• Skip to main content
  • Skip to secondary menu
  • Skip to primary sidebar
  • Skip to footer
  • RBSE Model Papers
    • RBSE Class 12th Board Model Papers 2022
    • RBSE Class 10th Board Model Papers 2022
    • RBSE Class 8th Board Model Papers 2022
    • RBSE Class 5th Board Model Papers 2022
  • RBSE Books
  • RBSE Solutions for Class 10
    • RBSE Solutions for Class 10 Maths
    • RBSE Solutions for Class 10 Science
    • RBSE Solutions for Class 10 Social Science
    • RBSE Solutions for Class 10 English First Flight & Footprints without Feet
    • RBSE Solutions for Class 10 Hindi
    • RBSE Solutions for Class 10 Sanskrit
    • RBSE Solutions for Class 10 Rajasthan Adhyayan
    • RBSE Solutions for Class 10 Physical Education
  • RBSE Solutions for Class 9
    • RBSE Solutions for Class 9 Maths
    • RBSE Solutions for Class 9 Science
    • RBSE Solutions for Class 9 Social Science
    • RBSE Solutions for Class 9 English
    • RBSE Solutions for Class 9 Hindi
    • RBSE Solutions for Class 9 Sanskrit
    • RBSE Solutions for Class 9 Rajasthan Adhyayan
    • RBSE Solutions for Class 9 Physical Education
    • RBSE Solutions for Class 9 Information Technology
  • RBSE Solutions for Class 8
    • RBSE Solutions for Class 8 Maths
    • RBSE Solutions for Class 8 Science
    • RBSE Solutions for Class 8 Social Science
    • RBSE Solutions for Class 8 English
    • RBSE Solutions for Class 8 Hindi
    • RBSE Solutions for Class 8 Sanskrit
    • RBSE Solutions

RBSE Solutions

Rajasthan Board Textbook Solutions for Class 5, 6, 7, 8, 9, 10, 11 and 12

  • RBSE Solutions for Class 7
    • RBSE Solutions for Class 7 Maths
    • RBSE Solutions for Class 7 Science
    • RBSE Solutions for Class 7 Social Science
    • RBSE Solutions for Class 7 English
    • RBSE Solutions for Class 7 Hindi
    • RBSE Solutions for Class 7 Sanskrit
  • RBSE Solutions for Class 6
    • RBSE Solutions for Class 6 Maths
    • RBSE Solutions for Class 6 Science
    • RBSE Solutions for Class 6 Social Science
    • RBSE Solutions for Class 6 English
    • RBSE Solutions for Class 6 Hindi
    • RBSE Solutions for Class 6 Sanskrit
  • RBSE Solutions for Class 5
    • RBSE Solutions for Class 5 Maths
    • RBSE Solutions for Class 5 Environmental Studies
    • RBSE Solutions for Class 5 English
    • RBSE Solutions for Class 5 Hindi
  • RBSE Solutions Class 12
    • RBSE Solutions for Class 12 Maths
    • RBSE Solutions for Class 12 Physics
    • RBSE Solutions for Class 12 Chemistry
    • RBSE Solutions for Class 12 Biology
    • RBSE Solutions for Class 12 English
    • RBSE Solutions for Class 12 Hindi
    • RBSE Solutions for Class 12 Sanskrit
  • RBSE Class 11

RBSE Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.4

April 7, 2022 by Safia Leave a Comment

Rajasthan Board RBSE Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.4 Textbook Exercise Questions and Answers.

RBSE Class 10 Maths Solutions Chapter 2 Polynomials Exercise 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, \(\frac{1}{2}\), 1, – 2
(ii) x3 – 4x2 + 5x – 2; 2, 1, 1
Solution:
(i) Comparing the given polynomial with the polv-nomial ax3 + bx2 + ex + d, a = 2, b = 1, c = – 5 and d = 2
RBSE Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.4 1
p(1) = 2(1)3 + (1)2 – 5(1) + 2
= 2 + 1 – 5 + 2 = 0
p(- 2) = 2(- 2)3 + (-2)2 – 5(-2) + 2
= 2(- 8) + 4 + 10 + 2
= – 16 + 16 = 0
Therefore, \(\frac{1}{2}\), 1 and – 2 are the zeroes of the given polynomial 2x3 + x2 – 5x + 2
i.e. α = \(\frac{1}{2}\), β = 1 and γ = – 2
Then,
RBSE Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.4 2
Hence the sum and product by the zeroes \(\frac{1}{2}\). 1 and – 2 are the same as these.
∴ The relationship between the zeroes and the coefficients is correct.

(ii) Comparing the given polynomial with ax3 + bx2 + cx + d, a = 1, b = – 4, c = 5 and d = -2.
p(2) = (2)3 – 4(2)2 + 5(2) – 2 = 8 – 16 + 10 – 2 = 0
p(1) = (1)3 – 4(1)2 + 5(1) – 2 = 1 – 4 + 5 – 2 = 0
∴ 2, 1 and 1 arc the zeroes of the polynomial x3 – 4x2 + 5x – 2
So, α = 2, β = 1 and γ = 1
Thus,
α + β + γ = 2 + 1 + 1 = 4
= \(\frac{-(-4)}{1}\) = \(\frac{-b}{a}\)
αβ + βγ + γα = (2) (1) + (1) (1) + (1) (2)
= 2 + 1 + 2 = 5 = \(\frac{5}{1}=\frac{c}{a}\)
and αβγ = (2) (1) (1) = 2 = \(\frac{-(-2)}{1}=\frac{-d}{a}\)
∴ The sum and product obtained by zeroes 2, 1. 1 are also these. Therefore the above relationship of the zeroes of the polynomial with its coefficient is correct.

RBSE Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.4

Question 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 a cubic polynomial be ax3 + bx2 + cx + d and its zeroes are α, β and γ.
Then
α + β + γ = \(\frac{-(-2)}{1}=\frac{-b}{a}\)
αβ + βγ + γα = – 7 = \(\frac{-7}{1}=\frac{c}{a}\)
and αβγ = – 14 = \(\frac{-14}{1}=\frac{-d}{a}\)
If a = 1 then b = – 2, c = – 7 and d = 14
Thus the polynomial is x3 – 2x2 – 7x + 14

Question 3.
If the zeroes of the polynomial x3 – 3x2 + x + 1 are a – b, a, a + b find a and b.
Solution:
∵ (a – b), a, (a + b) are the zeroes of the polynomial x3 – 3x2 + x + 1.
Therefore
(a – b) + a + (a – b) = \(\frac{-(-3)}{1}\) = 3
∵ α + β + γ = \(-\frac{b}{a}\)
⇒ 3a = 3
⇒ a = 1
(a – b) a + a (a + b) + (a + b) (a – b) = \(\frac{1}{1}\) = 1
∵ αβ + βγ + γα = \(\frac{c}{a}\)
⇒ a2 – ab + a2 + ab + a2 – b2 = 1
⇒ 3a2 – b2 = 1
⇒ 3(1)2 – b2 = 1 [∵ a = 1]
⇒ 3 – b2 = 1
⇒ b2 = 2
⇒ b = ±\(\sqrt{2}\)
Hence, a = 1 and b = ±\(\sqrt{2}\)

RBSE Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.4

Question 4.
If two -zeroes of the polynomial x4 – 6x3 – 26x2 + 138x – 35 are 2 ± \(\sqrt{3}\), find other zeroes.
Solution:
∵ Two zeroes of the polynomial p(x) = x4 – 6x3 – 26x2 + 138x – 35 are 2 ± \(\sqrt{3}\)
Therefore, x = 2 ± \(\sqrt{3}\)
⇒ x – 2 = ± \(\sqrt{3}\)
Squaring both sides.
x2 – 4x + 4 = 3
⇒ x2 – 4x + 1 = 0
Now let us divide the polynomial p(x) by x2 – 4x + 1
so that other zeroes may be obtained.
RBSE Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.4 3
∴ 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) will be the other factors. Hence – 5 and 7 will be the other zeroes.

RBSE Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.4

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:
Dividing the polynomial x4 – 6x3 + 16x2 – 25x + 10 by the polynomial x2 – 2 x + k
RBSE Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.4 4
∴ Remaidner = (2k – 9) x – (8 – k) k + 10
But remainder = x + a
Therefore, Comparing the coefficients
2k – 9 = 1
⇒ 2k= 10
⇒ k = 5
and – (8 – k) k + 10 = a
⇒ a = – (8 – 5) 5 + 10
= – 3 × 5 + 10
= – 15 + 10 = – 5
Hence, k = 5 and a = – 5

Share this:

  • Click to share on WhatsApp (Opens in new window)
  • Click to share on Twitter (Opens in new window)
  • Click to share on Facebook (Opens in new window)

Related

Filed Under: Class 10

Reader Interactions

Leave a Reply Cancel reply

Your email address will not be published. Required fields are marked *

Primary Sidebar

Recent Posts

  • RBSE Solutions for Class 6 Maths Chapter 6 Decimal Numbers Additional Questions
  • RBSE Solutions for Class 11 Psychology in Hindi Medium & English Medium
  • RBSE Solutions for Class 11 Geography in Hindi Medium & English Medium
  • RBSE Solutions for Class 3 Hindi
  • RBSE Solutions for Class 3 English Let’s Learn English
  • RBSE Solutions for Class 3 EVS पर्यावरण अध्ययन अपना परिवेश in Hindi Medium & English Medium
  • RBSE Solutions for Class 3 Maths in Hindi Medium & English Medium
  • RBSE Solutions for Class 3 in Hindi Medium & English Medium
  • RBSE Solutions for Class 4 Hindi
  • RBSE Solutions for Class 4 English Let’s Learn English
  • RBSE Solutions for Class 4 EVS पर्यावरण अध्ययन अपना परिवेश in Hindi Medium & English Medium

Footer

RBSE Solutions for Class 12
RBSE Solutions for Class 11
RBSE Solutions for Class 10
RBSE Solutions for Class 9
RBSE Solutions for Class 8
RBSE Solutions for Class 7
RBSE Solutions for Class 6
RBSE Solutions for Class 5
RBSE Solutions for Class 12 Maths
RBSE Solutions for Class 11 Maths
RBSE Solutions for Class 10 Maths
RBSE Solutions for Class 9 Maths
RBSE Solutions for Class 8 Maths
RBSE Solutions for Class 7 Maths
RBSE Solutions for Class 6 Maths
RBSE Solutions for Class 5 Maths
RBSE Class 11 Political Science Notes
RBSE Class 11 Geography Notes
RBSE Class 11 History Notes

Copyright © 2023 RBSE Solutions

 

Loading Comments...