Factorization of Polynomials by Taking out the Greatest Common factors

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

Firstly, write each term as a product of prime factors.

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

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

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

Now, compare the expansions of the terms and identify the commonly appearing factors.

$(1).\,\,$ $2x^3$ $\,=\,$ $\boxed{2} \times \boxed{x} \times x \times x$

$(2).\,\,$ $4x^2y$ $\,=\,$ $\boxed{2} \times 2 \times \boxed{x} \times x \times y$

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

Now, let’s talk about our observations.

1. There is number $2$ commonly in each term and there is literal $x$ commonly in every term. So, they are considered as common factors.
2. There is literal $x$ commonly in first and second terms, but it is not in the third term. Similarly, there is literal $y$ in both second and third terms, but it is not in the first term. Therefore, they cannot be considered as common factors.

Now, use commutative property to write the common factors closely in the product of every term.

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

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

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

Finally, find the product of common factors and it is called greatest (or) highest common factor. Similarly, find the product of remaining factors in every term.

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

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

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

Now, substitute the equivalent value of every term in the given example algebraic expression.

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

Look at the expression on the right-hand side of the equation, the GCF or GCD or HCF is $2x$ in this factoring problem. Now, it can be taken out, from the terms by using distributive property.

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

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