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Class 8 RD Sharma Solutions – Chapter 7 Factorization – Exercise 7.4
  • Last Updated : 12 Nov, 2020
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Question 1. Factorize: qr – pr + qs – ps

Solution:

Given: qr- pr + qs – ps

Grouping similar terms together we get = qr + qs – pr -ps

Taking the similar terms common we get = q(r + s) – p[r + s]

Therefore, as (r + s) is common = (r + s) (q – p) 



Question 2. Factorize: p²q – pr² – pq + r²

Solution:

Given: p²q – pr² – pq + r2

Grouping similar terms together we get = p2q – pq – pr2 + r2

Taking the similar terms common we get = pq(p – 1)-r2(p – 1)

Therefore, as (p – 1) is common = (p – 1) (pq – r2)

Question 3. Factorize: 1 + x + xy + x2y

Solution:

Given: 1 + x + xy + x2y

Taking the similar terms common we get = 1 (1 + x) + xy(1 + x)



Therefore, as (1 + x) is common = (1 + x) (1 + xy) 

Question 4. Factorize: ax + ay – bx – by

Solution:

Given: ax + ay – bx – by

Taking the similar terms common we get = (1 + x) (1 + xy) 

Therefore, as (x + y) is common = (x + y) (a – b)

Question 5. Factorize: xa2 + xb2 – ya2 – yb2

Solution:

Given: xa2 + xb2 – ya2 – yb2

Taking the similar terms common we get = x (a2 + b2) – y (a2 + b2)

Therefore, as (a2 + b2) is common = (a2 + b2) (x – y)

Question 6. Factorize: x2 + xy + xz + yz

Solution:

Given: x2 + xy + xz + yz

Taking the similar terms common we get = x (x + y) + z(x + y)

Therefore, as (x + y) is common = (x + y) (x + z)

Question 7. Factorize: 2ax + bx + 2ay + by

Solution:

Given: 2ax + bx + 2ay + by

Taking the similar terms common we get = x(2a + b) + y (2a + b)

Therefore, as (2a + b) is common = (2a + b) (x + y)

Question 8. Factorize: ab- by- ay +y2

Solution:

Given: ab – by – ay + y2

Taking the similar terms common we get = b(a – y) – y(a – y)

Therefore, as (a – y) is common = (a – y) (b – y)

Question 9. Factorize: axy + bcxy – az – bcz

Solution:

Given: axy + bcxy – az – bcz

Taking the similar terms common we get = xy (a + bc) – z (a + bc)

Therefore, as (a + bc) is common = (a + bc) (xy – z)

Question 10. Factorize: lm2 – mn2 – lm + n2

Solution:

Given: lm2 – mn2 – lm + n2

Taking the similar terms common we get = m (lm – n2) – 1 (lm – n2)

therefore, as (lm – n2) is common = (lm – n2) (m – 1)

Question 11. Factorize: x3 – y2 + x – x2y2

Solution:

Given: x3 – y2 + x – x2y2

Grouping similar terms together we get = x3 + x – x2y2 – y2

Taking the similar terms common we get = x(x2 + 1) – y2(x2+ 1)

Therefore, as (x2 + 1) is common = (x2 + 1) (x – y2)

Question 12. Factorize: 6xy + 6 – 9y- 4x

Solution:

Given: 6xy + 6 – 9y – 4x

Grouping similar terms together we get = 6xy – 4x – 9y + 6

Taking the similar terms common we get = 2x (3y – 2) – 3 (3y – 2)

Therefore, as (3y – 2) is common = (3y – 2) (2x – 3)

Question 13. Factorize: x2 – 2ax – 2ab + bx

Solution:

Given: x2 – 2ax – 2ab + bx

Grouping similar term together we get = x2 – 2ax + bx – 2ab

Taking the similar terms common we get = x (x – 2a) + b (x – 2a)

Therefore, as (x – 2a) is common = (x – 2a) (x + b)

Question 14. Factorize: x3 – 2x2y + 3xy2 – 6y3

Solution:

Given: x3 – 2x2y + 3xy2 – 6y3

Taking the similar terms common we get = x2 (x – 2y) + 3y2 (x – 2y)

Therefore, as (x – 2y) is common = (x – 2y) (x2 + 3y2)

Question 15. Factorize: abx2 + (ay – b) x – y

Solution:

Given: abx2 + (ay – b) x – y

After solving the bracket we get = abx2 + ayx – bx – y  

Taking the similar terms common we get = ax (bx + y) – 1 (bx + y)   

Therefore, as (bx + y) is common = (bx + y) (ax – 1)

Question 16. Factorize: (ax + by)2 + (bx – ay)2

Solution:

Given: (ax + by)2 + (bx – ay)2

After solving the bracket by using the formula ((a + b)2 = a2 + b2 + 2ab) 

we get = a2x2 + b2y2 + 2abxy + b2x2 + a2y2 – 2abxy

Grouping similar terms together we get = a2x2 + a2y2+ b2x2 + b2y

Taking the similar terms common we get = x2 (a2 + b2) + y2 (a2 + b2)

Therefore, as (a2 + b2) is common = (a2 + b2) (x2 + y2)

Question 17. Factorize: 16 (a – b)3 – 24 (a – b)2

Solution:

Given: 16 (a – b)3 – 24 (a – b)2  

Taking the similar terms common we get = 8 (a – b)2 {2 (a – b) – 3}

Therefore, as (8(a – b)2) is common = 8 (a – b)2 (2a – 2b – 3)

Question 18. Factorize: ab (x2 + 1) + x (a2 + b2)

Solution:

Given: ab (x2 + 1) + x(a2 + b2)

After solving the bracket we get = abx2 + ab + a2x + b2x

Grouping similar terms together we get = abx2 + b2x + a2x + ab

Taking the similar terms common we get = bx (ax + b) + a (ax + b)

Therefore, as (ax + b) is common = (ax + b) (bx + a)

Question 19. Factorize: a2x2 + (ax2 + 1) x + a

Solution:

Given: a2x2 + (ax2 + 1) x + a

After solving the bracket we get = a2x2 + ax3 + x + a

Grouping similar terms together we get = ax3 + a2x2 + x + a

Taking the similar terms common we get = ax2 (x + a) + 1 (x + a)

Therefore, as (x + a) is common = (x + a) (ax2 + 1)

Question 20. Factorize: a(a – 2b – c) + 2bc

Solution:

Given: a(a – 2b – c) + 2bc

After solving the bracket we get = a2 2ab – ac + 2bc

Taking the similar terms common we get = a (a – 2b) – c (a – 2b) 

Therefore, as (a – 2b) is common = (a – 2b) (a – c)

Question 21. Factorize: a (a + b – c) – bc

Solution:

Given: a (a + b – c) – bc

After solving the bracket we get = a2 + ab – ac – bc

Taking the similar terms common we get = a (a + b) – c (a + b)

Therefore, as (a + b) is common = (a + b) (a – c)

Question 22. Factorize: x2 – 11xy – x + 11y

Solution:

Given: x2 – 11xy – x + 11y

Grouping similar terms together we get = x2 – x – 11 xy + 11 y

Taking the similar terms common we get = x (x – 1) – 11y (x – 1)

Therefore, as (x – 1) is common = (x – 1) (x – 11y)

Question 23. Factorize: ab – a – b + 1

Solution:

Given: ab – a – b + 1

Taking the similar terms common we get = a (b – 1) – 1 (b – 1)

Therefore, as (b – 1) is common = (b – 1) (a – 1)

Question 24. Factorize: x2 + y – xy – x

Solution:

Given: x2 + y – xy – x

Grouping similar terms together we get = x2 – x – xy + y

Taking the similar terms common we get = x (x – 1) – y (x – 1)

Therefore, as (x – 1) is common = (x – 1) (x – y)

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