You can Download KSEEB Solutions for Class 8 Maths Chapter 14 Factorization Ex 14.2 Questions and Answers helps you to revise the complete syllabus.

## KSEEB Solutions for Class 8 Maths Chapter 14 Factorization Ex 14.2

Question 1.

Factorize the following expressions.

(i) a^{2} + 8a + 16

(ii) p^{2} – 10p + 25

(iii) 25m^{2} + 30m + 9

(iv) 49y^{2} + 84yz + 36z^{2}

(v) 4x^{2} – 8x + 4

(vi) 121b^{2} – 88bc + 16c^{2}

(vii) (l + m)^{2} – 4lm (Hint: Expand (l + m)^{2} first)

(viii) a^{4} + 2a^{2}b^{2} + b^{4}

Solution:

(i) a^{2} + 8a + 16

= a^{2} + 4a + 4a + 16

= a(a + 4) + 4(a + 4)

= (a + 4) (a + 4)

(ii) p^{2} – 10p + 25

= p^{2} – 5p – 5p + 25

= p(p – 5) – 5(p – 5)

= (p – 5) (p – 5)

(iii) 25m^{2} + 30m + 9

= (5m)^{2} + 2 (5m)(3) + (3)^{2}

Since a^{2} + 2ab + b^{2} = (a + b)^{2}

= (5m + 3)^{2}

= (5m + 3) (5m + 3)

(iv) 4ay^{2} + 84yz + 36z^{2}

= (7y)^{2} + 2(7y)(6z) + (6z)^{2}

Since a^{2} + 2ab + b^{2} = (a + b)^{2}

= (7y + 6z)^{2}

= (7y + 6z) (7y + 6z)

(v) 4x^{2} – 8x + 4

= (2x)^{2} – 2(2x) (2) + (2)^{2}

Since a^{2} – 2ab + b^{2} = (a – b)^{2}

= (2x – 2)^{2}

= (2x – 2) (2x – 2)

(vi) 121b^{2} – 88bc + 16c^{2}

= (11b)^{2} – 2(11b) (4c) + (4c)^{2}

Since a^{2} – 2ab + b^{2} = (a – b)^{2}

= (11b – 4c)^{2}

= (11b – 4c) (11b – 4c)

(vii) (l + m)^{2} – 4lm

= l^{2} + 2lm + m^{2} – 4lm

= l^{2} – 2lm + m^{2}

But a^{2} – 2ab + b^{2} = (a – b)^{2}

(l – m)^{2} = (l – m) (l – m)

(viii) a^{4} + 2a^{2}b^{2} + b^{4}

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

But a^{2} + 2ab + b^{2} = (a + b)^{2}

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

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

Question 2.

Factorise.

(i) 4p^{2} – 9q^{2}

(ii) 63a^{2} – 112b^{2}

(iii) 49x^{2} – 36

(iv) 16x^{5} – 144x^{3}

(v) (l + m)^{2} – (l – m)^{2}

(vi) 9x^{2}y^{2} – 16

(vii) (x^{2} – 2xy + y^{2}) – z^{2}

(viii) 25a^{2} – 4b^{2} + 28bc – 49c^{2}

Solution:

(i) 4p^{2} – 9q^{2}

= (2p)^{2} – (3q)^{2}

= (2p – 3q) (2p + 3q)

(ii) 63a^{2} – 112b^{2}

= 7(9a^{2} – 16b^{2})

= 7[(3a)^{2} – (4b)^{2}]

∴ (a^{2} – b^{2}) = (a – b) (a + b)

= 7(3a – 4b) (3a + 4b)

(iii) 49x^{2} – 36

= (7x)^{2} – (6)^{2}

∴ (a^{2} – b^{2}) = (a – b) (a + b)

= (7x – 6) (7x + 6)

(iv) 16x^{5} – 144x^{3}

= 16x^{3 }(x^{2} – 9)

= 16x^{3} [(x)^{2} – (3)^{2}]

∴ (a^{2} – b^{2}) = (a – b) (a + b)

= 16x^{3} (x – 3) (x + 3)

(v) (l + m)^{2} – (l – m)^{2}

Let (l + m) = a, (l – m) = b

∴ (a^{2} – b^{2}) = (a – b) (a + b)

= [(l + m) – (l – m)] [l + m + l – m]

= [l + m – l + m] [2l]

= [2m] [2l]

= 4lm

(vi) 9x^{2}y^{2} – 16

= (3xy)^{2} – (4)^{2}

= (3xy – 4) (3xy + 4)

(vii) (x^{2} – 2xy + y^{2}) – z^{2}

= (x – y)^{2} – z^{2}

∴ (a^{2} – b^{2}) = (a + b) (a – b)

= (x – y + z) (x – y – z)

(viii) 25a^{2} – 4b^{2} + 28bc – 49c^{2}

= (5c)^{2} – [4b^{2} – 28bc + 49c^{2}]

= (5c)^{2} – [(2b)^{2} – 2(2b)(7c) + (7c)^{2}]

= (5c)^{2} – [(2b – 7c)^{2}]

= [5c – (2b – 7c)] [5c + 2b – 7c]

= (5c – 2b + 7c) (5c + 2b – 7c)

Question 3.

Factorize the expressions.

(i) ax^{2} + bx

(ii) 7p^{2} + 21q^{2}

(iii) 2x^{3} + 2xy^{2} + 2xz^{2}

(iv) am^{2} + bm^{2} + bn^{2} + an^{2}

(v) (lm + l) + m + 1

(vi) y(y + z) + 9(y + z)

(vii) 5y^{2} – 20y – 8z + 2yz

(viii) 10ab + 4a + 5b + 2

(ix) 6xy – 4y + 6 – 9x

Solution:

(i) ax^{2} + bx = x (ax + b)

(ii) 7p^{2} + 21q^{2} = 7(p^{2} + 3q^{2})

(iii) 2x^{3} + 2xy^{2} + 2xz^{2} = 2x(x^{2} + y^{2} + z^{2})

(iv) am^{2} + bm^{2} + bn^{2} + an^{2}

= m^{2}(a + b) + n^{2}(b + a)

= (a + b) (m^{2} + n^{2})

(v) (lm + l) + m + 1

= l(m + 1) + 1(m + 1)

= (m + 1) (l + 1)

(vi) y(y + z) + 9(y + z) = (y + z) (y + 9)

(vii) 5y^{2} – 20y – 8z + 2yz

= 5y(y – 4) – 2z(4 – y)

= 5y(y – 4) + 2z(y – 4)

= (y – 4) (5y + 2z)

(viii) 10ab + 4a + 5b + 2

= 10ab + 5b + 4a + 2

= 5b(2a + 1) + 2 (2a + 1)

= (2a + 1) (5b + 2)

(ix) 6xy – 4y + 6 – 9x

= 6xy – 9x – 4y + 6

= 3x (2y – 3) – 2(2y – 3)

= (2y – 3) (3x – 2)

Question 4.

Factorise.

(i) a^{4} – b^{4}

(ii) p^{4} – 81

(iii) x^{4} – (y + z)^{4}

(iv) x^{4} – (x – z)^{4}

(v) a^{4} – 2a^{2}b^{2} + b^{4}

Solution:

(i) a^{4} – b^{4}

= (a^{2})^{2} – (b^{2})^{2}

∴ (x^{2} – y^{2}) = (x – y) (x + y)

= (a^{2} – b^{2}) (a^{2} + b^{2})

= (a – b) (a + b) (a^{2} + b^{2})

(ii) p^{4} – 81

= (p^{2})^{2} – (9)^{2}

∴ (a^{2} – b^{2}) = (a – b) (a + b)

= (p^{2} – 9) (p^{2} + 9)

= (p^{2} – 3^{2}) (p^{2} + 9)

= (p – 3) (p + 3) (p^{2} + 9)

(iii) x^{4} – (y + z)^{4}

= (x^{2})^{2} – [(y + z)^{2}]^{2}

∴ (a^{2} – b^{2}) = (a – b) (a + b)

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

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

= (x – y – z) (x + y + z) (x^{2} + y^{2} + z^{2} + 2yz)

(iv) x^{4} – (x – z)^{4}

= (x^{2})^{2} – [(x – z)^{2}]^{2}

∴ (a^{2} – b^{2}) = (a – b) (a + b)

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

= [x – (x – z)] [x + (x – z)] [x^{2} + x^{2} + z^{2} + 2xz]

= (x – x + z) (x + x – z) [2x^{2} + z^{2} + 2xz]

= (z) (2x – z) (2x^{2} + z^{2} + 2xz)

(v) a^{4} – 2a^{2}b^{2} + b^{4}

= (a^{2})^{2} – 2a^{2}b^{2} + (b^{2})^{2}

∴ (a^{2} – 2ab + b^{2}) = (a – b)^{2}

= (a^{2} – b^{2})^{2}

= [(a – b) (a + b)]^{2}

= (a – b)^{2} (a + b)^{2}

= (a – b) (a – b) (a + b) (a + b)

Question 5.

Factorize the following expressions.

(i) p^{2} + 6p + 8

(ii) q^{2} – 10q + 21

(iii) p^{2} + 6p – 16

Solution:

(i) p^{2} + 6p + 8

= p^{2} + 2p + 4p + 8

= p(p + 2) + 4(p + 2)

= (p + 2) (p + 4)

(ii) q^{2} – 10q + 21

= q^{2} – 3q – 7q + 21

= q(q – 3) – 7(q – 3)

= (q – 3) (q – 7)

(iii) p^{2} + 6p – 16

= p^{2} + 8p – 2p – 16

= p(p + 8) – 2(p + 8)

= (p + 8) (p – 2)