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 x^{2} + 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) x^{2} + 6x + 9

Answer:

x^{2} + 6x + 9

= x^{2} + 2.x.3 + 3^{2}

= (x + 3)^{2}

(ii) 1 – 8x + 16x^{2}

Answer:

1 – 8x + 16x^{2}

= 1^{2} – 2.1.4x + (4x)^{2}

= (1 – 4x)^{2}

(iii) 4x^{2 }– 81y^{2}

Answer:

4x^{2} – 81 y^{2}

= (2x)^{2} – ( 9y)^{2}

= (2x + 9y) (2x – 9y)

(iv) 4a^{2} + 4ab + b^{2}

Answer:

4a^{2} + 4ab + b^{2}

= (2a)^{2} +2.2a.b + b^{2}

= (2a + b)^{2}

(v) a^{2}b^{2} + c^{2}d^{2} – a^{2}c^{2} – b^{2}d^{2}

Answer:

a^{2}b^{2} + c^{2}d^{2} – a^{2}c^{2} – b^{2}d^{2}

= a^{2}b^{2} – a^{2}c^{2} + c^{2}d^{2} – b^{2}d^{2}

= a^{2} (b^{2} – c^{2}) – d^{2} (b^{2} – c^{2})

= (b^{2} – c^{2}) (a^{2} – d^{2})

Question 3.

Factorise the following

(i) x^{2} + 7x + 12

Answer:

x^{2} + 7x + 12 12xz

x^{2} + 4x + 3x + 12

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

(x + 4) (x + 3)

(ii) x^{2} + x – 12

Answer:

x^{2} + 4x – 3x – 12

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

(x + 4) (x – 3)

(iii) x^{2} – 3x – 18

Answer:

x^{2} – 6x + 3x – 18

x(x – 6) +3 (x -6)

(x – 6) (x + 3)

(iv) x^{2} + 4x – 21

Answer:

x^{2} + 7x – 3x – 12

x(x + 7) -3 (x + 7)

(x + 7) (x – 3)

(v) x^{2} – 4x – 192

Answer:

x^{2} – 16x + 12x – 192

x(x – 16) +12 (x – 16)

(x -16) (x +12)

(vi) x^{4} – 5x^{2} + 4

Answer:

x^{4} – 4x^{2} – x^{2} + 4

x^{2}(x^{2} – 4) -1 (x^{2} – 4)

(x^{2} – 4) (x^{2} – 1) = (x + 2) (x – 2)

(x + 1) (x – 1)

(vii) x^{4} – 13x^{2}y^{2} + 36y^{4}

Answer:

x^{4} -9x^{2}y^{2} – 4x^{2}y^{2} +36y^{4}

x^{2}(x^{2} – 9y^{2}) -4y^{2}

(x^{2} – 9y^{2})

(x^{2} – 9y^{2}) (x^{2} – 4y^{2})

= [x^{2} – (3y)^{2}] [x^{2} – (2y)^{2}]

= (x + 3y) (x – 3y) (x + 2y) (x – 2y)

Question 4.

Factorise the following

(i) 2x^{2} + 7x + 6

Answer:

2x^{2} + 4x + 3x + 6

2x (x +2) + 3 (x + 2)

(x + 2) (2x + 3)

(ii) 3x^{2} – 17x + 20

Answer:

3x^{2} – 12x – 5x + 20

3x (x – 4) – 5 (x – 4)

(x – 4) (3x – 5)

(iii) 6x^{2} – 5x – 14

Answer:

6x^{2} – 12x + 7x – 14

6x (x – 2) + 7 (x – 2)

(x – 2) (6x +7)

(iv) 4x^{2} + 12xy – 5y^{2}

Answer:

4x^{2} – lOxy + 2xy + 5y^{2} – 20x^{2}y^{2}

2x (2x + 5y) + y

(2x + 5y)

(2x + 5y) (2x + y)

(v) 4x^{4} – 5x^{2} + 1

Answer:

4x^{2} – 4x^{2} – x^{2} + 1 4×4

4x^{2} (x^{2} – 1) – 1

(x^{2} – 1)

(x^{2} – 1) (4x^{2} – 1)

(x^{2} – 1^{2}) – ((2x)^{2} – 1^{2})

(x^{2} – l^{2}) – ((2x)^{2} – l^{2})

(x + 1) (x – 1) (2x + 1) (2x – 1)

Question 5.

Factorise the following

(i) x^{8} – y^{8}

Answer:

(x^{4})^{2} – (y^{4})^{2}

= (x^{4} + y^{4}) (x^{4} – y^{4})

= (x^{4} + y^{4}) [(x^{2})^{2} – (y^{2})^{2}]

= (x^{4} + y^{4}) (x^{2} + y^{2}) (x^{2} – y^{2})

= (x^{4} + y^{4}) (x^{2} + y^{2}) (x + y)

(x – y)

(ii) a^{12}x^{4} – a^{4}x^{12}

Answer:

[(a^{6})^{2}(x^{2})^{2} – (a^{2})^{2}(x^{6})^{2}]

= [(a6x^{2})2 – (a^{2}x6)2]

= (a6x^{2} + a2x6) (a6x2 – a2x6)

= a^{2}x^{2} (a^{4} + x^{4})

[(a^{3})^{2} x^{2} – a^{2} (x^{3})^{2}]

= a^{2}x^{2} (a^{4} + x^{4}) [(a^{3}x)^{2} – (ax3)2] = a2x2 (a^{4} + x^{4}) [(a3x + ax^{3})

(a^{3}x – ax^{3})]

= a^{2}x^{2} (a^{4} + x^{4}) [ax(a^{2} + x^{2}).ax (a^{2} – x^{2})]

= a^{2}x^{2} (a^{4} + x^{4}) [a^{2}x^{2} (a^{2} + x^{2}) (a^{2} – x^{2})]

= a^{4}x^{4} (a^{4} + x^{4}) (a^{2} + x^{2}) (a + x) (a – x)

(iii) x^{4} + x^{2} + 1

Answer:

X^{4} + X^{2} + 1 + X^{2} – X^{2}

= x^{4} + 2x^{2} + 1 – x^{2}

= (x^{2})^{2} + 2.x^{2}.1 + l^{2} – x^{2}

= (x^{2} + l)^{2} – X^{2}

= (x^{2} + 1 + x) (x^{2} + 1 – x)

= (x^{2} + x + 1) (x^{2} – X + 1)

(iv) x^{4} + 5x^{2} + 9

Answer:

x^{4} + 5x^{2} + 9 + x^{2} – x^{2}

= x^{4} + 6x^{2} + 9 – x^{2}

= (x^{2})^{2} + 2.x^{2}.3 + 3^{2} – x^{2} = (x^{2} + 3)2 – x^{2}

= (x^{2} + 3 + x) (x^{2} + 3 – x)

= (x^{2} + x + 3) (x^{2} – x + 3)

Question 6.

Factorise x^{4} + 4y^{4} use this to prove that 2011 4 + 64 is a composite number Answer:

x^{4} + 4y^{4}

= (x^{2})^{2} + (2y^{2})^{2}

= (x^{2} + 2y^{2})^{2} – 2x^{2}.2y^{2}

[v a^{2} + b^{2} = (a + b)^{2} – 2ab]

= (x^{2} + 2y^{2}) – (4x^{2}y^{2})

= (x^{2} + 2y^{2}) – (2xy)^{2}

(x^{2} + 2y^{2} + 2xy) (x^{2} + 2y^{2} – 2xy) [a^{2} – b^{2} = (a + b) (a – b)]

∴ x^{4} + 4y^{4} = (x^{2} + 2y^{2} + 2xy)

(x^{2} + 2y^{2} – 2xy)

Similarly 2011^{4} + 64 can be written as

2011^{4} + 4 x 16 = (2011)^{4} + 4.2^{4}

= (2011^{2} + 2.2^{2} + 2.2011.2) (2011^{2} + 2.2^{2} – 2.2011.2)

∴ 2011^{4} + 64 is a composite number.