Math Doubts

Factoring the expressions by Taking out the common factors

A mathematical approach of taking the common factors out from the terms of an algebraic expression is called the factorization or factorisation of algebraic expressions by taking out the common factors.

Introduction

There are two or more common factors in the terms of algebraic expressions and they are taken common from the algebraic terms as per distributive property for factorising the algebraic expression. The method of factorization is called factoring algebraic expressions by taking out the common factors.

Required knowledge

There are three mathematical concepts, used in factoring the algebraic expressions.

  1. Exponents Rules
  2. Distributive property of multiplication across addition
  3. Distributive property of multiplication across subtraction

Example

$x^3-2x^2y+3xy^2-6y^3$ is an algebraic expression in terms of variables $x$ and $y$.

In the first two terms, $x^2$ is a common factor and it can be taken out common from them by the distributive property of multiplication over subtraction.

$= \,\,\,$ $x^{2+1}-2 \times x^2 \times y+3xy^2-6y^3$

$= \,\,\,$ $x^2 \times x^1-x^2 \times 2 \times y+3xy^2-6y^3$

$= \,\,\,$ $x^2 \times x-x^2 \times 2y+3xy^2-6y^3$

$= \,\,\,$ $x^2(x-2y)+3xy^2-6y^3$

$= \,\,\,$ $x^2(x-2y)+3 \times x \times y^2-3 \times 2 \times y^{2+1}$

$= \,\,\,$ $x^2(x-2y)+3 \times y^2 \times x-3 \times 2 \times y^2 \times y^1$

$= \,\,\,$ $x^2(x-2y)+3y^2 \times x-3 \times y^2 \times 2 \times y$

$= \,\,\,$ $x^2(x-2y)+3y^2 \times x-3y^2 \times 2y$

Now, $3y^2$ is a common factor in the third and fourth terms of the algebraic expression and it can be taken common from them by using distributive property of multiplication across subtraction.

$= \,\,\,$ $x^2(x-2y)+3y^2(x-2y)$

Now, the factor $x-2y$ is a common factor in both terms of the algebraic expression and it can be taken out from them by the distributive property of multiplication over addition.

$= \,\,\,$ $(x-2y)(x^2+3y^2)$

$\therefore \,\,\,\,\,\,$ $x^3-2x^2y+3xy^2-6y^3$ $\,=\,$ $(x-2y)(x^2+3y^2)$

The given algebraic expression $x^3-2x^2y+3xy^2-6y^3$ is factored as $(x-2y)(x^2+3y^2)$. Thus, the algebraic expressions are factored by taking out common factors from the terms of the algebraic expressions in this method.



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