In this chapter, we provide RD Sharma Class 8 Solutions Chapter 7 Factorization Ex 7.2 for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 8 Solutions Chapter 7 Factorization Ex 7.2 Maths pdf, free RD Sharma Class 8 Solutions Chapter 7 Factorization Ex 7.2 Maths book pdf download. Now you will get step by step solution to each question.
Textbook | NCERT |
Class | Class 8 |
Subject | Maths |
Chapter | Chapter 7 |
Chapter Name | Factorization |
Exercise | Ex 7.2 |
RD Sharma Solutions for Class 8 Chapter 7 Factorization Ex 7.2 Download PDF
Factorize the following :
Question 1.
3x-9
Solution:
3x – 9 = 3 (x – 3) (HCF of 3, 9 = 3)
Question 2.
5x – 15x2
Solution:
5x- 15x2 = 5x (1 – 3x)
{HCF of 5, 15 = 5 and of x, x2 = x}
Question 3.
20a12b2 – 15a8b4
Solution:
20a12b2 – 15a8b4
{HCF of 20, 15 = 5, a12, a8 = a8, b2, b4 = b2}
= 5a8 b2(4a4 – 3b2)
Question 4.
72xy – 96x7y6
Solution:
72xy – 96x7y6
HCF of 72, 96 = 24 of x6x7 = x6, y7,y6 = y6
∴ 72x7y6 – 96x7y6 = 24x6y6 (3y – 4x)
Question 5.
20X3 – 40x2 + 80x
Solution:
20x3 – 40x2 + 80x
HCF of 20, 40,80 = 20
HCF of x3, x2, x = x
∴ 20x3 – 40x2 + 80x = 20x (x2 – 2x + 4)
Question 6.
2x3y2 – 4x2y3 + 8xy4
Solution:
2x3y2 – 4x2y3 + 8xy4
HCF of 2, 4, 8 = 2
HCF of x3, x2, x = 1
and HCF of y2, y3, y4 = y2
∴ 2x3y2 – 4x2y3 + 8xy4
= 2xy2 (x2 – 2xy + 4y2)
Question 7.
10m3n2 + 15m4n – 20m2n3
Solution:
10m3n2 + 15m4n – 20m2n3
HCF of 10, 15, 20 = 5
HCF of m3, m4, m2 = m2
HCF of n2, n, n3 = n
10m3n2 + 15m4n – 20m2n3
5m2n(2mn + 3m2– 4n2)
Question 8.
2a4b4 – 3a3b5 + 4a2b5
Solution:
2a4b4 – 3a3b5 + 4a2b5
HCF of 2, 3, 4= 1
HCF of a4, a3, a2 = a2
HCF of b4, b5 b5 = b4
∴ 2a4b4 – 3a3b5 + 4a2b5 = a2b4
(2a2 – 3ab + 4b)
Question 9.
28a2 + 14a2b2 – 21a4
Solution:
28a2 + 14a2b2 – 21a4
HCF of 28, 14,21 =7
HCF of a2, a2, a4 = a2
HCF of 1, b2, 1 = 1
∴ 28a2 + 14a2b2-21a4 = 7a2
(4 + 2b2 – 3a2)
Question 10.
a4b – 3a2b2 – 6ab3
Solution:
a4b – 3a2b2 – 6ab3
HCF of 1,3,6 = 1
HCF of a4, a2, a = a
HCF of b, b2, b3 = b
∴ a4b – 3a2b2 – 6ab3 = ab (a3 – 3ab – 6b2)
Question 11.
2l2mn – 3lm2n + 4lmn2
Solution:
2l2mn – 3lm2n + 4lmn2
HCF 2, 3,4 = 1,
HCF of l2,l,l = l
HCF of m, m2, m = m
HCF of n, n, n2 = n
∴ 2l2 mn – 3lm2n + 4lmn2
= lmn (21 -3m + 4n)
Question 12.
x4y2 – x2y4 – x4y4
Solution:
x4y2 – x2y4 – x4y4
HCF of x4, x2, x4 = x2
HCF of y2, y4, y4 =y2
∴ x4y2 – x2y4 – x4y4 = x2y2 (x2 -y2 -x2y2)
Question 13.
9 x2y + 3 axy
Solution:
9 x2y + 3 axy
HCF of 9, 3 = 3
HCF of x2, x = x
HCF of y,y = y
HCF of 1,a = 1
∴ 9x2y + 3axy = 3xy (3x + a)
Question 14.
16m – 4m2
Solution:
16m – 4m2
HCF of 16, 4 = 4
HCF of m, m2 = m
∴ 16m – 4m2 = 4m (4 – m)
Question 15.
-4a2 + 4ab – 4ca
Solution:
-4a2 + 4ab – 4ca
HCF of 4, 4, 4 = 4
HCF of a2, a, a = a
∴ -4a2 + 4ab – 4ca = -4a (a – b + c)
Question 16.
x2yz + xy2z + xyz2
Solution:
x2yz + xy2z + xyz2
HCF of x2, x, x = x
HCF of y,y2,y=y
HCF of z, z,z2 = z
∴ x2yz + xy2z + xyz2 = xyz (x + y + z)
Question 17.
ax2y + bxy2 + cxyz
Solution:
ax2y + bxy2 + cxyz
HCF of x2, x, x = x,
HCF of y,y2,y = y
ax2y + bxy2 + cxyz = xy (ax + by + cz)
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