KSEEB Solutions for Class 8 Maths Chapter 4 Factorisation Additional Questions

Students can Download Maths Chapter 4 Factorisation Additional Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 8 Maths Chapter 4 Factorisation Additional Questions

Question 1.
Choose the correct answer
(a) 4a + 12b is equal to
A. 4a
B. 12b
C. 4(a + 3b)
D. 3a
Answer:
(C) 4 (a + 3b)

KSEEB Solutions for Class 8 Maths Chapter 4 Factorisation Additional Questions

(b) The product of two numbers is positive and their sum is negative only when
A. both are positive
B. both are negative
C. one positive other negative
D. one of them is equal to zero
Answer:
(B) both are negative

(c) Factorising x2 + 6x + 8 we get
A. (x + 1) (x + 8)
B. (x + 6) (x + 2)
C. (x + 10) (x – 2)
D. (x + 4) (x + 2)
Answer:
(D) (x + 4) (x + 2)

(d) The denominator of an algebraic fraction should not be
A. 1
B. 0
C. 4
D. 7
Answer:
(B) 0

(e) If the sum of two integers is – 2 and their proiduct is – 24, the numbers are
A. 6 and 4
B. – 6 and 4
C. – 6 and -4
D. 6 and – 4
Answer:
(B) – 6 and 4

(f) The difference (0.7)2 – (0.3)2 simplifies to
A. 0.4
B. 0.04
C. 0.49
D. 0.56
Answer:
(A) 0.4

KSEEB Solutions for Class 8 Maths Chapter 4 Factorisation Additional Questions

Question 2.
Factorise the following
(i) x2 + 6x + 9
Answer:
x2 + 6x + 9
= x2 + 2.x.3 + 32
= (x + 3)2

(ii) 1 – 8x + 16x2
Answer:
1 – 8x + 16x2
= 12 – 2.1.4x + (4x)2
= (1 – 4x)2

(iii) 4x2 – 81y2
Answer:
4x2 – 81 y2
= (2x)2 – ( 9y)2
= (2x + 9y) (2x – 9y)

(iv) 4a2 + 4ab + b2
Answer:
4a2 + 4ab + b2
= (2a)2 +2.2a.b + b2
= (2a + b)2

(v) a2b2 + c2d2 – a2c2 – b2d2
Answer:
a2b2 + c2d2 – a2c2 – b2d2
= a2b2 – a2c2 + c2d2 – b2d2
= a2 (b2 – c2) – d2 (b2 – c2)
= (b2 – c2) (a2 – d2)

KSEEB Solutions for Class 8 Maths Chapter 4 Factorisation Additional Questions

KSEEB Solutions for Class 8 Maths Chapter 4 Factorisation Additional Questions 1

Question 3.
Factorise the following
(i) x2 + 7x + 12
Answer:
x2 + 7x + 12 12xz
x2 + 4x + 3x + 12

x (x + 4) + 3 (x + 4)
(x + 4) (x + 3)

KSEEB Solutions for Class 8 Maths Chapter 4 Factorisation Additional Questions 2

(ii) x2 + x – 12
Answer:
x2 + 4x – 3x – 12
x(x + 4) -3 (x + 4)
(x + 4) (x – 3)

KSEEB Solutions for Class 8 Maths Chapter 4 Factorisation Additional Questions 3

(iii) x2 – 3x – 18
Answer:
x2 – 6x + 3x – 18
x(x – 6) +3 (x -6)
(x – 6) (x + 3)

KSEEB Solutions for Class 8 Maths Chapter 4 Factorisation Additional Questions 4

(iv) x2 + 4x – 21
Answer:
x2 + 7x – 3x – 12
x(x + 7) -3 (x + 7)
(x + 7) (x – 3)

KSEEB Solutions for Class 8 Maths Chapter 4 Factorisation Additional Questions 5

(v) x2 – 4x – 192
Answer:
x2 – 16x + 12x – 192
x(x – 16) +12 (x – 16)
(x -16) (x +12)

KSEEB Solutions for Class 8 Maths Chapter 4 Factorisation Additional Questions 6

(vi) x4 – 5x2 + 4
Answer:
x4 – 4x2 – x2 + 4
x2(x2 – 4) -1 (x2 – 4)
(x2 – 4) (x2 – 1) = (x + 2) (x – 2)
(x + 1) (x – 1)

KSEEB Solutions for Class 8 Maths Chapter 4 Factorisation Additional Questions 7

(vii) x4 – 13x2y2 + 36y4
Answer:
x4 -9x2y2 – 4x2y2 +36y4
x2(x2 – 9y2) -4y2
(x2 – 9y2)
(x2 – 9y2) (x2 – 4y2)
= [x2 – (3y)2] [x2 – (2y)2]
= (x + 3y) (x – 3y) (x + 2y) (x – 2y)

KSEEB Solutions for Class 8 Maths Chapter 4 Factorisation Additional Questions 8

Question 4.
Factorise the following
(i) 2x2 + 7x + 6
Answer:
2x2 + 4x + 3x + 6
2x (x +2) + 3 (x + 2)
(x + 2) (2x + 3)

KSEEB Solutions for Class 8 Maths Chapter 4 Factorisation Additional Questions 9

(ii) 3x2 – 17x + 20
Answer:
3x2 – 12x – 5x + 20
3x (x – 4) – 5 (x – 4)
(x – 4) (3x – 5)

(iii) 6x2 – 5x – 14
Answer:
6x2 – 12x + 7x – 14
6x (x – 2) + 7 (x – 2)
(x – 2) (6x +7)
KSEEB Solutions for Class 8 Maths Chapter 4 Factorisation Additional Questions 10

(iv) 4x2 + 12xy – 5y2
Answer:
4x2 – lOxy + 2xy + 5y2 – 20x2y2
2x (2x + 5y) + y
(2x + 5y)
(2x + 5y) (2x + y)
KSEEB Solutions for Class 8 Maths Chapter 4 Factorisation Additional Questions 11

(v) 4x4 – 5x2 + 1
Answer:
4x2 – 4x2 – x2 + 1 4×4
4x2 (x2 – 1) – 1
(x2 – 1)
(x2 – 1) (4x2 – 1)
(x2 – 12) – ((2x)2 – 12)
(x2 – l2) – ((2x)2 – l2)
(x + 1) (x – 1) (2x + 1) (2x – 1)
KSEEB Solutions for Class 8 Maths Chapter 4 Factorisation Additional Questions 12

KSEEB Solutions for Class 8 Maths Chapter 4 Factorisation Additional Questions

Question 5.
Factorise the following
(i) x8 – y8
Answer:
(x4)2 – (y4)2
= (x4 + y4) (x4 – y4)
= (x4 + y4) [(x2)2 – (y2)2]
= (x4 + y4) (x2 + y2) (x2 – y2)
= (x4 + y4) (x2 + y2) (x + y)
(x – y)

(ii) a12x4 – a4x12
Answer:
[(a6)2(x2)2 – (a2)2(x6)2]
= [(a6x2)2 – (a2x6)2]
= (a6x2 + a2x6) (a6x2 – a2x6)
= a2x2 (a4 + x4)
[(a3)2 x2 – a2 (x3)2]
= a2x2 (a4 + x4) [(a3x)2 – (ax3)2] = a2x2 (a4 + x4) [(a3x + ax3)
(a3x – ax3)]
= a2x2 (a4 + x4) [ax(a2 + x2).ax (a2 – x2)]
= a2x2 (a4 + x4) [a2x2 (a2 + x2) (a2 – x2)]
= a4x4 (a4 + x4) (a2 + x2) (a + x) (a – x)

(iii) x4 + x2 + 1
Answer:
X4 + X2 + 1 + X2 – X2
= x4 + 2x2 + 1 – x2
= (x2)2 + 2.x2.1 + l2 – x2
= (x2 + l)2 – X2
= (x2 + 1 + x) (x2 + 1 – x)
= (x2 + x + 1) (x2 – X + 1)

(iv) x4 + 5x2 + 9
Answer:
x4 + 5x2 + 9 + x2 – x2
= x4 + 6x2 + 9 – x2
= (x2)2 + 2.x2.3 + 32 – x2 = (x2 + 3)2 – x2
= (x2 + 3 + x) (x2 + 3 – x)
= (x2 + x + 3) (x2 – x + 3)

KSEEB Solutions for Class 8 Maths Chapter 4 Factorisation Additional Questions

Question 6.
Factorise x4 + 4y4 use this to prove that 2011 4 + 64 is a composite number Answer:
x4 + 4y4
= (x2)2 + (2y2)2
= (x2 + 2y2)2 – 2x2.2y2
[v a2 + b2 = (a + b)2 – 2ab]
= (x2 + 2y2) – (4x2y2)
= (x2 + 2y2) – (2xy)2
(x2 + 2y2 + 2xy) (x2 + 2y2 – 2xy) [a2 – b2 = (a + b) (a – b)]
∴ x4 + 4y4 = (x2 + 2y2 + 2xy)
(x2 + 2y2 – 2xy)
Similarly 20114 + 64 can be written as
20114 + 4 x 16 = (2011)4 + 4.24
= (20112 + 2.22 + 2.2011.2) (20112 + 2.22 – 2.2011.2)
∴ 20114 + 64 is a composite number.

KSEEB Solutions for Class 8 Maths Chapter 4 Factorisation Additional Questions

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