³-3x²+5x–3

Divisor = g(x) = x²– 2

Therefore, upon division we get,

Quotient = x–3

Remainder = 7x–9

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

Solution:

Given,

Dividend = p(x) = x⁴ – 3x² + 4x +5

Divisor = g(x) = x² +1-x

Therefore, upon division we get,

Quotient = x² + x–3

Remainder = 8

(iii) p(x) =x⁴–5x+6, g(x) = 2–x²

Solution:

Given,

Dividend = p(x) =x⁴ – 5x + 6 = x⁴ +0x²–5x+6

Divisor = g(x) = 2–x² = –x²+2

Therefore, upon division we get,

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

Solutions:

Given,

First polynomial = t²-3

Second polynomial = 2t⁴ +3t³-2t² -9t-12

As we can see, the remainder is left as 0. Therefore, we say that, t²-3 is a factor of 2t⁴ +3t³-2t² -9t-12.

(ii)x²+3x+1 , 3x⁴+5x³-7x²+2x+2

Solutions:

Given,

First polynomial = x²+3x+1

Second polynomial = 3x⁴+5x³-7x²+2x+2

As we can see, the remainder is left as 0. Therefore, we say that, x² + 3x + 1 is a factor of 3x⁴+5x³-7x²+2x+2.

(iii) x³-3x+1, x⁵-4x³+x²+3x+1

Solutions:

Given,

First polynomial = x³-3x+1

Second polynomial = x⁵-4x³+x²+3x+1

As we can see, the remainder is not equal to 0. Therefore, we say that, x³-3x+1 is not a factor of x⁵-4x³+x²+3x+1 .

3. Obtain all other zeroes of 3x⁴+6x³-2x²-10x-5, if two of its zeroes are √(5/3) and – √(5/3).

Solutions:

Since this is a polynomial equation of degree 4, hence there will be total 4 roots.

√(5/3) and – √(5/3) are zeroes of polynomial f(x).

∴ (x –√(5/3)) (x+√(5/3) = x²-(5/3) = 0

(3x²−5)=0, is a factor of given polynomial f(x).

Now, when we will divide f(x) by (3x²−5) the quotient obtained will also be a factor of f(x) and the remainder will be 0.

Therefore, 3x⁴ +6x³ −2x² −10x–5 = (3x² –5)(x²+2x+1)

Now, on further factorizing (x²+2x+1) we get,

x²+2x+1 = x²+x+x+1 = 0

x(x+1)+1(x+1) = 0

(x+1)(x+1) = 0

So, its zeroes are given by: x= −1 and x = −1.

Therefore, all four zeroes of given polynomial equation are:

√(5/3),- √(5/3) , −1 and −1.

Hence, is the answer.

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).

Solution:

Given,

Dividend, p(x) = x³-3x²+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

∴ x³-3x²+x+2 = g(x)×(x-2) + (-2x+4)

x³-3x²+x+2-(-2x+4) = g(x)×(x-2)

Therefore, g(x) × (x-2) = x³-3x²+3x-2

Now, for finding g(x) we will divide x³-3x²+3x-2 with (x-2)

Therefore, g(x) = (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

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, p(x) = 3x²+3x+3 is a polynomial to be divided by g(x) = 3.

So, (3x²+3x+3)/3 = x²+x+1 = q(x)

Thus, you can see, the degree of quotient q(x) = 2, which also equal to the degree of dividend p(x).

Hence, division algorithm is satisfied here.

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

Let us take an example, p(x) = x² + 3 is a polynomial to be divided by g(x) = x – 1.

So, x² + 3 = (x – 1)×(x) + (x + 3)

Hence, quotient q(x) = x

Also, remainder r(x) = x + 3

Thus, you can see, the degree of quotient q(x) = 1, which is also equal to the degree of remainder r(x).

Hence, division algorithm is satisfied here.

(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) = x² + 1 is a polynomial to be divided by g(x) = x.

So, x² + 1 = (x)×(x) + 1

Hence, quotient q(x) = x

And, remainder r(x) = 1

Clearly, the degree of remainder here is 0.

Hence, division algorithm is satisfied here.

Exercise 2.4 Page: 36

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:

(i) 2x³+x²-5x+2; -1/2, 1, -2

Solution:

Given, p(x) = 2x³+x²-5x+2

And zeroes for p(x) are = 1/2, 1, -2

∴ p(1/2) = 2(1/2)³+(1/2)²-5(1/2)+2 = (1/4)+(1/4)-(5/2)+2 = 0

p(1) = 2(1)³+(1)²-5(1)+2 = 0

p(-2) = 2(-2)³+(-2)²-5(-2)+2 = 0

Hence, proved 1/2, 1, -2 are the zeroes of 2x³+x²-5x+2.

Now, comparing the given polynomial with general expression, we get;

∴ ax³+bx²+cx+d = 2x³+x²-5x+2

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

As we know, if α, β, γ are the zeroes of the cubic polynomial ax³+bx²+cx+d , then;

α +β+γ = –b/a

αβ+βγ+γα = c/a

α βγ = – d/a.

Therefore, putting the values of zeroes of the polynomial,

α+β+γ = ½+1+(-2) = -1/2 = –b/a

αβ+βγ+γα = (1/2×1)+(1 ×-2)+(-2×1/2) = -5/2 = c/a

α β γ = ½×1×(-2) = -2/2 = -d/a

Hence, the relationship between the zeroes and the coefficients are satisfied.

(ii) x³-4x²+5x-2 ;2, 1, 1

Solution:

Given, p(x) = x³-4x²+5x-2

And zeroes for p(x) are 2,1,1.

∴ p(2)= 2³-4(2)²+5(2)-2 = 0

p(1) = 1³-(4×1² )+(5×1)-2 = 0

Hence proved, 2, 1, 1 are the zeroes of x³-4x²+5x-2

Now, comparing the given polynomial with general expression, we get;

∴ ax³+bx²+cx+d = x³-4x²+5x-2

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

As we know, if α, β, γ are the zeroes of the cubic polynomial ax³+bx²+cx+d , then;

α + β + γ = –b/a

αβ + βγ + γα = c/a

α β γ = – d/a.

Therefore, putting the values of zeroes of the polynomial,

α +β+γ = 2+1+1 = 4 = -(-4)/1 = –b/a

αβ+βγ+γα = 2×1+1×1+1×2 = 5 = 5/1= c/a

αβγ = 2×1×1 = 2 = -(-2)/1 = -d/a

Hence, the relationship between the zeroes and the coefficients are satisfied.

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.

Solution:

Let us consider the cubic polynomial is ax³+bx²+cx+d and the values of the zeroes of the polynomials be α, β, γ.

As per the given question,

α+β+γ = -b/a = 2/1

αβ +βγ+γα = c/a = -7/1

α βγ = -d/a = -14/1

Thus, from above three expressions we get the values of coefficient of polynomial.

a = 1, b = -2, c = -7, d = 14

Hence, the cubic polynomial is x³-2x²-7x+14

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

Solution:

We are given with the polynomial here,

p(x) = x³-3x²+x+1

And zeroes are given as a – b, a, a + b

Now, comparing the given polynomial with general expression, we get;

∴px³+qx²+rx+s = x³-3x²+x+1

p = 1, q = -3, r = 1 and s = 1

Sum of zeroes = a – b + a + a + b

-q/p = 3a

Putting the values q and p.

-(-3)/1 = 3a

a=1

Thus, the zeroes are 1-b, 1, 1+b.

Now, product of zeroes = 1(1-b)(1+b)

-s/p = 1-b²

-1/1 = 1-b²

b² = 1+1 = 2

b = ±√2

Hence,1-√2, 1 ,1+√2 are the zeroes of x³-3x²+x+1.

4. If two zeroes of the polynomial x⁴-6x³-26x²+138x-35 are 2 ±√3, find other zeroes.

Solution:

Since this is a polynomial equation of degree 4, hence there will be total 4 roots.

Let f(x) = x⁴-6x³-26x²+138x-35

Since 2 +√3 and 2-√3 are zeroes of given polynomial f(x).

∴ [x−(2+√3)] [x−(2-√3)] = 0

(x−2−√3)(x−2+√3) = 0

On multiplying the above equation we get,

x²-4x+1, this is a factor of a given polynomial f(x).

Now, if we will divide f(x) by g(x), the quotient will also be a factor of f(x) and the remainder will be 0.

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

Now, on further factorizing (x²–2x−35) we get,

x²–(7−5)x −35 = x²– 7x+5x+35 = 0

x(x −7)+5(x−7) = 0

(x+5)(x−7) = 0

So, its zeroes are given by:

x= −5 and x = 7.

Therefore, all four zeroes of given polynomial equation are: 2+√3 , 2-√3, −5 and 7.

Q.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.

Solution:

Let’s divide x⁴ – 6x³ + 16x² – 25x + 10 by x² – 2x + k.

Given that the remainder of the polynomial division is x + a.

(4k – 25 + 16 – 2k)x + [10 – k(8 – k)] = x + a

(2k – 9)x + (10 – 8k + k²) = x + a

Comparing the coefficients of the above equation, we get;

2k – 9 = 1

2k = 9 + 1 = 10

k = 10/2 = 5

And

10 – 8k + k² = a

10 – 8(5) + (5)² = a [since k = 5]

10 – 40 + 25 = a

a = -5

Therefore, k = 5 and a = -5.

NCERT Solutions for Class 10 Maths Chapter 2 – Polynomials

As this is one of the important topics in Maths, it comes under the unit – Algebra which has a weightage of 20 marks in the Class 10 Maths CBSE exams. The average number of questions asked from this chapter is usually 1. This chapter talks about the following,

• Introduction to Polynomials
• Geometrical Meaning of the Zeros of Polynomial
• Relationship between Zeros and Coefficients of a Polynomial
• Division Algorithm for Polynomials

Polynomials are introduced in Class 9, where we discussed polynomials in one variable and their degrees in the previous class. This is discussed in further detail in Class 10. The NCERT Solutions for Class 10 Maths for this chapter discusses the answers to various types of questions related to polynomials and their applications. We study the division algorithm for polynomials of integers, and also whether the zeroes of quadratic polynomials are related to their coefficients.

The chapter starts with the introduction of polynomials in section 2.1, followed by two very important topics in sections 2.2 and 2.3

• Geometrical Meaning of the zeroes of a Polynomial – It includes 1 question having 6 different cases.
• Relationship between Zeroes and Coefficients of a Polynomial – Explore the relationship between zeroes and coefficients of a quadratic polynomial through solutions to 2 problems in Exercise 2.2, having 6 parts in each question.

Next, it discusses the following topics, which were introduced in Class 9.

• Division Algorithm for Polynomials – In this, the solutions for 5 problems in Exercise 2.3 is given, having three long questions.

Key Features of NCERT Solutions for Class 10 Maths Chapter 2 – Polynomials

• It covers the CBSE syllabus for 2023-24 of Class 10 Maths.
• After studying these NCERT Solutions prepared by our subject experts, you will be confident of scoring well in the exams.
• It follows NCERT guidelines which help in preparing the students accordingly.
• It contains all the important questions from the examination point of view.

For a strong grip over the concepts, students can also make use of the other reference materials which are present at BYJU’S.

• RD Sharma Solutions for Class 10 Maths Chapter 2 Polynomials

Disclaimer – 

Dropped Topics – 2.4 Division algorithm for polynomials