## NCERT Solutions for Class 10 Maths Chapter 2 Polynomials

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Page No: 28

Exercises 2.1

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) 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

Exercise 2.2

1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.
(i) x2 – 2x – 8
(ii) 4s2 – 4s + 1
(iii) 6x2 – 3 – 7x
(iv) 4u2 + 8u
(v) t2 – 15
(vi) 3x2 x – 4

(i) x2 – 2x – 8
= (x - 4) (x + 2)
The value of x2 – 2x – 8 is zero when x - 4 = 0 or x + 2 = 0, i.e., when x = 4 or x = -2
Therefore, the zeroes of x2 – 2x – 8 are 4 and -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
= (2s-1)2
The value of 4s2 - 4s + 1 is zero when 2s - 1 = 0, i.e., s = 1/2
Therefore, the zeroes of 4s2 - 4s + 1 are 1/2 and 1/2.

Sum of zeroes = 1/2 + 1/2 = 1 = -(-4)/4 = -(Coefficient of s)/Coefficient of s2
Product of zeroes = 1/2 × 1/2 = 1/4 = Constant term/Coefficient of s2.

(iii) 6x2 – 3 – 7x
= 6x2 – 7x – 3
= (3x + 1) (2x - 3)
The value of 6x2 - 3 - 7x is zero when 3x + 1 = 0 or 2x - 3 = 0, i.e., x = -1/3 or x = 3/2
Therefore, the zeroes of 6x2 - 3 - 7x are -1/3 and 3/2.

Sum of zeroes = -1/3 + 3/2 = 7/6 = -(-7)/6 = -(Coefficient of x)/Coefficient of x2
Product of zeroes = -1/3 × 3/2 = -1/2 = -3/6 = Constant term/Coefficient of x2.

(iv) 4u2 + 8u
4u2 + 8u + 0
= 4u(u + 2)
The value of 4u2 + 8u is zero when 4u = 0 or u + 2 = 0, i.e., u = 0 or u = - 2
Therefore, the zeroes of 4u2 + 8u are 0 and - 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
t- 0.t - 15
= (t - √15) (t + √15)
The value of t2 - 15 is zero when t - √15 = 0 or t + √15 = 0, i.e., when t = √15 or = -√15
Therefore, the zeroes of t2 - 15 are √15 and -√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 u2.

(vi) 3x2 – x – 4
= (3x - 4) (x + 1)
The value of 3x2 – x – 4 is zero when 3x - 4 = 0 and x + 1 = 0,i.e., when x = 4/3 or x = -1
Therefore, the zeroes of 3x2 – x – 4 are 4/3 and -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
(ii) √2 , 1/
(iii) 0, √5
(iv) 1,1
(v) -1/4 ,1/
(vi) 4,1

(i) 1/4 , -1
Let the polynomial be ax2 + bx + c, and its zeroes be Î± and ÃŸ
Î± + ÃŸ = 1/4 = -b/a
Î±ÃŸ = -1 = -4/4 = c/a
If a = 4, then b = -1, c = -4
Therefore, the quadratic polynomial is 4x2 - x -4.

(ii) √2 , 1/3
Let the polynomial be ax2 + bx + c, and its zeroes be Î± and ÃŸ
Î± + ÃŸ = √2 = 3√2/3 = -b/a
Î±ÃŸ = 1/3 = c/a
If a = 3, then b = -3√2c = 1
Therefore, the quadratic polynomial is 3x2 -3√2x +1.

(iii) 0, √5
Let the polynomial be ax2 + bx + c, and its zeroes be Î± and ÃŸ
Î± + ÃŸ = 0 = 0/1 = -b/a
Î±ÃŸ = √5 = √5/1 = c/a
If a = 1, then b = 0, c = √5
Therefore, the quadratic polynomial is x2 + √5.

(iv) 1, 1
Let the polynomial be ax2 + bx + c, and its zeroes be Î± and ÃŸ
Î± + ÃŸ = 1 = 1/1 = -b/a
Î±ÃŸ = 1 = 1/1 = c/a
If a = 1, then b = -1, c = 1
Therefore, the quadratic polynomial is x2 - x +1.

(v) -1/4 ,1/4
Let the polynomial be ax2 + bx + c, and its zeroes be Î± and ÃŸ
Î± + ÃŸ = -1/4 = -b/a
Î±ÃŸ = 1/4 = c/a
If a = 4, then b = 1, c = 1
Therefore, the quadratic polynomial is 4x2 + x +1.

(vi) 4,1
Let the polynomial be ax2 + bx + c, and its zeroes be Î± and ÃŸ
Î± + ÃŸ = 4 = 4/1 = -b/a
Î±ÃŸ = 1 = 1/1 = c/a
If a = 1, then b = -4, c = 1
Therefore, the quadratic polynomial is x2 - 4x +1.

Page No: 36

Exercise 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) = x3 – 3x2 + 5x – 3, g(x) = x2 – 2
Quotient = x-3 and remainder 7x - 9

(ii) p(x) = x4 – 3x2 + 4x + 5, g(x) = x2 + 1 – x
Quotient = x2- 3 and remainder 8

(iii) p(x) = x4 – 5x + 6, g(x) = 2 – x2
Quotient = -x2 -2 and 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
t2 – 3 exactly divides  2t4 + 3t3 – 2t2 – 9t – 12 leaving no remainder. Hence, it is a factor of  2t4 + 3t3 – 2t2 – 9t – 12.

(ii) x2 + 3x + 1, 3x4 + 5x3 – 7x2 + 2x + 2
x2 + 3x + 1 exactly divides 3x4 + 5x3 – 7x2 + 2x + 2 leaving no remainder. Hence, it is factor of 3x4 + 5x3 – 7x2 + 2x + 2.

(iii) x3 – 3x + 1, x5 – 4x3 + x2 + 3x + 1
x3 – 3x + 1 didn't divides exactly x5 – 4x3 + x2 + 3x + 1 and leaves 2 as remainder. Hence, it 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).

p(x) = 3x4 + 6x3 – 2x2 – 10x – 5
Since the two zeroes are √(5/3) and - √(5/3).
We factorize x2 + 2+ 1
= (+ 1)2
Therefore, its zero is given by x + 1 = 0
x = -1
As it has the term (+ 1)2 , therefore, there will be 2 zeroes at x = - 1.
Hence, the zeroes of the given polynomial are √(5/3) and - √(5/3), - 1 and - 1.

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).

Here in the given question,
Dividend = x3 - 3x2 + x + 2
Quotient = x - 2
Remainder = -2x + 4
Divisor = g(x)
We know that,
Dividend = Quotient × Divisor + Remainder
⇒ x3 - 3x2 + x + 2 = (x - 2) × g(x) + (-2x + 4)⇒ x3 - 3x2 + x + 2 - (-2x + 4) = (x - 2) × g(x)
⇒ x3 - 3x2 + 3x - 2 = (x - 2) × g(x)
g(x) =  (x3 - 3x2 + 3x - 2)/(x - 2)

∴ g(x) = (x2 - x + 1)

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

(i) Let us assume the division of 6x2 + 2x + 2 by 2
Here, p(x) = 6x2 + 2x + 2
g(x) = 2
q(x) = 3x2 + x + 1
r(x) = 0
Degree of p(x) and q(x) is same i.e. 2.
Checking for division algorithm,
p(x) = g(x) × q(x) + r(x)
Or, 6x2 + 2x + 2 = 2x (3x2 + x + 1)
Hence, division algorithm is satisfied.

(ii) Let us assume the division of x3+ x by x2,
Here, p(x) = x3 + x
g(x) = x2
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)
x3 + x = (x2 ) × x + x
x3 + x = x3 + x
Thus, the division algorithm is satisfied.

(iii) Let us assume the division of x3+ 1 by x2.
Here, p(x) = x3 + 1
g(x) = x2
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)
x3 + 1 = (x2 ) × x + 1
x3 + 1 = x3 + 1
Thus, the division algorithm is satisfied.

Page No: 36

Exercise 2.4 (Optional)

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 - 5+ 2; 1/2, 1, -2
(ii) x3 - 4x2 + 5x - 2; 2, 1, 1

(i) p(x) = 2x3 + x2 - 5+ 2
Now for zeroes, putting the given value in x.

p(1/2) = 2(1/2)3 + (1/2)2 - 5(1/2) + 2
= (2×1/8) + 1/4 - 5/2 + 2
= 1/4 + 1/4 - 5/2 + 2
= 1/2 - 5/2 + 2 = 0

p(1) = 2(1)3 + (1)2 - 5(1) + 2
= (2×1) + 1 - 5 + 2
= 2 + 1 - 5 + 2 = 0

p(-2) = 2(-2)3 + (-2)2 - 5(-2) + 2
= (2 × -8) + 4 + 10 + 2
= -16 + 16 = 0

Thus, 1/2, 1 and -2 are the zeroes of the given polynomial.

Comparing the given polynomial with ax3 + bx2 + c+ d, we get a=2, b=1, c=-5, d=2
Also, Î±=1/2, Î²=1 and Î³=-2
Now,
-b/a = Î±+Î²+Î³
⇒ 1/2 = 1/2 + 1 - 2
⇒ 1/2 = 1/2

c/a = Î±Î²+Î²Î³+Î³Î±
⇒ -5/2 = (1/2 × 1) + (1 × -2) + (-2 × 1/2)
⇒ -5/2 = 1/2 - 2 - 1
⇒ -5/2 = -5/2

-d/a = Î±Î²Î³
⇒ -2/2 = (1/2 × 1 × -2)
⇒ -1 = 1

Thus, the relationship between zeroes and the coefficients are verified.

(ii)  p(x) = x3 - 4x2 + 5x - 2
Now for zeroes, putting the given value in x.

p(2) = 23 - 4(2)2 + 5(2) - 2
= 8 - 16 + 10 - 2
= 0

p(1) = 13 - 4(1)2 + 5(1) - 2
= 1 - 4 + 5 - 2
= 0

p(1) = 13 - 4(1)2 + 5(1) - 2
= 1 - 4 + 5 - 2
= 0

Thus, 2, 1 and 1 are the zeroes of the given polynomial.

Comparing the given polynomial with ax3 + bx2 + c+ d, we get a=1, b=-4, c=5, d=-2
Also, Î±=2, Î²=1 and Î³=1
Now,
-b/a = Î±+Î²+Î³
⇒ 4/1 = 2 + 1 + 1
⇒ 4 = 4

c/a = Î±Î²+Î²Î³+Î³Î±
⇒ 5/1 = (2 × 1) + (1 × 1) + (1 × 2)
⇒ 5 = 2 + 1 + 2
⇒ 5 = 5

-d/a = Î±Î²Î³
⇒ 2/1 = (2 × 1 × 1)
⇒ 2 = 2

Thus, the relationship between zeroes and the coefficients are verified.

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.

Let the polynomial be ax3 + bx+ cx + d and the zeroes be Î±, Î² and Î³
Then, Î± + Î² + Î³ = -(-2)/1 = 2 = -b/a
Î±Î² + Î²Î³ + Î³Î± = -7 = -7/1 = c/a
Î±Î²Î³ = -14 = -14/1 = -d/a

∴ a = 1, b = -2, c = -7 and d = 14
So, one cubic polynomial which satisfy the given conditions will be 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.

Since, (a - b), a, (a + b) are the zeroes of the polynomial x3 – 3x2 + x + 1.
Therefore, sum of the zeroes = (a - b) + a + (a + b) = -(-3)/1 = 3

⇒ 3a = 3 ⇒ a =1

∴ Sum of the products of is zeroes taken two at a time = a(a - b) + a(a + b) + (a + b) (a - b) =1/1 = 1
a2 - ab + a2 + ab + a2 - b= 1
⇒ 3a2 - b2 =1

Putting the value of a,

⇒ 3(1)2 - b2 = 1
⇒ 3 - b2 = 1
⇒ b2 = 2
⇒ b = ±√2
Hence, a = 1 and b = ±√2

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

2+√3 and 2-√3 are two zeroes of the polynomial p(x) = x4 – 6x3 – 26x2 + 138x – 35.
Let x = 2±√3
So, x-2 = ±√3
On squaring, we get x2 - 4x + 4 = 3,
⇒ x2 - 4x + 1= 0

Now, dividing p(x) by x2 - 4x + 1

∴ p(x) = x4 - 6x3 - 26x2 + 138x - 35
= (x2 - 4x + 1) (x2 - 2x - 35)
= (x2 - 4x + 1) (x2 - 7x + 5x - 35)
= (x2 - 4x + 1) [x(x - 7) + 5 (x - 7)]
= (x2 - 4x + 1) (x + 5) (x - 7)

∴ (x + 5) and (x - 7) are other factors of p(x).
∴ - 5 and 7 are other zeroes of the given polynomial.

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.

On dividing x4 – 6x3 + 16x2 – 25x + 10 by x2 – 2x + k

∴ Remainder = (2k - 9)x - (8 - k)k + 10
But the remainder is given as x+ a.

On comparing their coefficients,
2k - 9 = 1
⇒ k = 10
⇒ k = 5 and,

-(8-k)k + 10 = a
⇒ a = -(8 - 5)5 + 10 =- 15 + 10 = -5

Hence, k = 5 and a = -5

Go Back To NCERT Solutions for Class 10th Maths

NCERT Solutions for Class 10 Maths Chapter 2 Polynomials

Chapter 2 Polynomials is mainly concerned about finding zeroes of given polynomials. You will also get to know about division algorithm which is very important. We have already studied in Class 9 that an expression of the form p(x) = a0 + a1x + a2x2 + .... + anxn where an ≠ 0 is called a polynomial in variable x of degree n. where; a0, a2 .... an are real numbers and each power of x is a non negative integer. A polynomial which contains only constant term, is called a zero polynomial.

This chapter is divided into four sections:
• First section is about introduction in which you will get to know about degrees of polynomials.
• The second section is Geometrical Meaning of the zeroes of a Polynomial on which one exercise has been provided which only has one question.
• The third section is named Relationship between zeroes and coefficient of a Polynomial. There are two questions in this exercise, in the first you have to find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients and in the another one you have to find a quadratic polynomial with the given numbers.
• The fourth section is division algorithm for Polynomials in which you have to divide the polynomials and find remainder and quotient.

Other than NCERT Solutions for Class 10 Maths Chapter 2 Polynomials, you can also check Notes of Chapter 2 Polynomials Class 10, MCQ of Class 10 Chapter 2 Poynomials.

There are total four exercises in this chapter in which last one is optional. We have provided Class 10 Maths Polynomials Solutions of every exercises step by step. For finding the exercise wise solutions you can click the links below.

### NCERT Solutions for Class 10 Maths Chapters:

 Chapter 1 Real Numbers Chapter 3 Linear Equations In Two Variables Chapter 4 Quadratic Equations Chapter 5 Arithmetic Progressions Chapter 6 Triangles Chapter 7 Coordinate Geometry Chapter 8 Introduction To Trigonometry Chapter 9 Some Applications Of Trigonometry Chapter 10 Circles Chapter 11 Constructions Chapter 12 Areas Related To Circles Chapter 13 Surface Areas And Volumes Chapter 14 Statistics Chapter 15 Probability

FAQ on Chapter 2 Polynomials

#### How many exercises in Chapter 2 Polynomials

There are only 4 exercises in the chapter. You will get detailed solutions of every questions and grasp the concepts embedded in it. These NCERT Solutions for Chapter 2 Polynomials Class 10 Maths will help you a lot in board exams.

#### What is Division Algorithm?

If we divide a polynomial p(x) by a polynomial g(x), then their exists polynomials q(x) and r(x) such that, p(x) = g(x) x q(x) + r(x), where r(x) = 0 or deg r(x) < deg g(x). It is known as Division Algorithm.

#### What is Linear Polynomial.

A polynomial of degree 1 is called a linear polynomial. A linear polynomial is generally written as ax + b (a ≠ 0), where a, b are real coefficients.

#### Find the zeroes of 2x2 - 8x + 6.

We have:
2x2 - 8x + 6 = 2x2 - 6x - 2x + 6
= 2x (x - 3) - 2 (x - 3)
= (2x - 2) (x - 3)
= 2 (x - 1) (x - 3)
For 2x2 - 8x + 6 to be zero,
Either, x - 1 = 0 ⇒ x = 1
or x - 3 = 0 ⇒ x = 3
∴ The zeroes of 2x2 - 8x + 6 are 1 and 3.