KSEEB Solutions for Class 8 Maths Chapter 4 Factorisation Ex 4.2

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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All Subject KSEEB Solutions For Class 9

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