Divide the given polynomial by the given monomial. 

(i) (5x ² – 6x) ÷ 3x 

(ii) (3y ⁸ – 4y ⁶ + 5y ⁴ ) ÷ y ⁴ 

(iii) 8(x 3y ² z ² + x ²y ³ z ² + x ²y ² z ³ ) ÷ 4x 2y ² z ² 

(iv) (x ³ + 2x ² +3x) ÷ 2x 

(v) (p ³q ⁶ – p ⁶q ³ ) ÷ p ³q ³

We know that, Factorization means finding out the factors of the given algebraic expression

(i) (5x ² – 6x) ÷ 3x :

let (5x ² – 6x)  can be written as,

 (5x ² – 6x)  = x( 5x $-$ 6)

Thus, we get 

(5x ² – 6x) ÷ 3x =  x( 5x $-$ 6) /  3x

= (5x $-$ 6) /  3

(ii) (3y ⁸ – 4y ⁶ + 5y ⁴ ) ÷ y ⁴ 

let  (3y ⁸ – 4y ⁶ + 5y ⁴ ) can be written as,

 (3y ⁸ – 4y ⁶ + 5y ⁴ ) = y⁴( 3y ⁴ – 4y ² + 5)

Thus, we get 

 (3y ⁸ – 4y ⁶ + 5y ⁴ ) ÷ y ⁴ = y⁴( 3y ⁴ – 4y ² + 5) / y⁴

= 3y ⁴ – 4y ² + 5

(iii) 8(x 3y ² z ² + x ²y ³ z ² + x ²y ² z ³ ) ÷ 4x 2y ² z ² :

let 8(x 3y ² z ² + x ²y ³ z ² + x ²y ² z ³ ) can be written as,

8(x 3y ² z ² + x ²y ³ z ² + x ²y ² z ³ )= 8 x ²y²z ²( x $+$ y $+$ z)

Thus, we get 

 8(x 3y ² z ² + x ²y ³ z ² + x ²y ² z ³ ) ÷ 4x 2y ² z ²  =8 x ²y²z ²( x $+$ y $+$ z) /  4x 2y ² z ²

= 2 ( x $+$ y $+$ z)

(iv) (x ³ + 2x ² +3x) ÷ 2x :

let  (x ³ + 2x ² +3x ) can be written as,

 (x ³ + 2x ² +3x ) = x ( x ² $+$ 2x $+$ 3)

Thus, we get 

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

=  ( x ² $+$ 2x $+$ 3) / 2

(v) (p ³q ⁶ – p ⁶q ³ ) ÷ p ³q ³:

let  (p ³q ⁶ – p ⁶q ³ ) can be written as,

 (p ³q ⁶ – p ⁶q ³ ) = p ³q ³ ( p ³$-$q ³)

Thus, we get 

(p ³q ⁶ – p ⁶q ³ ) ÷ p ³q ³ =  p ³q ³ ( q ³$-$p ³) / p ³q ³ 

=  q ³$-$p ³