In this chapter, we provide RD Sharma Class 8 Solutions Chapter 7 Factorization Ex 7.4 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 7.4 Maths pdf, free RD Sharma Class 8 Solutions Chapter 7 Factorization Ex 7.4 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 7.4 |

**RD Sharma Solutions for Class 8 Chapter 7 Factorization Ex 7.4** **Download PDF**

**Factorize each of the following expressions :****Question 1.****qr-pr + qs – ps****Solution:**

qr- pr + qs-ps

Arranging in suitable groups = r(q-p) +s (q-p) {(q – p) is common}

= (q-p) (r + s)

**Question 2.****p ^{2}q -pr^{2}-pq + r^{2}**

**Solution:**

p

^{2}q -pr

^{2}-pq + r2

= p

^{2}q -pq-pr

^{2}+ r

^{2}(Arranging in group)

= pq(p- 1)-r

^{2}(p-1) {(p – 1) is common}

= (p – 1) (pq – r

^{2})

**Question 3.**

**1 + x + xy + x**

^{2}y**Solution:**

1 + x + xy + x 2y

= 1 (1 + x) +xy (1 +x)

= (1 + x) (1 + xy) {(1 + x) is common}

**Question 4.****ax + ay – bx – by****Solution:**

ax + ay – bx – by

= a (x + y) – b (x + y) {(x + y) is coinmon}

= (x+y) (a- b)

**Question 5.****xa ^{2} + xb^{2} -ya^{2} – yb^{2}**

**Solution:**

xa

^{2}+ xb

^{2}– ya

^{2}– yb

^{2}= x (a

^{2}+ b

^{2}) -y (a

^{2}+ b

^{2}) {(a

^{2}+ b

^{2}) is common}

= {a

^{2}+ b

^{2}) (x -y)

**Question 6.x ^{2} + xy + xz + yz**

**Solution:**

x

^{2}+ xy + xz + yz

= x (x + y) + z(x + y) {(x + y) is common}

= (x + y) (x + z)

**Question 7.**

**2ax + bx + 2ay**

*+*by**Solution:**

2ax + bx + 2ay + by

= x {2a + b) + y (2a + b) {(2a + b) is common}

= (2a + b) (x + y)

**Question 8.**

**ab- by- ay +y**

^{2}

**Solution:**

ab – by – ay + y

^{2}= b(a-y)-y(a-y) {(a -y) is common}

= (a-y) (b – y)

**Question 9.**

**axy + bcxy -az- bcz**

**Solution:**

axy + bcxy – az – bcz

= xy (a + bc) – z (a + bc) {(a + bc) is common}

= (a + bc) (xy – z)

**Question 10.lm ^{2} – mn^{2} – lm + n**

^{2}

**Solution:**

lm

^{2}– mn

^{2}– lm + n

^{2}= m (lm – n

^{2})- 1 (lm – n

^{2}) {(lm – n

^{2}) is common}

= (lm – n

^{2}) (m – 1)

**Question 11.**

**x**

^{3 }– y^{2 }+ x – x^{2}y^{2}

**Solution:**

x

^{3}-y

^{2}+ x – x

^{2}y

^{2}⇒ x

^{3}+ x – x

^{2}y

^{2}– y

^{2}= x(x

^{2}+ 1)-y

^{2}(x

^{2}+ 1) {(x

^{2}+ 1) is common}

= (x

^{2}+ 1) (x -y

^{2})

**Question 12.**

**6xy +**

*6 –*9y – 4x**Solution:**

6xy + 6 – 9y – 4x

= 6 xy – 4x – 9y + 6

= 2x (3y – 2) – 3 (3y – 2) {(3y – 2) is common}

= (3y-2) (2x – 3)

**Question 13. x^{2} – 2ax – 2ab + bxSolution:**x

^{2}– 2ax – 2ab + bx

⇒ x

^{2}– 2ax + bx – 2ab

= x (x – 2a) + b (x – 2a) {(x – 2a) is common}

= (x – 2a) (x + b)

**Question 14.**

**x**

^{3}– 2x^{2}y + 3xy^{2}– 6y^{3}

**Solution:**

x

^{3}– 2x

^{2}y + 3xy

^{2}– 6y

^{3}= x

^{2}(x – 2y) + 3y

^{2}(x – 2y) {(x – 2y) is common}

= (x – 2y) (x

^{2}+ 3y

^{2})

**Question 15.**

**abx**

^{2}*+*(ay – b) x-y**Solution:**

abx

^{2}+ (ay – b) x-y

= abx

^{2}+ ayx – bx -y = ax (bx + y) – 1 (bx + y) {(bx +y) is common}

= (bx + y) (ax – 1)

**Question 16.**

**(ax + by)**

^{2}+ (bx – ay)^{2}

**Solution:**

(ax + by)

^{2}+ (bx – ay)

^{2}= a

^{2}x

^{2}+ b

^{2}y

^{2}+ 2abxy + b

^{2}x

^{2}+ a

^{2}y

^{2}– 2abxy

= a

^{2}x

^{2}+ b

^{2}y

^{2}+ b

^{2}x

^{2}+ a

^{2}y

^{2}= a

^{2}x

^{2}+ b

^{2}x

^{2}+ a

^{2}y

^{2}+ by

^{2}= x

^{2}(a

^{2}+ b

^{2}) + y

^{2}(a

^{2}+ b

^{2}) {(a

^{2}+ b

^{2}) is common}

*=*(a

^{2}+ b

^{2})

*(x*y

^{2}+^{2})

**Question 17. 16 (a – b)^{3} -24 (a- b)**

^{2}

**Solution:**

16 (a – b)

^{3}-24 (a- b)

^{2}

HCF of 16, 24 = 8

and HCF of (a – b)

^{3}, (a – b)

^{2}= (a – b)

^{2}∴16 (a – b)

^{3}– 24 (a – b)

^{2}= 8 (a-b)

^{2}{2 (a-b)- 3}

{8 (a – b)

^{2}is common}

= 8 (a – b)

^{2}(2a – 2b – 3)

**Question 18.**

**ab (x**

^{2}+ 1) + x (a^{2}+ b^{2})**Solution:**

ab (x

^{2}+ 1) + x (a

^{2}+ b

^{2})

= abx

^{2}+ ab + a

^{2}x + b

^{2}x

= abx

^{2}+ b

^{2}x + a

^{2}x + ab

= bx (ax + b) + a (ax + b) {(ax + b) is common}

= (ax + b) (bx + a)

**Question 19.**

**a**

^{2}x^{2}*+ (*ax^{2}*+ 1)*x + a**Solution:**

a

^{2}x

^{2}+ (ax

^{2}+ 1) x + a

= a

^{2}x

^{2}+ ax

^{3}+ x + a

= ax

^{3}+ a

^{2}x

^{2}+ x + a

= ax

^{2}(x + a) + 1 (x + a) {(x + a) is common}

= (x + a) (ax

^{2}+ 1)

**Question 20.**

**a(a- 2b -c) + 2bc**

**Solution:**

a(a- 2b -c) + 2bc= a

^{2}– 2ab -ac +2bc= a (a – 2b) – c (a – 2b) {(a – 2b) is common}

= (a – 2b) (a – c)

**Question 21.**

**a (a + b – c)- bc**

**Solution:**

a (a + b – c) – bc

= a

^{2}+ ab – ac – bc

= a (a + b) – c (a + b) {(a + b) is common}

= (a + b) (a – c)

**Question 22.x ^{2} – 11xy – x +11y**

**Solution:**

x

^{2}– 11xy-x + 11y

= x

^{2}-x – 11 xy + 11 y

= x (x – 1) – 11y (x – 1) {(x – 1) is common}

= (x- 1) (x- 11y)

**Question 23.**

**ab – a – b +**

**1****Solution:**

ab – a-b + 1

= a (b – 1) – 1 (b – 1) {(b – 1) is common}

= (b – 1) (a – 1)

**Question 24.x ^{2} + y – xy – x**

**Solution:**

x

^{2}+ y – xy – x

= x

^{2}– x- xy + y

= x (x – 1) – y (x – 1) {(x – 1) is common}

*=*(x-

*1)*(x-y)

**All Chapter RD Sharma Solutions For Class 8 Maths**

—————————————————————————–**All Subject NCERT Exemplar Problems Solutions For Class 8**

**All Subject NCERT Solutions For Class 8**

*************************************************

I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment on the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.

If these solutions have helped you, you can also share rdsharmasolutions.in to your friends.