You are here
Home > CBSE > NCERT Solutions For Class 10th Maths Chapter 2 : Polynomials

NCERT Solutions For Class 10th Maths Chapter 2 : Polynomials

CBSE NCERT Solutions For Class 10th Maths Chapter 2 : Polynomials. NCERT Solutins For Class 10 Mathematics. Exercise 2.1, Exercise 2.2, Exercise 2.3, Exercise 2.4.


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.

http://2.bp.blogspot.com/-SUcpdLViueI/VS5bdp3SdCI/AAAAAAAAFHs/gmL0A6hHy-s/s1600/fig-2.10-polynomials-maths-class-10th.PNG

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

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
Answer

(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 ofx)/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 ofs)/Coefficient of s2
Product of zeroes = 1/2 × 1/2 = 1/4 = Constant term/Coefficient ofs2.

(iii) 6x2 – 3 – 7x
6x– 7– 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

ALSO READ:  NCERT Solutions for Class 6th Civics Chapter 1 : Understanding Diversity

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 + 
= 4u(u + 2)
The value of 4u2 + 8u is zero when 4u = 0 or u + 2 = 0, i.e., u = 0 oru = – 2
Therefore, the zeroes of 4u2 + 8u are 0 and – 2.
Sum of zeroes = 0 + (-2) = -2 = -(8)/4 = -(Coefficient ofu)/Coefficient of u2
Product of zeroes = 0 × (-2) = 0 = 0/4 = Constant term/Coefficient of u2.

(v) t2 – 15
t– 0.t – 15
= (– √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 , 1/3

(iii) 0, √5

(iv) 1,1

(v) -1/4 ,1/4

(vi) 4,1
Answer

(i) 1/4 , -1
Let the polynomial be ax2 + bx + c, and its zeroes be α and ß
α + ß = 1/4 = –b/a
αß = -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 α and ß
α + ß = √2 = 3√2/3 = –b/a
αß = 1/3 = c/a
If a = 3, then b = -3√2, c = 1
Therefore, the quadratic polynomial is 3x2 -3√2x +1.

(iii) 0, √5
Let the polynomial be ax2 + bx + c, and its zeroes be α and ß
α + ß = 0 = 0/1 = –b/a
αß = √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 α and ß
α + ß = 1 = 1/1 = –b/a
αß = 1 = 1/1 = c/a
If a = 1, then b = -1, c = 1
Therefore, the quadratic polynomial is x2 – x +1.

ALSO READ:  NCERT Solutions for Class 6th Hindi Chapter 11 : जो देखकर भी नहीं देखते

(v) -1/4 ,1/4
Let the polynomial be ax2 + bx + c, and its zeroes be α and ß
α + ß = -1/4 = –b/a
αß = 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 α and ß
α + ß = 4 = 4/1 = –b/a
αß = 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:


Answer
(i) p(x) = x3 – 3x2 + 5x – 3, g(x) = x2 – 2

http://1.bp.blogspot.com/-gFWKfihspGU/VS8fETs8j5I/AAAAAAAAFIA/zPCayvJC_TI/s1600/solutions-exercise-2.3-polynomials-maths-class-10th.PNG

Quotient = x-3 and remainder 7x – 9

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

http://4.bp.blogspot.com/-ADKD8ytYX-I/VS8jP1cTzFI/AAAAAAAAFIM/YApTWZAiF1I/s1600/solutions-exercise-2.3-polynomials-maths-class-10th-1.PNG

Quotient = x2 + – 3 and remainder 8

(iii) p(x) = x4 – 5x + 6, g(x) = 2 – x2

http://1.bp.blogspot.com/-Rvxih0Y3Sko/VS8lO7oJhCI/AAAAAAAAFIY/IM-yJV8ONcs/s1600/solutions-exercise-2.3-polynomials-maths-class-10th-2.PNG

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:

Answer

(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

http://2.bp.blogspot.com/-loIF5YHaj04/VS9cdeCVqBI/AAAAAAAAFIw/dapxCQfujH4/s1600/solutions-exercise-2.3-polynomials-maths-class-10th-4.PNG

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

http://3.bp.blogspot.com/-NqwVDuPgF8Q/VS9dOIz2xCI/AAAAAAAAFJA/-oCDArLNkbM/s1600/solutions-exercise-2.3-polynomials-maths-class-10th-5.PNG

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).

Answer

p(x) = 3x4 + 6x3 – 2x2 – 10x – 5
Since the two zeroes are √(5/3) and – √(5/3).

http://3.bp.blogspot.com/-Sp6Ive9E2rM/VS-JsM4MKXI/AAAAAAAAFJg/PttL-ZkLSts/s1600/solutions-exercise-2.3-polynomials-maths-class-10th-7.PNG

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).

ALSO READ:  NCERT Solutions for Class 6th Civics Chapter 3 : What Is Government?

Answer

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)

http://1.bp.blogspot.com/-3rYTMsBv8fg/Vbh8HGaHREI/AAAAAAAAGPk/do_V7ydPERo/s1600/exercise-2.3-question-4-class-10th-polynomials.JPG

∴ 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

Answer

(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 + + 1)
Hence, division algorithm is satisfied.
(ii) Let us assume the division of x3x 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 ) × + 1
x3 + 1 = x3 + 1
Thus, the division algorithm is satisfied.

NCERT Solutions for Class 10th Maths Chapter1 1NCERT Solutions for Class 10th Maths Chapter1 2NCERT Solutions for Class 10th Maths Chapter1 3NCERT Solutions for Class 10th Maths Chapter1 4NCERT Solutions for Class 10th Maths Chapter1 5NCERT Solutions for Class 10th Maths Chapter1 6NCERT Solutions for Class 10th Maths Chapter1 7

Do You Liked Our Contents? If Yes! Then Please Spare Us Some Time By Commenting Below. Or To Get Daily Minute by Minute Updates On Facebook Twitter and Google+ From Us (Indiashines.in) Please Like Us On Facebook , Follow Us On Twitter and Follow Us On Google+ . If You also Want To Ask Us/Experts Any Questions Then Please Join Our Forum Here and Be Our Exclusive Member.


GO BACK TO CLASS X MATHS ALL CHAPTERS SOLUTION


Indiashines.in
IMPORTANT : All the contents of this website is for educational and informational purpose only. The data is collected from various websites, links, pages spread all over internet and owner self-made.We take utmost care to keep the information on this webpage as complete and accurate as possible but we do not claim our site to be error free. And in no way Indiashines.In will be liable for any losses & damages arises from its contents/text. Indiashines.In in good faith, makes an effort to become a helpful platform for students, jobseekers etc. This website is not endorsed by any Govt. bodies. Before using any information of ours please READ OUR DISCLAIMER.

Leave a Reply

Top