Maharashtra State BoardSSC (English Medium) 8th Standard
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Division of Algebraic Expressions - Dividing a Polynomial by a Monomial

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Dividing a Polynomial by a Monomial:

In the case of division of a polynomial by a monomial, we cannot proceed by dividing each term in the dividend polynomial by the divisor polynomial. Instead, we factorise both the polynomials and cancel their common factors.

Let us consider the division of the trinomial 4y3 + 5y2 + 6y by the monomial 2y. 

4y3 + 5y2 + 6y = (2 × 2 × y × y × y) + (5 × y × y) + (2 × 3 × y)

We find that 2 × y is common in each term. Therefore, separating 2 × y from each term. We get,

4y3 + 5y2 + 6y = `2 xx y xx (2 xx y xx y) + 2 xx y xx (5/2 xx y) + 2 xx y xx 3`.

`= 2y (2y^2) + 2y (5/2  y) + 2y(3)`

`= 2y (2y^2 + 5/2  y +3)`    ...(the common factor 2y is shown separately).

Therefore, 4y3 + 5y2 + 6y ÷ 2y

`= (4y^3 + 5y^2 + 6y)/(2y)= (2y(2y^2 + 5/2 y + 3))/(2y) = 2y^2 + 5/2 y + 3`

Alternatively, we could divide each term of the trinomial by the monomial using the cancellation method.

`(4y^3 + 5y^2 + 6y) ÷ 2y = (4y^3 + 5y^2 + 6y)/(2y)`

`= (4y^3)/(2y) + (5y^2)/(2y) + (6y)/(2y)`

` = 2y^2 + 5/2 y + 3`

Example

Divide 24(x2yz + xy2z + xyz2) by 8xyz using both the methods.

24(x2yz + xy2z + xyz2

`= 2 xx 2 xx 2 xx 3 xx[(x xx x xx y xx z) + (x xx y xx y xx z) + (x xx y xx z xx z)]`

`= 2 xx 2 xx 2 xx 3 xx x xx y xx z  xx ( x + y + z) = 8 xx 3 xx xyz xx (x + y + z)` .....(By taking out the common factor)

Therefore,

24(x2yz + xy2z + xyz2) ÷ 8xyz

`= (8 xx 3 xx xyz xx (x + y + z))/(8 xx xyz)`

= 3 × (x + y + z)

= 3(x + y + z)

Alternately,
24(x2yz + xy2z + xyz2) ÷ 8xyz
`= (24x^2yz)/(8xyz) + (24xy^2z)/(8xyz) + (24xyz^2)/(8xyz)`
= 3x + 3y + 3z
= 3(x + y + z)
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