Class 8 RD Sharma Solutions - Chapter 7 Factorization - Exercise 7.4

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 + b²y² 

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)