In this chapter, we provide KSEEB SSLC Class 8 Maths Chapter 4 Factorisation Ex 4.2 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 4.2 pdf, free KSEEB SSLC Class 8 Maths Chapter 4 Factorisation Ex 4.2 pdf download. Now you will get step by step solution to each question.

**Karnataka State Syllabus Class 8 Maths Chapter 4 Factorisation Ex 4.2**

## Karnataka Board Class 8 Maths Chapter 4 Factorisation Ex 4.2

1. In the following, you are given the product pq and the sum p + q. Determine p and q.

Question i.

pq = 18, p + q = 11

Answer:

p = 9, q = 2

Question ii.

pq = 32 and p + q = – 12

Ans.

p = -8, q = – 4

Question iii.

pq = -24 and p + q = 2

Answer:

p = 6, q = – 4

Question iv.

pq = -12 and p + q = 11

Answer:

p = 12, q = – 1

Question v.

pq = – 4 and p + q = – 5

Answer:

p = – 6 and q = 1

Question vi.

pq = – 44 and p + q = – 7

Answer:

p = -11, q = 4

2. Factorise.

Question i.

x² + 6x + 8

Answer:

x² + 4x + 2x + 8

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

(x + 4)(x + 2)

Question ii.

x² + 4x + 3

Answer:

x² + 3x + x + 3

x(x + 3) + 1 (x + 3)

(x + 3)(x + 1)

Question iii.

a² + 5a + 6

Answer:

a² + 3a + 2a + 6

a(a + 3) + 2(a + 3)

(a + 3)(a + 2)

Question iv.

a² – 5a + 6

Answer:

a² – 3a – 2a + 6

a(a – 3) – 2(a – 3)

(a – 3)(a – 2)

Question v.

a² – 3a – 40

Answer:

a² – 8a + 5a – 40

a(a – 8) + 5(a – 8)

(a – 8)(a + 5)

Question vi.

x² – x – 72

Answer:

x² – 9x + 8x – 72

x(x – 9) + 8(x – 9)

(x – 9)(x + 8)

3. Factorise :

Question i.

x² + 14x + 49

Answer:

x² + 14x + 49

Identity a² + 2ab + b² = (a + b)²

= x² + 2 × x × 7 + 7²

= (x + 7 )²

Question ii.

4x² + 4x + 1

Answer:

4x² + 4x + 1

Identity a² +2ab + b² =(a + b)²

= (2x)² + (2)(x)(l) +1²

= (2x +1)²

Question iii.

a²-10a+ 25

Answer:

a² – 10a + 25

Identity a² – 2ab + b² = (a – b)²

= a² – (2)(a)(5) + 5²

= (a-5)²

Question iv.

2×2 – 24x + 72

Answer:

2x² – 24x + 72

= 2(x² – 12x + 36)

Identity a² – 2ab + b² = (a – b)²

= 2(X² – (2)(x)(6) + 6²)

= 2(x-6)²

Question v.

p² – 24p + 144

Answer:

p2 – 24p + 144

Identity a² – 2ab + b² = (a – b)²

= p² – (2)(p)(12²) + 12²

= (p – 12)²

Question vi.

x3-12×2 + 36x

Answer:

x3 – 12×2 + 36x

Identify a² – 2ab + b2 =(a – b)²

=x [x² -12.x + 36]

= x[x² – (2)(x)(6) + 6²]

= x(x – 6)²

**All Chapter KSEEB Solutions For Class 8 maths**

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