**1.Carry out the following divisions.**

**(i) 28x ^{4 }÷ 56x**

**Solution:**

**(ii) –36y ^{3} ÷ 9y^{2}**

**Solution:**

**(iii) 66pq ^{2}r^{3} ÷ 11qr^{2}**

**Solution:**

**(iv) 34x ^{3}y^{3}z^{3} ÷ 51xy^{2}z^{3}**

**Solution:**

**(v) 12a ^{8}b^{8} ÷ (– 6a^{6}b^{4})**

**Solution:**

**2. Divide the given polynomial by the given monomial.**

**(i)(5x ^{2}–6x) ÷ 3x**

**Solution:**

**(ii)(3y ^{8}–4y^{6}+5y^{4}) ÷ y^{4}**

**Solution:**

**(iii) 8(x ^{3}y^{2}z^{2}+x^{2}y^{3}z^{2}+x^{2}y^{2}z^{3}) ÷ 4x^{2} y^{2} z^{2}**

**Solution:**

**(iv)(x ^{3}+2x^{2}+3x) ÷2x**

**Solution:**

**(v) (p ^{3}q^{6}–p^{6}q^{3}) ÷ p^{3}q^{3}**

**Solution:**

**3.Work out the following divisions.**

**(i)(10x–25) ÷ 5**

**Solution:**

**(ii) (10x–25) ÷ (2x–5)**

**Solution:**

**(iii) 10y(6y+21) ÷ 5(2y+7)**

**Solution:**

**(iv) 9x ^{2}y^{2}(3z–24) ÷ 27xy(z–8)**

**Solution:**

= 9x^{2}y^{2}×3(z-8)/27xy(z-8)

= xy

**(v) 96abc(3a–12)(5b–30) ÷ 144(a–4)(b–6)**

**Solution:**

**4. Divide as directed.**

**(i) 5(2x+1)(3x+5)÷ (2x+1)**

**Solution:**

**(ii) 26xy(x+5)(y–4)÷13x(y–4)**

**Solution:**

**(iii) 52pqr(p+q)(q+r)(r+p) ÷ 104pq(q+r)(r+p)**

**Solution:**

**(iv) 20(y+4) (y2+5y+3) ÷ 5(y+4)**

**Solution:**

**(v) x(x+1) (x+2)(x+3) ÷ x(x+1)**

**Solution:**

**5. Factorise the expressions and divide them as directed.**

**(i) (y ^{2}+7y+10) ÷ (y+5)**

**Solution:**

Firstly, we solve equation, (y^{2}+7y+10)

= y^{2}+2y+5y+10

= y(y+2)+5(y+2)

= (y+2)(y+5)

Now,

(y^{2}+7y+10) ÷ (y+5)

= (y+2) (y+5)/(y+5)

= y+2

**(ii) (m ^{2}–14m–32) ÷ (m+2)**

**Solution:**

Firstly, we solve equation: m^{2}–14m–32

m^{2}–14m–32

= m^{2}+2m – 16m–32

= m(m+2) – 16(m+2)

= (m–16) (m+2)

Now,

= (m–16) (m+2)/(m+2)

= m-16

**(iii) (5p ^{2}–25p+20) ÷ (p–1)**

**Solution:**

Firstly, we solve equation: (5p^{2}–25p+20)

Taking 5 common

= 5(p^{2}–5p+4)

Now, factorize equation:

= p^{2}–5p+4

= p^{2}–p – 4p+4

= (p–1) (p–4)

Now, we solve the original equation by substituting the value we get

(5p^{2}–25p+20) ÷ (p–1)

= 5(p–1) (p–4)/(p-1)

= 5(p–4)

**(iv) 4yz (z ^{2} + 6z–16) ÷ 2y(z+8)**

**Solution:**

First, we factorize equation: z^{2}+6z–16,

= z^{2}-2z+8z–16

= (z–2) (z+8)

Now, solve the original equation

= 4yz(z^{2}+6z–16) ÷ 2y(z+8)

= 4yz(z–2) (z+8)/2y(z+8)

= 2z(z-2)

**(v) 5pq(p ^{2}–q^{2}) ÷ 2p (p + q)**

**Solution:**

[By using identity (a^{2} – b^{2}) = (a + b) (a – b)]

We can write p^{2}–q^{2} as (p–q) (p+q)

Now,

= 5pq(p^{2}–q^{2}) ÷ 2p(p + q)

= 5pq(p–q) (p+q)/2p(p+q)

= 5q(p–q)/2

**(vi) 12xy(9x ^{2}–16y^{2}) ÷ 4xy(3x+4y)**

**Solution:**

Firstly, we factorize 9x^{2}–16y^{2}

[By using identity (a^{2} – b^{2}) = (a + b) (a – b)]

= (3x)^{2}–(4y)^{2}

= (3x+4y) (3x-4y)

Now,

= 12xy(9x^{2}–16y^{2}) ÷ 4xy(3x+4y)

= 12xy(3x+4y) (3x-4y) /4xy(3x+4y)

= 3(3x-4y)

**(vii) 39y3(50y2–98) ÷ 26y2(5y+7)**

**Solution:**

First, we solve equation 50y^{2}–98

= 2(25y^{2}–49)

= 2((5y)^{2}–72)

= 2(5y–7) (5y+7)

Now,

= 39y^{3}(50y^{2}–98) ÷ 26y2(5y+7)

**👍👍👍**