Divide as directed. 

(i) 5(2x + 1) (3x + 5) ÷ (2x + 1) 

(ii) 26xy(x + 5)(y – 4) ÷ 13x(y – 4) 

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

(iv) 20(y + 4) (y ² + 5y + 3) ÷ 5(y + 4) 

(v) x(x + 1) (x + 2) (x + 3) ÷ x(x + 1)

In this question we simply divide the given expression and delete the same part and then we get the correct solution

(i) 5(2x + 1) (3x + 5) ÷ (2x + 1) :

so here, (2x $+$1) is common in both 

Thus, we get

5(2x + 1) (3x + 5) ÷ (2x + 1) = 5(2x + 1) (3x + 5) / (2x + 1)

=5  (3x $+$ 5)

(ii) 26xy(x + 5)(y – 4) ÷ 13x(y – 4) :

so here, (y $-$4), 13x is common in both 

Thus, we get

26xy(x + 5)(y – 4) ÷ 13x(y – 4) = 26xy(x + 5)(y – 4) / 13x(y – 4)

= 2y (x $+$ 5)

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

so here,  52pq(q + r) (r + p)  is common in both 

Thus, we get

 52pqr (p + q) (q + r) (r + p) ÷ 104pq(q + r) (r + p) =  52pqr (p + q) (q + r) (r + p) / 104pq(q + r) (r + p)

= r (p $+$ q) /2

(iv) 20(y + 4) (y ² + 5y + 3) ÷ 5(y + 4) :

so here, 5 (y $+$ 4), is common in both 

Thus, we get

 20(y + 4) (y ² + 5y + 3) ÷ 5(y + 4)  = 20(y + 4) (y ² + 5y + 3) / 5(y + 4) 

= 4 (y ² $+$ 5y $+$ 3)

(v) x(x + 1) (x + 2) (x + 3) ÷ x(x + 1) :

so here, x (x $+$ 1), is common in both 

Thus, we get

 x(x + 1) (x + 2) (x + 3) ÷ x(x + 1) = x(x + 1) (x + 2) (x + 3) / x(x + 1) 

= ( x $+$ 2) ( x $+$ 3)