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

In this chapter, we provide KSEEB SSLC Class 8 Maths Chapter 4 Factorisation Additional Questions for English medium students, Which will very helpful for every student in their exams. Students can download the latest KSEEB SSLC Class 8 Maths Chapter 4 Factorisation Additional Questions pdf, free KSEEB SSLC Class 8 Maths Chapter 4 Factorisation Additional Questions pdf download. Now you will get step by step solution to each question.

## Karnataka State Syllabus Class 8 Maths Chapter 4 Factorisation Additional Questions

Question 1.
(a) 4a + 12b is equal to
A. 4a
B. 12b
C. 4(a + 3b)
D. 3a
(C) 4 (a + 3b)

(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
(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)
(D) (x + 4) (x + 2)

(d) The denominator of an algebraic fraction should not be
A. 1
B. 0
C. 4
D. 7
(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
(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
(A) 0.4

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

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

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

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

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

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

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

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

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

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

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

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

(vii) x4 – 13x2y2 + 36y4
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)

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

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

Question 5.
Factorise the following
(i) x8 – y8
(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
[(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
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
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)

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.

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