# 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 + r

^{2}Grouping similar terms together we get = p

^{2}q – pq – pr^{2 }+ r^{2}Taking the similar terms common we get = pq(p – 1)-r

^{2}(p – 1)Therefore, as (p – 1) is common = (p – 1) (pq – r

^{2})

**Question 3. Factorize: **1 + x + xy + x^{2}y

**Solution:**

Given: 1 + x + xy + x

^{2}yTaking 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: **xa^{2} + xb^{2 }– ya^{2} – yb^{2}

**Solution:**

Given: xa

^{2}+ xb^{2}– ya^{2}– yb^{2}Taking the similar terms common we get = x (a

^{2}+ b^{2}) – y (a^{2}+ b^{2})Therefore, as (a

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

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

**Solution:**

Given: x

^{2}+ xy + xz + yzTaking 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 +y^{2}

**Solution:**

Given: ab – by – ay + y

^{2}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: **lm^{2} – mn^{2} – lm + n^{2}

**Solution:**

Given: lm

^{2}– mn^{2}– lm + n^{2}Taking the similar terms common we get = m (lm – n

^{2}) – 1 (lm – n^{2})therefore, as (lm – n

^{2}) is common = (lm – n^{2}) (m – 1)

**Question 11. Factorize: **x^{3} – y^{2} + x – x^{2}y^{2}

**Solution:**

Given: x

^{3}– y^{2}+ x – x^{2}y^{2}Grouping similar terms together we get = x

^{3}+ x – x^{2}y^{2}– y^{2}Taking the similar terms common we get = x(x

^{2 }+ 1) – y^{2}(x^{2}+ 1)Therefore, as (x

^{2}+ 1) is common = (x^{2}+ 1) (x – y^{2})

**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: **x^{2} – 2ax – 2ab + bx

**Solution:**

Given: x

^{2}– 2ax – 2ab + bxGrouping similar term together we get = x

^{2}– 2ax + bx – 2abTaking 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: **x^{3} – 2x^{2}y + 3xy^{2} – 6y^{3}

**Solution:**

Given: x

^{3}– 2x^{2}y + 3xy^{2}– 6y^{3}Taking the similar terms common we get = x

^{2}(x – 2y) + 3y^{2}(x – 2y)Therefore, as (x – 2y) is common = (x – 2y) (x

^{2 }+ 3y^{2})

**Question 15. Factorize: **abx^{2} + (ay – b) x – y

**Solution:**

Given: abx

^{2}+ (ay – b) x – yAfter solving the bracket we get = abx

^{2}+ ayx – bx – yTaking 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}= a^{2}+ b^{2}+ 2ab)we get = a

^{2}x^{2}+ b^{2}y^{2}+ 2abxy + b^{2}x^{2 }+ a^{2}y^{2}– 2abxyGrouping similar terms together we get = a

^{2}x^{2}+ a^{2}y^{2}+ b^{2}x^{2}+ b^{2}y^{2 }Taking the similar terms common we get = x

^{2}(a^{2}+ b^{2}) + y^{2}(a^{2}+ b^{2})Therefore, as (a

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

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

**Solution:**

Given: ab (x

^{2}+ 1) + x(a^{2}+ b^{2})After solving the bracket we get = abx

^{2}+ ab + a^{2}x + b^{2}xGrouping similar terms together we get = abx

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

**Solution:**

Given: a

^{2}x^{2}+ (ax^{2}+ 1) x + aAfter solving the bracket we get = a

^{2}x^{2}+ ax^{3}+ x + aGrouping similar terms together we get = ax

^{3}+ a^{2}x^{2}+ x + aTaking the similar terms common we get = ax

^{2}(x + a) + 1 (x + a)Therefore, as (x + a) is common = (x + a) (ax

^{2}+ 1)

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

**Solution:**

Given: a(a – 2b – c) + 2bc

After solving the bracket we get = a

^{2 }–^{ }2ab – ac + 2bcTaking 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 = a

^{2}+ ab – ac – bcTaking 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: **x^{2} – 11xy – x + 11y

**Solution:**

Given: x

^{2}– 11xy – x + 11yGrouping similar terms together we get = x

^{2}– x – 11 xy + 11 yTaking 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: **x^{2} + y – xy – x

**Solution:**

Given: x

^{2}+ y – xy – xGrouping similar terms together we get = x

^{2}– x – xy + yTaking the similar terms common we get = x (x – 1) – y (x – 1)

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

## Please

Loginto comment...