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

Karnataka State Syllabus Class 8 Maths Chapter 4 Factorisation Additional Questions

Question 1.
Choose the correct answer
(a) 4a + 12b is equal to
A. 4a
B. 12b
C. 4(a + 3b)
D. 3a
Answer:
(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
Answer:
(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)
Answer:
(D) (x + 4) (x + 2)

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

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

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

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

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

(v) a²b² + c²d² – a²c² – b²d²
Answer:
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
Answer:
x² + 7x + 12 12xz
x² + 4x + 3x + 12

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

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

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

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

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

(vi) x⁴ – 5x² + 4
Answer:
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
Answer:
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
Answer:
2x² + 4x + 3x + 6
2x (x +2) + 3 (x + 2)
(x + 2) (2x + 3)

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

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