RBSE Solutions for Class 8 Maths Chapter 10 Factorization Ex 10.1 is part of RBSE Solutions for Class 8 Maths. Here we have given Rajasthan Board RBSE Class 8 Maths Chapter 10 Factorization Exercise 10.1.
| Board | RBSE |
| Textbook | SIERT, Rajasthan |
| Class | Class 8 |
| Subject | Maths |
| Chapter | Chapter 10 |
| Chapter Name | Factorization |
| Exercise | Exercise 10.1 |
| Number of Questions | 3 |
| Category | RBSE Solutions |
Rajasthan Board RBSE Class 8 Maths Chapter 10 Factorization Ex 10.1
Question 1.
Find the common factors of the given terms
(i) 12x, 36
(ii) 14pq, 28p²q²
(iii) 6abc, 24ab², 12a²b
(iv) 16x³, – 4x², 32x
(v) 10pq, 20qr, 30rp
(vi) 3x²y³, 10x²y2, 6x²y²z
Solution:
(i) 12x, 36
12x = 2 × 2 × 3 × x
36 = 2 × 2 × 3 × 3
∴ Common(RBSESolutions.com)factor = 2 × 2 × 3 = 12
(ii) 14pq, 28p²q²
14pq = 2 × 7 × p × q
28p²q² = 2 × 2 × 1 × p × p × q × q
∴ Common factor = 2 × 7 × p × q = 14pq
(iii) 6abc, 24ab², 12a²b
6abc = 2 × 3 × a × b × c
24 ab² = 2 × 2 × 2 × 3 × a × b × b
12 a²b = 2 × 2 × 3 × a × a × b
∴ Common(RBSESolutions.com)factor = 2 × 3 × a × b = 6ab
(iv) 16x³, – 4x², 32x
16x³ = 2 × 2 ×2 × 2 × x × x × x
– 4x² = (- 1) × 2 × 2 × x × x
32x = 2 × 2 × 2 × 2 × 2 × x
∴ Common factor = 2 × 2 × x = 4x
(v) 10pq, 20qr, 30rp
10pq = 2 × 5 × p × q
20qr = 2 × 2 × 5 × q × r
30rp = 2 × 3 × 5 × r × p
∴ Common factor = 2 × 5 = 10
(vi) 3x²y³, 10x²y², 6x²y²z
3x²y³ = 3 × x × x × y × y × y
10x²y² = 2 × 5 × x × x × y × y
6x²y²z = 2 × 3 × x × x × y × y × z
∴ Common factor = x × x × y × y = x²y²
Question 2.
Factorize the following expressions. (by common factors method)
(i) 6p – 12q
(ii) 7a² + 14a
(iii) 10a² – 15b² + 20c²
(iv) ax²y + bxy² + cxyz
(v) x²yz + xy²z + xyz²
(vi) – 16z + 20z³
Solution:
(i) 6p – 12q
6p – 12q
= 6 (p – 2q)
(ii) 7a2 + 14a
7a² + 14a
= 7a (a + 2)
(iii) 10a² – 15b² + 20c²
10a² – 15b² + 20c²
= 5 (2a² – 3b² + 4c²)
(iv) ax²y + bxy² + cxyz
ax²y + bxy² + cxyz
= xy (ax + by + cz)
(v) x²yz + xy²z + xyz²
x²yz + xy²z + xyz²
= xyz (x + y + z)
(vi) – 16z + 20z³
– 16z + 20z³
= 4z (- 4 + 5z²)
Question 3.
Factorize (by regrouping method).
(i) 2xy + 3 + 2y + 3x
(ii) z – 7 – 7xy + xyz
(iii) 6xy – 4y + 6 – 9x
(iv) 15pq + 15 + 9q + 25p
Solution:
(i) 2xy + 3 + 2y + 3x
2xy + 3 + 2y + 3x
= 2xy + 2y + 3 + 3x
= 2y (x + 1) + 3(1 + x)
= 2y (x + 1) + 3 (x + 1)
= (x + 1) (2y + 3)
(ii) z – 1 – 7xy + xyz
z – 7 – 7xy + xyz.
= z – 7 – xy(7 – z)
= 1 (z – 7) + xy (z – 7)
= (1 + xy) (z – 7)
= (xy + 1) (z – 7)
(iii) 6xy – 4y + 6 – 9x
6xy – 4y + 6 – 9x
= 2y (3x – 2) + 3(2 – 3x)
= 2y (3x – 2) – 3(3x – 2)
= (3x – 2)(2y – 3)
(iv) 15pq + 15 + 9q + 25p
15pq + 15 + 9q + 25p
= 15pq + 9q + 15 + 25p
= 3q (5p + 3) + 5(3 + 5p)
= 3q (5p + 3) + 5 (5p + 3)
= (5p + 3) (3q + 5)
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