AP Board Class 10 Maths Chapter 3 Polynomials Ex 3.3 Textbook Solutions PDF: Download Andhra Pradesh Board STD 10th Maths Chapter 3 Polynomials Ex 3.3 Book Answers

AP Board Class 10 Maths Chapter 3 Polynomials Ex 3.3 Textbook Solutions PDF: Download Andhra Pradesh Board STD 10th Maths Chapter 3 Polynomials Ex 3.3 Book Answers

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Andhra Pradesh State Board Class 10th Maths Chapter 3 Polynomials Ex 3.3 Books Solutions

Board	AP Board
Materials	Textbook Solutions/Guide
Format	DOC/PDF
Class	10th
Subject	Maths
Chapters	Maths Chapter 3 Polynomials Ex 3.3
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10th Class Maths 3rd Lesson Polynomials Ex 3.3 Textbook Questions and Answers

Question 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) Given polynomial is x² – 2x – 8
We have x² – 2x – 8 = x² – 4x + 2x – 8
= x(x – 4) + 2(x – 4)
= (x – 4) (x + 2)
So, the value of x² – 2x – 8 is zero
when x – 4 = 0 or x + 2 = 0 i.e.,
when x = 4 or x = -2
So, the zeroes of x² – 2x – 8 are 4 and -2.
Sum of the zeroes = 4 – 2 = 2 Coefficient of ,x -(-2)
= – Coefficient of 𝑥 Coefficient of 𝑥2 = −(−2)1 = 2
And product of the zeroes = 4 × (-2) = -8
= Constant term Coefficient of 𝑥2 = −81 = -8

ii) Given polynomial is 4s² – 4s + 1
We have, 4s² – 4s + 1
= 4s² – 2s – 2s + 1
= 2s (2s – 1) – 1(2s – 1)
= (2s – 1) (2s – 1)
= (2s – 1)²
So, the value of 4s2 – 4s + 1 is zero
when 2s-1 = 0 or s = 12
∴ Zeroes of the polynomial are 12 and 12
∴ Sum of the zeroes = 12 + 12 = 1.
= – Coefficient of 𝑠 Coefficient of 𝑠2 = –−44 = 1
And product of the zeroes = (12)×(12) = 14
= Constant term Coefficient of 𝑥2 = 14

iii) Given polynomial is 6x² – 3 – 7x
We have, 6x² – 3 – 7x = 6x² – 7x – 3
= 6x² – 9x + 2x – 3
= 3x(2x – 3) + 1(2x – 3)
= (2x – 3) (3x + 1)
The value of 6x² – 3 – 7x is zero, when the value of (3x +1) (2x – 3) is 0
i.e., when 3x + 1 = 0 and 2x – 3 = 0
3x = -1 and 2x = 3
x = −13 and x = 32
∴ The zeroes of 6x² – 3 – 7x = −13 and 32
∴ Sum of the zeroes = 13 + 32 = 76.
= – Coefficient of 𝑥 Coefficient of 𝑥2 = −(−7)6 = 76
And product of the zeroes = (−13)×(32) = −12
= Constant term Coefficient of 𝑥2 = −36 = −12

iv) Given polynomial is 4u² + 8u
We have, 4u² + 8u = 4u (u + 2)
The value of 4u² + 8u is 0,
when the value of 4u(u + 2) = 0, i.e.,
when u = 0 or u + 2 = 0, i.e.,
when u = 0 (or) u = – 2
∴ The zeroes of 4u² + 8u are 0 and – 2.
Therefore, sum of the zeroes = 0 + (-2) = -2
= – Coefficient of 𝑢 Coefficient of 𝑢2 = −84 = -2
And product of the zeroes 0 . (-2) = 0
= Constant term Coefficient of 𝑢2 = 04 = 0

v) Given polynomial is t² – 15.
We have, t² – 15 = (t – √15 ) (t + √l5)
The value of t² – 15 is 0,
when the value of (t – √15 ) (t + √l5) = 0, i.e.,
when t – √15 = 0 or t + √15 = 0, i.e.,
when t = √15 (or) t = -√15
∴ The zeroes of t² – 15 are √15 and -√15.
Therefore, sum of the zeroes = √15 + (-√15) = 0
= – Coefficient of 𝑡 Coefficient of 𝑡2 = –01 = 0
And product of the zeroes √15 × (-√15) = -15
= Constant term Coefficient of 𝑡2 = −151 = -15

vi) Given polynomial is 3x² – x – 4
we have, 3x² – x – 4 = 3x² + 3x – 4x – 4
= 3x(x + 1) – 4(x + 1)
= (x + 1) (3x – 4)
The value of 3x² – x – 4 is 0 when the value of (x + 1) (3x – 4) is 0.
i.e., when x + 1 = 0 or 3x – 4 = 0
i.e., when x = -1 or x = 43
∴ The zeroes of 3x² – x – 4 are -1 and 43
Therefore, sum of the zeroes = -1 + 43 = −3+43 = 13
= – Coefficient of 𝑥 Coefficient of 𝑥2 = −(−1)3 = 13
And product of the zeroes -1 × 43 = −43
= Constant term Coefficient of 𝑥2 = −43

Question 2.
Find the quadratic polynomial in each case, with the given numbers as the sum and product of its zeroes respectively.
i) 14, -1
ii) √2, 13
iii) 0, √5
iv) 1, 1
v) –14, 14
vi) 4, 1
Answer:
Let the polynomial be ax² + bx + c
and its zeroes be α and β.
i) Here, α + β = 14 and αβ = -1
Thus, the polynomial formed = x² – (sum of the zeroes)x + product of the zeroes
= x² – (14)x – 1
= x² – 𝑥4 – 1
The other polynomials are (x² – 𝑥4 – 1)
then the polynomial is 4x² – x – 4.

ii) Here, α + β = √2 and αβ = 13
Thus, the polynomial formed = x² – (sum of the zeroes)x + product of the zeroes
= x² – (√2)x + 13
= x² – √2x + 13
The other polynomials are (x² – √2x + 13)
then the polynomial is 3x² – 3√2x + 1.

iii) Here, α + β = 0 and αβ = √5
Thus, the polynomial formed = x² – (sum of the zeroes)x + product of the zeroes
= x² – (0)x + √5
= x² + √5

iv) Let the polynomial be ax² + bx + c and its zeroes be α and β.
Then α + β = 1 = −(−1)1 = −𝑏𝑎 and
αβ = 1 = latex]\frac{1}{1}[/latex] = 𝑐𝑎
If a = 4, then b = 1 and c = 1
∴ One quadratic polynomial which satisfies the given conditions is 4x² + x + 1.

v) Let the polynomial be ax² + bx + c and its zeroes be α and β.
Then α + β = −14 = −𝑏𝑎 and
αβ = 14 = 𝑐𝑎
If a = 4, then b = 1 and c = 1
∴ One quadratic polynomial which satisfies the given conditions is 4x² + x + 1.

vi) Let the polynomial be ax² + bx + c and its zeroes be α and β.
Then α + β = 4 = −(−4)1 = −𝑏𝑎 and
αβ = 1 = 11 = 𝑐𝑎
If a = 1, then b = -4 and c = 1
∴ One quadratic polynomial which satisfies the given conditions is x² – 4x + 1.

Question 3.
Find the quadratic polynomial, for the zeroes α, β given in each case.
i) 2, -1
ii) √3, -√3
iii) 14, -1
iv) 12, 32
Answer:
i) Let the polynomial be ax² + bx + c, a ≠ 0 and its zeroes be α and β.
Here α = 2 and β = – 1
Sum of the zeroes = α + β = 2 + (-l) = 1
Product of the zeroes = αβ = 2 × (-1) = -2
Therefore the quadratic polynomial ax² + bx + c is x² – (α + β)x + αβ = [x² – x – 2]
the quadratic polynomial will be x² – x – 2.

ii) Let the zeroes be α = √3 and β = -√3
Sum of the zeroes = α + β
= √3 + (-√3) = 0
Product of the zeroes = αβ
= √3 × (-√3) = -3
∴ The quadratic polynomial
ax² + bx + c is [x² – (α + β)x + αβ]
= [x² – 0.x + (-3)] = [x² – 3]
the quadratic polynomial will be x² – 3.

iii) Let the zeroes be α = 14 and β = -1
Sum of the zeroes = α + β
= 14 + (-1) = 1+(−4)4 = −34
Product of the zeroes = αβ
= 14 × (-1) = −14
∴ The quadratic polynomial
ax² + bx + c is [x² – (α + β)x + αβ]
= [x² – (−34).x + (−14)]
the quadratic polynomial will be 4x² + 3x – 1.

iv) Let the zeroes be α = 12 and β = 32
Sum of the zeroes = α + β
= 12 + 32 = 1+32 = 42 = 2
Product of the zeroes = αβ
= 12 × 32 = 34
∴ The quadratic polynomial
ax² + bx + c is [x² – (α + β)x + αβ]
= [x² – 2x + (34)]
the quadratic polynomial will be 4x² – 8x + 3.

Question 4.
Verify that 1, -1 and -3 are the zeroes of the cubic polynomial x³ + 3x² – x – 3 and check the relationship between zeroes and the coefficients.
Answer:
Given cubic polynomial
p(x) = x³ + 3x² – x – 3
Comparing the given polynomial with ax³ + bx² + cx + d, we get a = 1, b = 3, c = -1, d = -3
Futher given zeroes are 1,-1 and – 3
p(1) = (1)³ + 3(1)² – 1 – 3
= 1 + 3 – 1 – 3 = 0
p(-1) = (-1)³ + 3(-1)² – 1 – 3
= -1 + 3 + 1 – 3 = 0
p(-3) = (-3)³ + 3(-3)² – (-3) – 3
= -27 + 27 + 3 – 3 = 0
Therefore, 1, -1 and -3 are the zeroes of x³ + 3x² – x – 3.
So, we take α = 1, β = -1 and γ = -3 Now,
α + β + γ = 1 + (-1) + (-3) = -3
αβ + βγ + γα = 1(-l) + (-1) (-3) + (-3)1
= -1 + 3 – 3 = -1
= 𝑐𝑎 = −11 = -1
αβγ = 1 (-1) (-3) = 3 = −𝑑𝑎 = −(−3)1 = 3

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