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) $