Factorize $4x^2y^3$ $-$ $6x^3y^2$ $-$ $12xy^2$

In the given algebraic expression, $4x^2y^3,$ $6x^3y^2$ and $12xy^2$ are three terms. Now, let’s write the three terms as a product of prime factors.

$(1).\,\,$ $4x^2y^3$ $\,=\,$ $2$ $\times$ $2$ $\times$ $x$ $\times$ $x$ $\times$ $y$ $\times$ $y$ $\times$ $y$

$(2).\,\,$ $6x^3y^2$ $\,=\,$ $2$ $\times$ $3$ $\times$ $x$ $\times$ $x$ $\times$ $x$ $\times$ $y$ $\times$ $y$

$(3).\,\,$ $12xy^2$ $\,=\,$ $2$ $\times$ $2$ $\times$ $3$ $\times$ $x$ $\times$ $y$ $\times$ $y$

Let’s begin the process of finding the commonly appearing factors in three terms $4x^2y^3,$ $6x^3y^2$ and $12xy^2$ by comparing their prime factors.

$(1).\,\,$ $4x^2y^3$ $\,=\,$ $\underline{2}$ $\times$ $2$ $\times$ $\underline{x}$ $\times$ $x$ $\times$ $\underline{y}$ $\times$ $\underline{y}$ $\times$ $y$

$(2).\,\,$ $6x^3y^2$ $\,=\,$ $\underline{2}$ $\times$ $3$ $\times$ $\underline{x}$ $\times$ $x$ $\times$ $x$ $\times$ $\underline{y}$ $\times$ $\underline{y}$

$(3).\,\,$ $12xy^2$ $\,=\,$ $\underline{2}$ $\times$ $2$ $\times$ $3$ $\times$ $\underline{x}$ $\times$ $\underline{y}$ $\times$ $\underline{y}$

In the factor form of each term, $2, x, y$ and $y$ are commonly appearing. So, use the commutative property to write the common factors closely in three terms.

$(1).\,\,$ $4x^2y^3$ $\,=\,$ $\underline{2}$ $\times$ $\underline{x}$ $\times$ $\underline{y}$ $\times$ $\underline{y}$ $\times$ $2$ $\times$ $x$ $\times$ $y$

$(2).\,\,$ $6x^3y^2$ $\,=\,$ $\underline{2}$ $\times$ $\underline{x}$ $\times$ $\underline{y}$ $\times$ $\underline{y}$ $\times$ $3$ $\times$ $x$ $\times$ $x$

$(3).\,\,$ $12xy^2$ $\,=\,$ $\underline{2}$ $\times$ $\underline{x}$ $\times$ $\underline{y}$ $\times$ $\underline{y}$ $\times$ $2$ $\times$ $3$

Now, multiply the common factors $2, x, y$ and $y$ to find their greatest common factor (GCF) or highest common factor (HCF), and also multiply the remaining factors separately in all three terms.

$(1).\,\,$ $4x^2y^3$ $\,=\,$ $2xy^2$ $\times$ $2xy$

$(2).\,\,$ $6x^3y^2$ $\,=\,$ $2xy^2$ $\times$ $3x^2$

$(3).\,\,$ $12xy^2$ $\,=\,$ $2xy^2$ $\times$ $6$

The polynomial $4$ times $x$ square $y$ cube minus $6$ times $x$ cube $y$ square minus $12x$ times $y$ square can be written in factor form by substituting their equivalent values.

$\implies$ $4x^2y^3$ $-$ $6x^3y^2$ $-$ $12xy^2$ $\,=\,$ $2xy^2$ $\times$ $2xy$ $-$ $2xy^2$ $\times$ $3x^2$ $-$ $2xy^2$ $\times$ $6$

The greatest common factor (GCF) is $2xy^2$ in the above algebraic expression and it can be taken out from the three terms by the distributive property.

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

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