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Class 10 Maths NCERT Solutions Chapter 2 Polynomials

Polynomials Class 10 Maths NCERT Solutions

NCERT Solutions for Class 10 Maths Chapter 2 Polynomials are part of NCERT Solutions for Class 10 Maths. Here we have given Maths NCERT Solutions Class 10 Chapter 2 Polynomials.

Class 10 Maths NCERT Solutions Chapter 2 Polynomials Ex 2.1

Question 1:
The graphs of y = p(x) are given below for some polynomials p(x). Find the number of zeroes of p(x) in each case.
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NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.1 Q1

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Polynomials Ex 2.1 NCERT Solutions for Class 10 Maths Chapter 2

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Polynomials Exercise 2.1 NCERT Solutions for Class 10 Maths Chapter 2

Solution:
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NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.1 Q1

Class 10 Maths NCERT Solutions Chapter 2 Polynomials Ex 2.2

Question 1.
Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and their coefficients:
(i) x2 – 2x – 8
(ii) 4s2 – 4s + 1
(iii) 6x2 – 3 – 7x
(iv) 4u2 + 8u
(v) t2 – 15
(vi) 3x2 – x – 4
Solution:
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NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.2 Q1

Question 2.
Find a quadratic polynomial each with the given numbers as the sum and product of zeroes respectively:
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Polynomials Ex 2.2 NCERT Solutions for Class 10 Maths Q2

Solution:
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Polynomials Class 10 Chapter 2 NCERT Solutions Ex 2.2 Q2

Class 10 Maths NCERT Solutions Chapter 2 Polynomials Ex 2.3

Question 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
(ii) p(x) = x4 – 3x2 + 4x + 5, g(x) = x2 + 1 – x
(iii) p(x) = x4– 5x + 6, g(x) = 2 – x2
Solution:
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NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.3 Q1

Question 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
(ii) x2 + 3x + 1, 3x4 + 5x3 – 7x2 + 2x + 2
(iii) x2 + 3x + 1, x5 – 4x+ x2 + 3x + 1
Solution:
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Polynomials class 10 Ex 2.3 Maths NCERT Solutions Q2

Question 3.
Obtain all other zeroes of 3x4 + 6x3 – 2x2 – 10x – 5, if two of its zeroes are  and Image may be NSFW.
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\sqrt { \frac { 5 }{ 3 } }
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\sqrt { \frac { 5 }{ 3 } }

Solution:
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Polynomials Class 10 Chapter 2 Maths NCERT Solutions Ex 2.3 Q3

Question 4.
On dividing x– 3x2 + x + 2bya polynomial g(x), the quotient and remainder were x – 2 and -2x + 4 respectively. Find g(x).
Solution:
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Polynomials NCERT Solutions for Class 10 Maths Chapter 2 Ex 2.3 Q4

Question 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
Solution:
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Polynomials Class 10 Maths NCERT Solutions Chapter 2 Ex 2.3 Q5

Class 10 Maths NCERT Solutions Chapter 2 Polynomials Ex 2.4

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:
(i) 2x3 + x2 – 5x + 2;  Image may be NSFW.
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\frac { 1 }{ 4 }
, 1, -2
(ii) x3 – 4x2 + 5x – 2; 2, 1, 1
Solution:
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NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.4 Q1

Question 2.
Find a cubic polynomial with the sum, some of the product of its zeroes taken two at a time, and the product of its zeroes as 2, -7, -14 respectively.
Solution:
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Polynomials Class 10 Maths NCERT Solutions Chapter 2 Ex 2.4 Q2

Question 3.
If the zeroes of the polynomial x3 – 3x2 + x + 1 are a-b, a, a + b, find a and b.
Solution:
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Polynomials Class 10 Maths NCERT Solutions Exercise 2.4 Q3

Question 4.
If two zeroes of the polynomial x4 – 6x3 – 26x2 + 138x – 35 are 2 ± √3, find other zeroes.
Solution:
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Polynomials Exercise 2.4 Class 10 Maths NCERT Solutions Q4

Question 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:
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Polynomials Ex 2.4 Class 10 Maths NCERT Solutions Q4

The post Class 10 Maths NCERT Solutions Chapter 2 Polynomials appeared first on Learn CBSE.


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