This topic gives an overview of;
We have learnt how to add and subtract algebraic expressions. We also know how to multiply two expressions. We have not however, looked at division of one algebraic expression by another. This is what we wish to do in this section.
We recall that division is the inverse operation of multiplication. Thus, 7 × 8 = 56 gives 56 ÷ 8 = 7 or 56 ÷ 7 = 8.
We may similarly follow the division of algebraic expressions. For example,
2x * 3x2 = 6x3
Therefore, 6x3 ÷ 2x= 3x2 and also,6x3 ÷ 3x2 = 2x.
5x (x + 4) = 5x2 + 20x
Therefore,(5x2 + 20x) ÷ 5x = x + 4 and also(5x2 + 20x) ÷ (x + 4) = 5x.
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.
Consider 6x3 ÷ 2x
We may write 2x and 6x3 in irreducible factor forms,
2x = 2 * x
6x3 = 2 * 3 × x *x * x
Now we group factors of 6x3 to separate 2x,
6x3 = 2 *x * (3 * x * x) = (2x) * (3x2) Therefore, 6x3 ÷ 2x = 3x2
A shorter way to depict cancellation of common factors is as we do in division of numbers:
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)
Here, we expressed each term of the polynomial in factor form) we find that 2 × y is common in each term. Therefore, separating 2×y from each term. We get
Therefore, (4y3 + 5y2 + 6y) ÷ 2y
Alternatively, we could divide each term of the trinomial by the monomial using the cancellation method.
We shall factorise (7x2 + 14x) first to check and match factors with the denominator:
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