Factoring a polynomial is just the opposite of multiplying polynomials. So when any polynomial is factored, what we are trying to do is to look for simpler polynomial, which can be multiplied together to get the polynomial we started.

When any polynomial is factored, it is mostly broken down into simpler polynomials which will contain coefficients and constant which are both integers. Each polynomial has its own way of factoring, lets check each of them as different cases.

Case:1- simplest form

The simplest form of factoring is to find a simple polynomial in each term of the given polynomial. It is the greatest common factor, also called the common factor. Hence, the common factor can be taken out using the reverse property of distribution law.

According to the distribution law, $a(b+c)= ab + ac$

So the reverse of distribution law is $ab +ac = a (b+c)$

For example:

$6y^{3}+2y$

Here first find the factor each term

$6y^{3} = (1*2*3*y*y*y)$

and

$2y=1*2*y$

So,

common factor = $(1*2*y) = 2y$

Hence by using the reverse of distribution law,

$(6y^{3}+2y) = (1*2*3*y*y*y + 1*2*y)$

= $(2y(3y^{2}+1))$

Case:2-difference between two square terms.

If any of the polynomial we use is in the form $x^{2}-y^{2}$, then it can be factored

using the formula $ x^{2}-y^{2}=(x-y)(x+y)$

For example,

Factor $4x^{2}-49$

Here

$4x^{2}= (2x)^{2}$

and $49=7^{2}$

so,

$4x^{2}-49= (2x)^{2}-7^{2}$

Now, this is in the difference of two square form

So $(2x)^{2}-7^{2}=(2x-7)(2x+7)$

Hence,

$4x^{2}-49=(2x-7)(2x+7)$

Case:3- quadratic trinomial form

A quadratic trinomial will have a general form $ax^{2}+bx+c=0$, where $a≠0$ and $a, b, c $

are constants. For this type of polynomial, the first and foremost thing is to check if its factorable.

Not all type of quadratic polynomial is factorable.

To do it, find the discriminant, $D = b^{2}-4ac$

If discriminant is a perfect square term, then this is factorable.

Lets take an example $6x^{2}+11x-35$

Here $a= 6$, $b=11$ and $c= -35$

First lets check if its factorable

$D = b^{2}-4ac= 11^{2}-4*6*(-35) $

= $121+840$

= $961$

Now $961$ is the square number, as $\sqrt{961}=31$

So this polynomial is factorable.

Now, to factor it, first set two parentheses, like this

$( )( )$

Each parenthesis will contain a binomial, with two terms each.

The first terms must be a multiple of $6x^{2}$, so we can take $2x$ as first term of first parenthesis and $3x$ as the first term of second parenthesis.

$(2x )(3x )$

Now we need two terms say $a$ and $b$ such that $ab= -35$ and $3a+2b=11$

Factoring $-35 = -1*3*7$

and $3*7+2*-5=11$

So the required term is -7 and 3

So $(2x+7)(3x -5)$

Hence,

$6x^ {2}+11x-35= (2x+7)(3x -5)$

When any polynomial is factored, it is mostly broken down into simpler polynomials which will contain coefficients and constant which are both integers. Each polynomial has its own way of factoring, lets check each of them as different cases.

Case:1- simplest form

The simplest form of factoring is to find a simple polynomial in each term of the given polynomial. It is the greatest common factor, also called the common factor. Hence, the common factor can be taken out using the reverse property of distribution law.

According to the distribution law, $a(b+c)= ab + ac$

So the reverse of distribution law is $ab +ac = a (b+c)$

For example:

$6y^{3}+2y$

Here first find the factor each term

$6y^{3} = (1*2*3*y*y*y)$

and

$2y=1*2*y$

So,

common factor = $(1*2*y) = 2y$

Hence by using the reverse of distribution law,

$(6y^{3}+2y) = (1*2*3*y*y*y + 1*2*y)$

= $(2y(3y^{2}+1))$

Case:2-difference between two square terms.

If any of the polynomial we use is in the form $x^{2}-y^{2}$, then it can be factored

using the formula $ x^{2}-y^{2}=(x-y)(x+y)$

For example,

Factor $4x^{2}-49$

Here

$4x^{2}= (2x)^{2}$

and $49=7^{2}$

so,

$4x^{2}-49= (2x)^{2}-7^{2}$

Now, this is in the difference of two square form

So $(2x)^{2}-7^{2}=(2x-7)(2x+7)$

Hence,

$4x^{2}-49=(2x-7)(2x+7)$

Case:3- quadratic trinomial form

A quadratic trinomial will have a general form $ax^{2}+bx+c=0$, where $a≠0$ and $a, b, c $

are constants. For this type of polynomial, the first and foremost thing is to check if its factorable.

Not all type of quadratic polynomial is factorable.

To do it, find the discriminant, $D = b^{2}-4ac$

If discriminant is a perfect square term, then this is factorable.

Lets take an example $6x^{2}+11x-35$

Here $a= 6$, $b=11$ and $c= -35$

First lets check if its factorable

$D = b^{2}-4ac= 11^{2}-4*6*(-35) $

= $121+840$

= $961$

Now $961$ is the square number, as $\sqrt{961}=31$

So this polynomial is factorable.

Now, to factor it, first set two parentheses, like this

$( )( )$

Each parenthesis will contain a binomial, with two terms each.

The first terms must be a multiple of $6x^{2}$, so we can take $2x$ as first term of first parenthesis and $3x$ as the first term of second parenthesis.

$(2x )(3x )$

Now we need two terms say $a$ and $b$ such that $ab= -35$ and $3a+2b=11$

Factoring $-35 = -1*3*7$

and $3*7+2*-5=11$

So the required term is -7 and 3

So $(2x+7)(3x -5)$

Hence,

$6x^ {2}+11x-35= (2x+7)(3x -5)$