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 State Syllabus Class 8 Maths Chapter 4 Factorisation Ex 4.1**

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

1. Resolve into factors.

Question i.

x² + xy

Answer:

x² + xy = x(x + y)

Question ii.

3x² – 6x

Answer:

3x² – 6x = 3x (x – 2)

Question iii.

(1.6)a² – (0.8)a

Answer:

(1.6)² – (0.8)a

= (0.8 x 2a²) – (0.8)a

= 0. 8a(2a- 1)

Question iv.

5 – 10m – 20n

Answer:

5 – 10m -20n = 5(1 – 2m – 4n)

2. Froctorise:

Question i.

a² + ax + ab + bx

Answer:

a² + ax + ab + bx a(a + x) + b(a + x)

(a + x)(a + b)

Question ii.

3ac + 7bc – 3ad – 7bd

Answer:

3ac + 7bc – 3ad – 7bd

c(3a + 7b – d(3a + 7b) (3a + 7b) (c-d)

Question iii.

3xy – 6zy – 3xt + 6zt

Answer:

3y (x – 2z) – 3t(x – 2z)

(x – 2z) (3y – 3t)

Question iv.

y3 + 3y² + 2y – 6 – xy + 3x

Answer:

y² (y – 3) + 2(y – 3) – x(y – 3)

(y- 3) (y² + 2 – x)

3. Factorise:

Question i.

4a² – 25

Answer:

4a² – 25

= (2a)² – 52

[a² – b²=(a+b)(a-b)]

= (2a + 5)(2a – 5)

Question ii.

x2−916

Ans.

Question iii.

x^{4} – y4

Answer:

x^{4} – y^{4}

= (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

Answer:

(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

Answer:

(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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