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 – 15x**

^{2}**Solution:**

5x- 15x

^{2}= 5x (1 – 3x)

{HCF of 5, 15 = 5 and of x, x

^{2}= x}

**Question 3.20a ^{12}b^{2} – 15a^{8}b^{4}**

**Solution:**

20a

^{12}b

^{2}– 15a

^{8}b

^{4}

{HCF of 20, 15 = 5, a

^{12}, a

^{8}= a

^{8}, b

^{2}, b

^{4}= b

^{2}}

= 5a

^{8 }b

^{2}(4a

^{4}– 3b

^{2})

**Question 4.**

**72xy – 96x**

^{7}y^{6}**Solution:**

72xy – 96x

^{7}y

^{6}

HCF of 72, 96 = 24 of x

^{6}x

^{7}= x

^{6}, y

^{7},y

^{6}= y

^{6}∴ 72x

^{7}y

^{6}– 96x

^{7}y

^{6}= 24x

^{6}y

^{6}(3y – 4x)

**Question 5.20X ^{3} – 40x^{2} + 80x**

**Solution:**

20x

^{3}– 40x

^{2}+ 80x

HCF of 20, 40,80 = 20

HCF of x

^{3}, x

^{2}, x = x

∴ 20x

^{3}– 40x

^{2}+ 80x = 20x (x

^{2}– 2x + 4)

**Question 6.**

**2x**

^{3}y^{2}– 4x^{2}y^{3}+ 8xy^{4}**Solution:**

2x

^{3}y

^{2}– 4x

^{2}y

^{3}+ 8xy

^{4}HCF of 2, 4, 8 = 2

HCF of x

^{3}, x

^{2}, x = 1

and HCF of y

^{2}, y

^{3}, y

^{4}= y

^{2}∴ 2x

^{3}y

^{2}– 4x

^{2}y

^{3}+ 8xy

^{4}= 2xy

^{2}(x

^{2}– 2xy + 4y

^{2})

**Question 7.10m ^{3}n^{2} + 15m^{4}n – 20m^{2}n^{3}**

**Solution:**

10m

^{3}n

^{2}+ 15m

^{4}n – 20m

^{2}n

^{3}

HCF of 10, 15, 20 = 5

HCF of m

^{3}, m

^{4}, m

^{2}= m

^{2}HCF of n

^{2}, n, n

^{3}= n

10m

^{3}n

^{2}+ 15m

^{4}n – 20m

^{2}n

^{3}5m

^{2}n(2mn + 3m

^{2}– 4n

^{2})

**Question 8.**

**2a**

^{4}b^{4}– 3a^{3}b^{5}+ 4a^{2}b^{5}**Solution:**

2a

^{4}b

^{4}– 3a

^{3}b

^{5}+ 4a

^{2}b

^{5}HCF of 2, 3, 4= 1

HCF of a

^{4}, a

^{3}, a

^{2}= a

^{2}HCF of b

^{4}, b

^{5}b

^{5}= b

^{4}∴ 2a

^{4}b

^{4}– 3a

^{3}b

^{5}+ 4a

^{2}b

^{5}= a

^{2}b

^{4}

(2a

^{2}– 3ab + 4b)

**Question 9.**

**28a**

^{2}+ 14a^{2}b^{2}– 21a^{4}**Solution:**

28a

^{2}+ 14a

^{2}b

^{2}– 21a

^{4}HCF of 28, 14,21 =7

HCF of a

^{2}, a

^{2}, a

^{4}= a

^{2}HCF of 1, b

^{2}, 1 = 1

∴ 28a

^{2}+ 14a

^{2}b

^{2}-21a

^{4}= 7a

^{2}(4 + 2b

^{2}– 3a

^{2})

**Question 10.a ^{4}b – 3a^{2}b^{2} – 6ab^{3}Solution:**a

^{4}b – 3a

^{2}b

^{2}– 6ab

^{3}HCF of 1,3,6 = 1

HCF of a

^{4}, a

^{2}, a = a

HCF of b, b

^{2}, b

^{3}= b

∴ a

^{4}b – 3a

^{2}b

^{2}– 6ab

^{3}= ab (a

^{3}– 3ab – 6b

^{2})

**Question 11.**

**2l**

^{2}mn – 3lm^{2}n + 4lmn^{2}**Solution:**

2l

^{2}mn – 3lm

^{2}n + 4lmn

^{2}HCF 2, 3,4 = 1,

HCF of l

^{2},l,l = l

HCF of m, m

^{2}, m = m

HCF of n, n, n

^{2}= n

∴ 2l

^{2 }mn – 3lm

^{2}n + 4lmn

^{2}

= lmn (21 -3m + 4n)

**Question 12.**

**x**

^{4}y^{2}– x^{2}y^{4}– x^{4}y^{4}**Solution:**

x

^{4}y

^{2}– x

^{2}y

^{4}– x

^{4}y

^{4}HCF of x

^{4}, x

^{2}, x

^{4}= x

^{2}HCF of y

^{2}, y

^{4}, y

^{4}=y

^{2}∴ x

^{4}y

^{2}– x

^{2}y

^{4}– x

^{4}y

^{4}= x

^{2}y

^{2}(x

^{2 }-y

^{2}-x

^{2}y

^{2})

**Question 13.9 x ^{2}y + 3 axy**

**Solution:**

9 x

^{2}y + 3 axy

HCF of 9, 3 = 3

HCF of x

^{2}, x = x

HCF of y,y = y

HCF of 1,a = 1

∴ 9x

^{2}y + 3axy = 3xy (3x + a)

**Question 14.**

**16m – 4m**

^{2}**Solution:**

16m – 4m

^{2}HCF of 16, 4 = 4

HCF of m, m

^{2}= m

∴ 16m – 4m

^{2}= 4m (4 – m)

**Question 15.**

**-4a**

^{2}+ 4ab – 4ca**Solution:**

-4a

^{2}+ 4ab – 4ca

HCF of 4, 4, 4 = 4

HCF of a

^{2}, a, a = a

∴ -4a

^{2}+ 4ab – 4ca = -4a (a – b + c)

**Question 16.**

^{x2yz + xy2z + xyz2}**Solution:**

x

^{2}yz + xy

^{2}z + xyz

^{2}HCF of x

^{2}, x, x = x

HCF of y,y

^{2},y=y

HCF of z, z,z

^{2}= z

∴ x

^{2}yz + xy

^{2}z + xyz

^{2}= xyz (x + y + z)

**Question 17.ax ^{2}y + bxy^{2} + cxyz**

**Solution:**

ax

^{2}y + bxy

^{2}+ cxyz

HCF of x

^{2}, x, x = x,

HCF of y,y

^{2},y = y

ax

^{2}y + bxy

^{2}+ cxyz = xy (ax + by + cz)

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