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Factorisation

    Division of Algebraic Expressions

    Division of Algebraic Expressions Continued ( Polynomial ÷ Polynomial)

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8th class > Factorisation > Division of Algebraic Expressions

Division of Algebraic Expressions

We have covered the addition, subtraction and multiplication of algebraic expressions. The only arithematic operation remaining is division of the expressions which we will be looking into in this section.

We know that division is the inverse operation of multiplication. Thus, for eg:

7 × 8 = 56 gives

568 = 7 (or) 567 = 8

We will now do the same for the division of algebraic expressions.

For example,

(i) 2x×3x2 = 6x^3

Therefore,

6x3÷2x = and also

6x3÷3x2 =

(ii) 5x(x + 4) = 5x2+20x

Therefore,

5x2+20x ÷ 5x = and also

5x2+20x ÷ (x + 4) =

We shall now look closely at how the division of one expression by another can be carried out. To begin with we shall consider the division of a monomial by another monomial.

Division of a monomial by another monomial

Take for example the division: 6x3÷2x

Like earlier we will write each of the involved terms in irreducible factor forms:

2x = 2 × x

6x3=2×3×x×x×x

Now we take a look at what are the common factors amongst them which cancel each other:

6x3 = 2 × x × (3 × x × x) = (2x) × (3x2)

Now,

6x3÷2x

  • Between the numerator and denominator, cancel out the common factors
  • We get:
  • We cancel out 2 and x
  • Which gives the above answer

Thus the answer for the division, 6x3 ÷ 2x is 3x2

Division of a polynomial by a monomial

Now, let's try doing the same method for longer and more complex equations. Consider the example consisiting of a trinomial divided by a monomial:

4y3+5y2+6y÷2y

We have:

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

We observe that is common amongst the first and the third term but not the second.

None the less we try to take 2y out as a common term. So,

4y3 + 5y2 + 6y = (2y) × (2 × y × y) + (2y) × (\frac{5}{2} × y) + (2y) × 3

= 2y(2y2+ \frac{5}{2} × y + 3 )

On dividing by 2y we get:

= 2y2 + \frac{5}{2} × y + 3

which is the correct result.

An alternate method: Another way of doing division involving multiple terms is to divide all the individual terms with the denominator.

= 4y3+5y2+6y2y

= 4y32y+5y22y+6y2y

= 2y2+52y+3

which gives us the same answer.

Using the same method try to solve the following:

(i) 24xy2z3 by 6yz2

24xy2z3÷6yz2

  • Between the numerator and denominator, cancel out the common factors
  • We get: keys="+ – × π ÷ brackets frac"
  • Cancelling out the common factors
  • Which gives us the above answer

(ii) 63a2b4c6 by 7a2b2c3

63a2b4c6÷7a2b2c3

  • Between the numerator and denominator, cancel out the common factors
  • We get:
  • Cancelling out the common factors
  • Which gives us the above answer

(iii) 24x2yz+xy2z+xyz2 by 8xyz

Method 1: Taking out the common factor from numerator and denominator

24x2yz+xy2z+xyz2÷8xyz

  • We have the following
  • Taking out the common factors
  • We get:
  • Which gives us the above answer

Method 2: Dividing each term by denominator

24x2yz+xy2z+xyz2÷8xyz

  • We have the following
  • Individually dividing each term by the denominator
  • We get:
  • Which gives us the same answer as earlier.