Factorisation Of Algebraic Expression: Algebra

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 = 1×12

12 = 2×6

12 = 3×4

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 × 3 × 2 × 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 \(3xyz\)\(- 16x^{2}\) and \(-yz\). Each term of this algebraic expression can be expressed in the form of its factors as:

\(3xyz = 3.x.y.z\), \(- 16x^{2} = -1.2.2.2.2.x.x\)  and \(-yz = -1 . y . z\).

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-  \(-3y^{2} + 18y\)

Solution- The algebraic expression can be re-written as

\(-3y^{2} + 18y = -3.y.y + 3.6.y\)

\(\Rightarrow -3y^{2} + 18y = -3.y(y + 6)\)

Factorization using common factors

Consider the algebraic expression -3y( y – 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 – 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 ‘12’ similarly second and third term has n as common factor. So the terms can be regrouped as:

⇒12a + n – na – 12= 12a – 12 + n – an

⇒12a – 12 – an + n = 12(a -1) –n(a -1)

After regrouping it can be seen that (a-1) is a common factor in each term,

⇒12a + n -na – 12=(a-1) (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:

\((a+b)^{2} = a^{2}+b^{2}+2ab\)

\((a-b)^{2} = a^{2}+b^{2}-2ab\)

\(a^{2}-b^{2} = (a+b) (a-b)\)

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 \((a+b){2} = a^{2}+b^{2}+2ab\) where a = 3x, b = 2m.

\(9x^{2} + 4m^{2} + 12mx = (3x)^{2} + (2m)^{2} +2. 3x . 2m\)

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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Practise This Question

10lm is one of the factors of
20l2m+30alm.
The other factor is: