# Factoring Polynomials

A polynomial is defined as a finite sum(or difference) of terms in which the variables will have non negative exponent. The standard form of a polynomial in one variable is,
${a}_{n}x^{n}+{a}_{n-1}x^{n-1}+\cdot \cdot \cdot \cdot+ {a}_{1}x+{a}_{0}$,   $n$ = non negative integer and ${a}_{i}$, $i=0, 1, 2\cdot \cdot\cdot\cdot n$ are constant.
By definition, one number can be called the factor of polynomial, if its possible to divide polynomial without leaving any remainder. Factoring is the process by which a polynomial can be written as the product of its factors.

A polynomial can be defined as mathematical expression which involves a sum of powers in one or more variables which is multiplied by coefficients.

## Polynomial Factoring

Factoring is the process by which a polynomial can be written as the product of two or more simpler polynomial. To understand the process of factoring a polynomial, first we need to know how to find the common factor, which is also called as greatest common factor (GCF).

Lets us see with the help of examples,

## Solved Examples

Question 1: Find the GCF of $18x^{4}y^{3}$ and $9xy^{5}$

Solution:
The factors of $18$ = $1, 2, 3, 6, 9$ and $18$

The factors of $9$ = $1, 3$ and $9$

So the GCF = $9$ = $9$

Now, lets move to the variable part

Factors of $x^ {4} y^ {3}$ and $xy^ {5}$

$x^ {4} y^ {3}$ = x * x * x * x * y * y * y

$xy^ {5}$ = x * y * y * y  * y * y

So the GCF of the variables is $x^ {1} * y^ {3}$

So, the GCF of the two terms is $9 * x^ {1} * y^ {3} = 9 xy^ {3}$.

Question 2: Solve $2x^{3}y + 6x^{2}y^{3}$

Solution:
Given polynomial is a binomial, with first term $2x^{3}y$ and second term $6x^{2}y^{3}$

Factoring each term, $2x^ {3}y = 1 * 2 * x * x * x * y$

$6x^ {2}y^{3} = 1 * 2 * 3 * x * x * y * y * y$

The GCD = $1 * 2 * x * x * y$ = $2x^ {2} y$

So we write it outside $2x^ {2} y * (x + 3y^ {2})$

Hence,  $2x^{3}y + 6x^{2}y^{3} = 2x^ {2} y (x + 3y^ {2})$

Question 3: Find the factors of $14y^ {2} + 21y + 7$ expression.

Solution:
Factoring each term,
$14y^{2} = 1 * 2 * 7 * y * y$

$21y = 1 * 3 * 7 * y$

$7 = 1 * 7$

So, GCD = $7$

Dividing each term by $7$

$\frac{14y^{2}}{7} = 2y^{2}$

$\frac{21y}{7} = 3y$

$\frac{7}{7} = 1$

Hence, $14y^ {2} + 21y + 7 = 7(2y^{2} + 3y + 1)$