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
We will now do the same for the division of algebraic expressions.
For example,
(i)
Therefore,
(ii) 5x(x + 4) =
Therefore,
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:
Like earlier we will write each of the involved terms in irreducible factor forms:
2x = 2 × x
Now we take a look at what are the common factors amongst them which cancel each other:
Now,
- Between the numerator and denominator, cancel out the common factors
- We get:
- We cancel out
2 andx - Which gives the above answer
Thus, the answer for the division,
Example 13: Do the following divisions.
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:
We have:
We observe that
None the less we try to take 2y out as a common term. So,
= 2y(
On dividing by 2y we get:
=
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.
=
=
=
which gives us the same answer.
Try These
Using the same method try to solve the following:
(i)
- 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)
- Between the numerator and denominator, cancel out the common factors
- We get:
- Cancelling out the common factors
- Which gives us the above answer
Example 14:
Method 1: Taking out the common factor from numerator and denominator
- We have the following
- Taking out the common factors
- We get:
- Which gives us the above answer
Method 2: Dividing each term by denominator
- We have the following
- Individually dividing each term by the denominator
- We get:
- Which gives us the same answer as earlier.