# Factoring Polynomials by Grouping

There are some cases, when polynomial satisfies the follow conditions
1. There are four terms.
2. There is no common term (GCF) for all the terms.
In such situation, its possible to try to factor by the process of grouping.
In it, the main process is to factor the polynomial by grouping the terms into two groups and then factoring the common term (GCF) out of each group.

Lets consider an example, and go through the steps that has to be followed when factoring the polynomial by grouping.
$x^{3}-9x^{2}-2x+18$
Separate the polynomial terms into two groups, by grouping the first two terms together and then the last two terms together.
$(x^ {3}-9x^ {2}) +(-2x+18)$
Identify a common factor in each of the separate groups.

Lets take $(x^ {3}-9x^ {2})$
$x^ {3} =x*x*x$
$9x^ {2} =1*3*3*x*x$
So common factor = $x*x=x^ {2}$
Lets take $(-2x +18)$
$-2x = -1*2*x$
$18 =1*3*3*2$
So common factor = $1*2= 2$
So, $(x^{3}-9x^{2})+(-2x^{2}+18)= x^{2}(x – 9)+(2)(-x+9)$

To make a similar term in both groups, make suitable rearrangements
$x^{2}(x – 9)+(2)(-x+9) = x^{2}(x – 9) - (2)(x-9)$
Identify the common term in each group
In $x^{2}(x – 9)$ and $-2(x – 9)$, the common term is $(x-9)$
Factor it out using reverse of distribution property
$(x^{2} – 2)(x-9)$
So, $x^ {3}-9x^ {2} -2x +18 =(x-9) (x^ {2} – 2)$

There can arise again a set of polynomial that has no common factors but contains on three terms.
The general form of this type of polynomial is $ax^ {2} +bx+c$ where a, b and c are constant. This type of polynomial is called  quadratic polynomial.

Any polynomial of this type need not be factorable.
So first a condition has to be checked.
A new value called discriminant has to be found using the formula,
D = b2-4ac
If this value gives a perfect square number, then factorization is possible
.
Depending on the value of the first coefficient of $x^ {2}$, that is $a$, it can be broadly divided into two types
1. When $a=1$.
So the general form is $x^ {2}+bx+c$
Lets check the steps followed by using the following example
$x^{2}+12x+35$
• First set up a product of two parentheses which will contain a binomial each.
Here $( )*( )$
• Determine the factors that will be placed in the first position of each parenthesis.
Here to get $x^ {2}$, place x in each of the first positions.
So, $(x )*(x )$
• Determine the factors that will be placed in the last position of each parenthesis.For that, two numbers are required, say u and v, such that its product is $c$, that is $u v = c$ and its sum is $b$, that is $u + v = b$.
Here we need $uv=35$ and $u + v =12$
Now $35$ can be factored as $1*5*7$
It can be seen that $5 + 7 = 12$ and $5*7=35$
So the values that can be put in the last position is $5$ and $7$
So, $(x+5)*(x+7)$
Hence,
$x^ {2} +12x+35=(x+5)*(x+7)$

2. When $a\neq 1$, that is $a$ can be any number other than 1.

So the general form is $ax^ {2} +bx+c$
Lets check the steps followed by using the following example
$5x^ {2} +8x+3$
• First set up a product of two parentheses which will contain a binomial each.
Here $( )*( )$
• Determine the factors that will be placed in the first position of each parenthesis.
Here to get $5x^ {2}$, place 5x in first position of the first parenthesis and x in the first position of the second parenthesis.
So, $(5x )*(x )$
• Determine the factors that will be placed in the last position of each parenthesis.
In the original question last term is 3, so factoring it $3= 3*1$
So the values that can be put in the last position is $3$ and $1$
So, $(5x+3)*(x+1)$
Hence,
$5x^ {2} +8x+3=(5x+3)*(x+1)$