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)

We hope the given RBSE Solutions for Class 8 Maths Chapter 10 Factorization Ex 10.1 will help you. If you have any query regarding Rajasthan Board RBSE Class 8 Maths Chapter 10 Factorization Exercise 10.1, drop a comment below and we will get back to you at the earliest.

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