10th Maths - Book Back Answers - Chapter 3 Exercise 3.2 - English Medium Guides

  

 

SSLC / 10th - Maths - Book Back Answers - Chapter 3 Exercise 3.2 - English Medium

Tamil Nadu Board 10th Standard Maths - Chapter 3 Exercise 3.2: Book Back Answers and Solutions

    This post covers the book back answers and solutions for Chapter 3 Exercise 3.2 – Maths from the Tamil Nadu State Board 10th Standard textbook. These detailed answers have been carefully prepared by our expert teachers at KalviTips.com.

    We have explained each answer in a simple, easy-to-understand format, highlighting important points step by step under the relevant subtopics. Students are advised to read and memorize these subtopics thoroughly. Once you understand the main concepts, you’ll be able to connect other related points with real-life examples and confidently present them in your tests and exams.

    By going through this material, you’ll gain a strong understanding of Chapter 3 Exercise 3.2 along with the corresponding book back questions and answers (PDF format).

Question Types Covered:

• 1 Mark Questions: Choose the correct answer, Fill in the blanks, Identify the correct statement, Match the following 
• 2 Mark Questions: Answer briefly 
• 3, 4, and 5 Mark Questions: Answer in detail

All answers are presented in a clear and student-friendly manner, focusing on key points to help you score full marks.

All the best, Class 10 students! Prepare well and aim for top scores. Thank you!

Chapter 3 Algebra Ex 3.2

1. Find the GCD of the given polynomials by Division Algorithm

(i) x⁴ + 3x³ – x – 3, x³ + x² – 5x + 3
Answer Key:
p(x) = x⁴ + 3x³ – x – 3
g(x) = x³ + x² – 5x + 3

3x² + 6x – 9 = 3(x² + 2x – 3)
Now dividing g(x) = x³ + x² – 5x + 3
by the new remainder
(leaving the constant 3)
we get x² + 2x – 3
G.C.F. = x² + 2x – 3

 

(ii) x⁴ – 1, x³ – 11x² + x – 11

p(x) = x⁴ – 1
g(x) = x³ – 11x² + x – 11

120x² + 120 = 120 (x² + 1)
Now dividing g(x) = x³ – 11x² + x – 11 by the new remainder (leaving the constant) we get x² + 1

G.C.D. = x² + 1

(iii) 3x⁴ + 6x³ – 12x² – 24x, 4x⁴ + 14x³ + 8x² – 8x
Answer Key:

p(x) = 3x⁴ + 6x³ – 12x² – 24x
= 3x (x³ + 2x² – 4x – 8)
g(x) = 4x⁴ + 14x³ + 8x² – 8x
= 2x (2x³ + 7x² + 4x – 4)
G.C.D. of 3x and 2x = x
Now g(x) is divide by p(x) we get

3x² + 12x + 12 = 3 (x² + 4x + 4)
Now dividing p(x) = x³ + 2x² – 4x – 8
by the new remainder
(leaving the constant)
x² + 4x + 4

G.C.D. = x(x² + 4x + 4) [Note x is common for p(x) and g(x)]

 

(iv) 3x³ + 3x² + 3x + 3, 6x³ + 12x² + 6x+12
p(x) = 3x³ + 3x² + 3x + 3
= 3(x³ + x² + x + 1)
g(x) = 6x³ + 12x² + 6x + 12
= 6(x³ + 2x² + x + 2)
G.C.D. of 3 and 6 = 3
Now g(x) is divided by p(x)

Now dividing p(x) by the remainder x² + 1
we get x + 1

∴ G.C.D. = 3(x² + 1) [3 is the G.C.D. of 3 and 6]

 

2. Find the LCM of the given polynomials
(i) 4x²y, 8x³y²
Answer Key:

4x² y = 2 × 2 × x² × y
8 x³ y² = 2 × 2 × 2 × x³ × y²
L.C.M. = 2³ × x³ × y²
= 8x³y²

Aliter: L.C.M of 4 and 8 = 8
L.C.M. of x²y and x³y² = x³y²
∴ L.C.M. = 8x³y²

 

(ii) -9a³b², 12a²b²c
Answer Key:
-9a³b² = -(3² × a³ × b²)
12a²b²c = 2² × 3 × a² × b² × c
L.C.M. = -(2² × 3² × a³ × b² × c)
= -36 a³b²c

 

(iii) 16m, -12m²n², 8n²
Answer Key:

16m = 24 × m
-12 m²n² = -(2² × 3 × m² × n²)
8n² = 2³ × n²
L.C.M. = -(2⁴ × 3 × m² × n²)
= -48 m²n²

 

(iv) p² – 3p + 2, p² – 4
Answer Key:

P² – 3p + 2 = p² – 2p – p + 2
= p(p – 2) – 1 (p – 2)
= (p – 2) (p – 1)

p² – 4 = p² – 2² (using a² – b² = (a + b) (a – b)]

= (p + 2) (p – 2)
L.C.M. = (p – 2) (p + 2) (p – 1)

 

(v) 2x² – 5x – 3,4.x² – 36
Answer Key:
2x² – 5x – 3 = 2x² – 6x + x – 3
= 2x (x – 3) + 1 (x – 3)
= (x – 3) (2x + 1)

= 4x² – 36 = 4 [x² – 9]
= 4 [x² – 3²]
= 4(x + 3) (x – 3)
L.C.M. = 4(x – 3) (x + 3) (2x + 1)

 

(vi) (2x² – 3xy)²,(4x – 6y)³,(8x³ – 27y³)
Answer Key:

(2x² – 3xy)² = x² (2x – 3y)²
(4x – 6y)³ = 2³ (2x – 3y)³
= 8 (2x – 3y)³
8x³ – 27y³ = (2x)³ – (3y)³
= (2x – 3y) [(2x)² + 2x × 3y + (3y²)]
[using a³ – b³ = (a – b) (a² + ab + b²)
(2x – 3y) (4x² + 6xy + 9y²)
L.C.M. = 8x² (2x – 3y)³ (4x² + 6xy + 9y)²