A number or quantity that when multiplied with another number produces a given number or expression. For example, the factors of 12 are 1, 2, 3, 4, 6 and 12.
12 = 1Ã—12
12 = 2Ã—6
12 = 3Ã—4
Any number can be expressed in the form of its factors as explained shown above.
In terms of its prime factors 12 can be expressed as:
12 = 2 Ã— 3 Ã— 2 Ã— 1
Similarly an algebraic expression can also be expressed in the form of its factors. An algebraic expression consists of variables, constants and operators. An algebraic expression consists of terms separated by addition operation. Consider the following algebraic expression:
\(3xyz – 16x^{2} yz\)
This expression consists of 3 terms \(3xyz\)
\(3xyz = 3.x.y.z\)
Algebraic Expressions can be factorized using many methods. The most common methods used for factorization of algebraic expressions are:
 Factorization using common factors

Factorization by regrouping terms

Factorization using identities
Let us discuss these methods one by one in detail:

Factorization using common factors
In order to factorize an algebraic expression, the highest common factors of the terms of the given algebraic expression are determined and then we group the terms accordingly. In simple terms the reverse process of expansion of an algebraic expression is its factorization.
To understand this more clearly let us take an example.
Example Â \(3y^{2} + 18y\)
Solution The algebraic expression can be rewritten as
\(3y^{2} + 18y = 3.y.y + 3.6.y\)
\(\Rightarrow 3y^{2} + 18y = 3.y(y + 6)\)
Consider the algebraic expression 3y( y â€“ 6), if we expand this we will obtain 3y2 + 18y.

Factorization by regrouping terms
In some algebraic expressions it is not possible that every term has a common factor. For instance consider the algebraic expression 12a + n na â€“ 12. The terms of this expression do not have a particular factor in common but the first and last term have a common factor of â€˜12â€™ similarly second and third term has n as common factor. So the terms can be regrouped as:
â‡’12a + n – na â€“ 12= 12a â€“ 12 + n – an
â‡’12a â€“ 12Â â€“ an + n = 12(a 1) â€“n(a 1)
After regrouping it can be seen that (a1) is a common factor in each term,
â‡’12a + n na â€“ 12=(a1) (12 – n)
Thus by regrouping terms we can factorize algebraic expressions.

Factorizing Expressions using standard identities
An equality relation which holds true for all the values of variables in mathematics is known as an identity. Consider the following identities:
\((a+b)^{2} = a^{2}+b^{2}+2ab\)
\((ab)^{2} = a^{2}+b^{2}2ab\)
\(a^{2}b^{2} = (a+b) (ab)\)
On substituting any value of a and b, both sides of the given equations remain the same. Therefore, these equations are identities
Example: Factorize \(9x^{2} + 4m^{2} + 12mx\)
Solution: Observe the given expression carefully. This expression has three terms and all the terms are positive. Moreover, the first and the second term are perfect squares. The expression fits the form \((a+b){2} = a^{2}+b^{2}+2ab\)
\(9x^{2} + 4m^{2} + 12mx = (3x)^{2} + (2m)^{2} +2. 3x . 2m\)
Therefore, \(9x^{2} + 4m^{2} + 12mx = (3x+2m)^{2}\)
Thus, the required factorization of 9x2 + 4m2 + 12mx is (3x + 2m)2 by using standard identities.
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