# KSEEB Solutions for Class 8 Maths Chapter 4 Factorisation Ex 4.1

In this chapter, we provide KSEEB SSLC Class 8 Maths Chapter 4 Factorisation Ex 4.1 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 Ex 4.1 pdf, free KSEEB SSLC Class 8 Maths Chapter 4 Factorisation Ex 4.1 pdf download. Now you will get step by step solution to each question.

## Karnataka Board Class 8 Maths Chapter 4 Factorisation Ex 4.1

1. Resolve into factors.

Question i.
x² + xy
x² + xy = x(x + y)

Question ii.
3x² – 6x
3x² – 6x = 3x (x – 2)

Question iii.
(1.6)a² – (0.8)a
(1.6)² – (0.8)a
= (0.8 x 2a²) – (0.8)a
= 0. 8a(2a- 1)

Question iv.
5 – 10m – 20n
5 – 10m -20n = 5(1 – 2m – 4n)

2. Froctorise:

Question i.
a² + ax + ab + bx
a² + ax + ab + bx a(a + x) + b(a + x)
(a + x)(a + b)

Question ii.
3ac + 7bc – 3ad – 7bd
3ac + 7bc – 3ad – 7bd
c(3a + 7b – d(3a + 7b) (3a + 7b) (c-d)

Question iii.
3xy – 6zy – 3xt + 6zt
3y (x – 2z) – 3t(x – 2z)
(x – 2z) (3y – 3t)

Question iv.
y3 + 3y² + 2y – 6 – xy + 3x
y² (y – 3) + 2(y – 3) – x(y – 3)
(y- 3) (y² + 2 – x)

3. Factorise:

Question i.
4a² – 25
4a² – 25
= (2a)² – 52
[a² – b²=(a+b)(a-b)]
= (2a + 5)(2a – 5)

Question ii.
x2−916
Ans. Question iii.
x4 – y4
x4 – y4
= (x²)² – (y²)²
Identity a² – b² = (a + b)(a – b)
= (x² + y²)(x² – y²)
= (x²+ y²)(x + y)(x – y)

Question iv.
(7310)2−(2110)2
Ans.  Question v.
(0.7)² – (0.3)²
(0.7)² – (0.3)²
Identity a² – b² = (a + b)(a – b)
= (0.7 + 0.3)(0.7 – 0.3)
= (1.0)(0.4)
= 0.4

Question vi.
(5a-2b)²-(2a-b)²
(5a-2b)² – (2a-b)²
Identity a² – b² = (a + b)(a – b)
= [5a – 2b + 2a – b]
= [5a – 2b – (2a – b)]
= (7a – 3b)[5a – 2b – 2a + b]
= (7a – 3b)(3a – b)

All Chapter KSEEB Solutions For Class 8 maths

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