AP Board Class 10 Maths Chapter 3 Polynomials Ex 3.4 Textbook Solutions PDF: Download Andhra Pradesh Board STD 10th Maths Chapter 3 Polynomials Ex 3.4 Book Answers ~ HSSlive: Plus One & Plus Two Notes & Solutions for Kerala State Board

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AP Board Class 10 Maths Chapter 3 Polynomials Ex 3.4 Textbook Solutions PDF: Download Andhra Pradesh Board STD 10th Maths Chapter 3 Polynomials Ex 3.4 Book Answers

AP Board Class 10th Maths Chapter 3 Polynomials Ex 3.4 Textbooks Solutions and answers for students are now available in pdf format. Andhra Pradesh Board Class 10th Maths Chapter 3 Polynomials Ex 3.4 Book answers and solutions are one of the most important study materials for any student. The Andhra Pradesh State Board Class 10th Maths Chapter 3 Polynomials Ex 3.4 books are published by the Andhra Pradesh Board Publishers. These Andhra Pradesh Board Class 10th Maths Chapter 3 Polynomials Ex 3.4 textbooks are prepared by a group of expert faculty members. Students can download these AP Board STD 10th Maths Chapter 3 Polynomials Ex 3.4 book solutions pdf online from this page.

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Andhra Pradesh State Board Class 10th Maths Chapter 3 Polynomials Ex 3.4 Books Solutions

Board	AP Board
Materials	Textbook Solutions/Guide
Format	DOC/PDF
Class	10th
Subject	Maths
Chapters	Maths Chapter 3 Polynomials Ex 3.4
Provider	Hsslive

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10th Class Maths 3rd Lesson Polynomials Ex 3.4 Textbook Questions and Answers

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 – 3×2 + 4x + 5, g(x) = x² + 1 – x
iii) p(x) = x⁴ – 5x + 6, g(x) = 2 – x²
Answer:
i) Given polynomials are
p(x) = x³ – 3x² + 5x – 3 and
g(x) = x² – 2
Here, dividend and divisor are both in standard forms.
So, we have

∴ The quotient is x – 3 and the remainder is 7x – 9.

ii) Given polynomials are
p{x) = x⁴ – 3x² + 4x + 5 and
g(x) = x² + 1 – x
Here, the dividend is already in the standard form and the divisor is not in the standard form. It can be written as x² – x + 1.
We have,

∴ The quotient is x² + x – 3 and the remainder is +8.

iii) Given polynomials are
p(x) = x⁴ – 5x + 6 and
g(x) = 2 – x²
Here, the dividend is already in the standard form and the divisor is not in the standard form. It can be written as -x² + 2.
So, we have

∴ The quotient is -x² – 2 and the remainder is -5x + 10.

Question 2.
Check in which case 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) x³ – 3x + 1, x⁵ – 4x³ + x² + 3x + 1
Answer:
i) Given first polynomial is t² – 3.
Second polynomial is
2t⁴ + 3t³ – 2t² – 9t – 12.
Let us divide 2t⁴ + 3t³ – 2t² – 9t – 12 by t² – 3, we get

Since the remainder is 0, therefore, t² – 3 is a factor of 2t⁴ + 3t³ – 2t² – 9t – 12.

ii) Given first polynomial is x² + 3x + 1
Second polynomial is 3x⁴ + 5x³ – 7x² + 2x + 2
Let us divide 3x⁴ + 5x³ – 7x² + 2x + 2 by x² + 3x + 1, we get

Since the remainder is 0, therefore x² + 3x + 1 is a factor of 3x⁴ + 5x³ – 7x² + 2x + 2.

iii) Given first polynomial = x³ – 3x + 1
Second polynomial = x⁵ – 4x³ + x² + 3x + 1
Let us divide x⁵ – 4x³ + x² + 3x + 1 by x³ – 3x + 1, we get

Here, remainder is 2(≠ 0).
Therefore, x³ – 3x + 1 is not a factor of x⁵ – 4x³ + x² + 3x + 1.

Question 3.
Obtain all other zeroes of 3x⁴ + 6x³ – 2x² – 10x – 5, if two of its zeroes are 53‾‾√ and –53‾‾√.
Answer:
Let the other two zeroes are α and β.
Now compare the given polynomial 3x⁴ + 6x³ – 2x² – 10x – 5 with the standard form ax⁴ + bx³ + cx² + dx + e we get a = 3, b = 6, c = -2, d = -10, e = -5

–53αβ = −53 ⇒ αβ = 1
now (α – β)² = (α + β)² – 4αβ
= (-2)² – 4(1)
= 4 – 4 = 0
α – β = 0 …. (2)
Now solving (1) and (2) we get

⇒ α = -1, β = -1
Then the remaining the zeroes are -1 and -1.
Hence all zeroes of it = –53‾‾√, 53‾‾√, -1, -1.

Question 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).
Answer:
Given, p(x) = x³ – 3x² + x + 2
q(x) = x – 2 and
r(x) = -2x + 4
By division algorithm, we know that Dividend = Divisor × Quotient + Remainder
p(x) = q(x) × g(x) + r(x)
Therefore, x³ – 3x² + x + 2
= (x – 2) × g(x) + (- 2x + 4)
⇒ x³ – 3x² + x + 2 + 2x – 4 = (x – 2) × g(x)
g(x) = 𝑥3−3𝑥2+3𝑥−2𝑥−2
On dividing x³ – 3x² + x + 2, by x – 2, we get

First term of g(x) = x3x = x²
Second term of g(x) = −𝑥2𝑥 = -x
Third term of g(x) = 𝑥𝑥 = 1
Hence, g(x) = x² – x + 1.

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
Answer:
Let q(x) = 3x² + 2x + 6, degree of q(x) = 2
p(x) = 12x² + 8x + 24, degree of p(x) = 2
Given degree p(x) = degree q(x)
i) Using division algorithm,
We gave, p(x) = q(x) × g(x) + r(x)
On dividing 12x² + 8x + 24 by 3x² + 2x + 6, we get

Since, the remainder is zero, therefore 3x² + 2x + 6 is a factor of 12x² + 8x + 24.
∴ g(x) = 4 and r(x)= 0

ii) Let p(x) = x⁵ + 2x⁴ + 3x³ + 5x² + 2
q(x) = x² + x + 1, degree q(x) = 2
Given degree q(x) = degree r(x)
On dividing x⁵ + 2x⁴ + 3x³ + 5x² + 2 by x² + x + 1, we get

Here, g(x) = x³ + x² + x + 1 and r(x) = 2x² – 2x + 1
degree of r(x) = 2.
∴ deg g(x) = deg r(x).

iii) Let p(x) = 2x⁴ + 8x³ + 6x² + 4x + 12, r(x) = 2
Here, degree r(x) = 0
On dividing 2x⁴ + 8x³ + 6x² + 4x + 12 by 2, we get

Here, g(x) = x⁴ + 4x³ + 3x² + 2x + 1 and r(x) = 10
so degree of r(x) = 0

AP Board Textbook Solutions PDF for Class 10th Maths

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Thursday, June 9, 2022