BSEB Class 10 Maths Chapter 2 Polynomials Ex 2.4 Textbook Solutions PDF: Download Bihar Board STD 10th Maths Chapter 2 Polynomials Ex 2.4 Book Answers |
Bihar Board Class 10th Maths Chapter 2 Polynomials Ex 2.4 Textbooks Solutions PDF
Bihar Board STD 10th Maths Chapter 2 Polynomials Ex 2.4 Books Solutions with Answers are prepared and published by the Bihar Board Publishers. It is an autonomous organization to advise and assist qualitative improvements in school education. If you are in search of BSEB Class 10th Maths Chapter 2 Polynomials Ex 2.4 Books Answers Solutions, then you are in the right place. Here is a complete hub of Bihar Board Class 10th Maths Chapter 2 Polynomials Ex 2.4 solutions that are available here for free PDF downloads to help students for their adequate preparation. You can find all the subjects of Bihar Board STD 10th Maths Chapter 2 Polynomials Ex 2.4 Textbooks. These Bihar Board Class 10th Maths Chapter 2 Polynomials Ex 2.4 Textbooks Solutions English PDF will be helpful for effective education, and a maximum number of questions in exams are chosen from Bihar Board.Bihar Board Class 10th Maths Chapter 2 Polynomials Ex 2.4 Books Solutions
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Format | DOC/PDF |
Class | 10th |
Subject | Maths Chapter 2 Polynomials Ex 2.4 |
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BSEB Class 10th Maths Chapter 2 Polynomials Ex 2.4 Textbooks Solutions with Answer PDF Download
Find below the list of all BSEB Class 10th Maths Chapter 2 Polynomials Ex 2.4 Textbook Solutions for PDF’s for you to download and prepare for the upcoming exams:BSEB Bihar Board Class 10th Maths Solutions Chapter 2 Polynomials Ex 2.4
Question 1.
Verify that the numbers given alongside of the cubic polynomials below are their zeros.
Also verify the relationship between the zeroes and the coefficients in each case:
- 2x3 + x2 – 5x + 2; 12, 1, – 2
- x3 – 4x2 + 5x – 2; 2, 1, 1
Solution:
1. Comparing the given polynomial with ax3 + bx2 + cx + d, we get
a = 2, b = 1, c = – 5 and d = 2.
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
∴ 12, 1 and – 2 are the zeroes of 2x3 + x2 – 5x + 2.
So, α = 12, β = 1 and γ = – 2.
2. Comparing the given polynomial with ax3 + bx2 + cx + d, we get
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 are the zeroes of x3 – 4x2 + 5x – 2.
So, α = 2, β = 1 and γ = 1.
Therefore, α + β + γ = 2 + 1 + 1 = 4 = −(−4)1 = −𝑏𝑎
αβ + βγ + γα = (2)(1) + (1)(1) + (1)(2)
= 2 + 1 + 2 + 5 = 51 = 𝑐𝑎
and αβγ = (2)(1)(1) = 2 = −(−2)1 = −𝑑𝑎.
Question 2.
Find a cubic polynomial with the sum, sum of the products of its zeroes taken two at a time, and the product of its zeroes as 2, – 7, – 14 respectively.
Solution:
Let the cubic polynomial be ax3 + bx2 + cx + d, and its zeroes be α, β and γ.
Then, α + β + γ = 2 = −(−2)1 = −𝑏𝑎
αβ + βγ + γα = – 7 = −71 = 𝑐𝑎
and αβγ = – 14 = −141 = −𝑑𝑎
If a = 1, then b = – 2, c = – 7 and d = 14.
So, one cubic polynomial which fits the given conditions x3 – 2x2 – 7x + 14.
Question 3.
If the zeroes of the polynomial x3 – 3x2 + x + 1 are a – b, a and a + b, find a and b.
Solution:
Since (a – b), a and (a + b) are the zeroes of the polynomial x3 – 3x2 + x + 1, therefore
So, (a – b) + a + (a + b) = −(−3)1 = 3
So, 3a = 3 or a = 1
(a – b)a + a(a + b) + (a + b)(a – b) = 11 = 1
or a2 – ab + a2 + ab + a2 – b2 = 1
or 3a2 – b2 = 1
So, 3(1)2 – b2 = 1 [∵ a = 1]
or 3 – b2 = 1
or b2 = 2 or b = ± 2‾√
Hence, a = 1 and b = ± 2‾√.
Question 4.
If two zeroes of the polynomial x4 – 6x3 – 26x2 + 138x – 35 are 2 ± 3‾√, find other zeroes.
Solution:
We have:
2 ± 3‾√ are two zeroes of the polynomial p(x) = x4 – 6x3 – 26x2 + 138x – 35
Let x = 2 ± 3‾√. So, x – 2 = ± 3‾√
Squaring, we get
x2 – 4x + 4 = 3, i.e; x2 – 4x + 1 = 0
Let us divide p(x) by x2 – 4x + 1 to obtain other zeroes.
∴ 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)
So, (x + 5) and (x – 7) are other factors of p(x).
∴ – 5 and 7 are other zeroes of the given polynomial.
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:
Let us divide 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, we have:
2k – 9 = 1 or 2k = 10 or k = 5
and – (8 – k)k + 10 = a
So, a = – (8 – 5)5 + 10
= – 3 × 5 + 10 = – 15 + 10 = – 5
Hence, k = 5 and a = – 5.
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