14.1.1 Factors of natural numbers

You will remember what you learnt about factors in Class VI. Let us take a natural number,
say 30, and write it as a product of other natural numbers, say Thus, 1, 2, 3, 5, 6, 10, 15 and 30 are the factors of 30.
Of these, 2, 3 and 5 are the prime factors of 30 (Why?)
A number written as a product of prime factors is said to
be in the prime factor form; for example, 30 written as
2 × 3 × 5 is in the prime factor form.
The prime factor form of 70 is 2 × 5 × 7.
The prime factor form of 90 is 2 × 3 × 3 × 5, and so on.
Similarly, we can express algebraic expressions as products of their factors. This is
what we shall learn to do in this chapter.

14.1.2 Factors of algebraic expressions

We have seen in Class VII that in algebraic expressions, terms are formed as products of
factors. For example, in the algebraic expression 5xy + 3x the term 5xy has been formed
by the factors 5, x and y,

When we factorise an algebraic expression, we write it as a product of factors. These
factors may be numbers, algebraic variables or algebraic expressions.
Expressions like 3xy, 5x2 y , 2x (y + 2), 5 (y + 1) (x + 2) are already in factor form.
Their factors can be just read off from them, as we already know.
On the other hand consider expressions like 2x + 4, 3x + 3y, x2 + 5x, x2 + 5x + 6.
It is not obvious what their factors are. We need to develop systematic methods to factorise
these expressions, i.e., to find their factors. This is what we shall do now

14.2.1 Method of common factors We begin with a simple example: Factorise 2x + 4.
We shall write each term as a product of irreducible factors;

14.2.3 Factorisation using identities

The following solved examples illustrate how to use these identities for factorisation. What
we do is to observe the given expression. If it has a form that fits the right hand side of one
of the identities, then the expression corresponding to the left hand side of the identity
gives the desired factorisation.

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

xpression enclosed in a bracket by a constant (or a variable) outside, each term of the expression has to be multiplied by the constant (or the variable). Coefficient 1 of a term is usually not shown. But while adding like terms, we include it in the sum.