KSEEB Solutions for Class 8 Maths Chapter 4 Factorisation Additional Questions

Students can Download Maths Chapter 4 Factorisation Additional Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

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)

(iii) 6x² – 5x – 14
Answer:
6x² – 12x + 7x – 14
6x (x – 2) + 7 (x – 2)
(x – 2) (6x +7)

(iv) 4x² + 12xy – 5y²
Answer:
4x² – lOxy + 2xy + 5y² – 20x²y²
2x (2x + 5y) + y
(2x + 5y)
(2x + 5y) (2x + y)

(v) 4x⁴ – 5x² + 1
Answer:
4x² – 4x² – x² + 1 4×4
4x² (x² – 1) – 1
(x² – 1)
(x² – 1) (4x² – 1)
(x² – 1²) – ((2x)² – 1²)
(x² – l²) – ((2x)² – l²)
(x + 1) (x – 1) (2x + 1) (2x – 1)

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.