You can Download KSEEB Solutions for Class 8 Maths Chapter 14 Factorization Ex 14.1 Questions and Answers to help you to revise the complete syllabus.
KSEEB Solutions for Class 8 Maths Chapter 14 Factorization Ex 14.1
Question 1.
Find the common factors of the given terms.
(i) 12x, 36
(ii) 2y, 22xy
(iii) 14pq, 28p2q2
(iv) 2x, 3x2, 4
(v) 6abc, 24ab2, 12a2b
(vi) 16x3, -4x2, 32x
(vii) 10pq, 20qr, 30rp
(viii) 3x2y3, 10x3y2, 6x2y2z
Solution:
(i) 12x, 36
Common factor 12
(ii) 2y, 22xy
Common factors 2, y
(iii) 14pq, 28p2q2
Common factors 14, p, q
(iv) 2x, 3x2, 4
Common factor 1
(v) 6abc, 24ab2, 12a2b
Common factors 6, a, b
(vi) 16x3, -4x2, 32x
Common factors 4, x
(vii) 10pq, 20qr, 30rp
Common factor 10
(viii) 3x2y3, 10x3y2, 6x2y2z
Common factors x2, y2
Question 2.
Factorize the following expressions.
(i) 7x – 42
(ii) 6p – 12q
(iii) 7a2 + 14a
(iv) -16z + 20z3
(v) 20l2m + 30alm
(vi) 5x2y – 15xy2
(vii) 10a2 – 15b2 + 20c2
(viii) -4a2 + 4ab – 4ca
(ix) x2yz + xy2z + xyz2
(x) ax2y + bxy2 + cxyz
Solution:
(i) 7x – 42 = 7(x – 6)
(ii) 6p – 12q = 6(p – 2q)
(iii) 7a2 + 14a = 7a(a + 2)
(iv) -16z + 20z3 = 4z(-4 + 5z2)
(v) 20l2m + 30alm = 10lm(2l + 3a)
(vi) 5x2y – 15xy2 = 5xy(x – 3y)
(vii) 10a2 – 15b2 + 20c2 = 5(2a2 – 3b2 + 4c2)
(viii) -4a2 + 4ab – 4ca = 4a(-a + b – c)
(ix) x2yz + xy2z + xyz2 = xyz (x + y + z)
(x) ax2y + bxy2 + cxyz = xy(ax + by + cz)
Question 3.
Factorise.
(i) x2 + xy + 8x + 8y
(ii) 15xy – 6x + 5y – 2
(iii) ax + bx – ay – by
(iv) 15pq + 15 + 9q + 25p
(v) z – 7 + 7xy – xyz
Solution:
(i) x2 + xy + 8x + 8y
= x(x + y) + 8(x + y)
= (x + y) (x + 8)
(ii) 15xy – 6x + 5y – 2
= 3x(5y – 2) + 1(5y – 2)
= (5y – 2) (3x + 1)
(iii) ax + bx – ay – by
= x(a + b) – y(a + b)
= (a + b) (x – y)
(iv) 15pq + 15 + 9q + 25p
= 15pq + 9q + 15 + 25p
= 3q(5p + 3) + 5(3 + 5p)
= (5p + 3) (3q + 5)
(v) z – 7 + 7xy – xyz
= 1(z – 7) + xy(7 – z)
= 1(z – 7) – xy(z – 7)
= (1 – xy) (z – 7)