The process of writing a given number or a polynomial expression as the product of two or more numbers or simple polynomial is called factorization. The numbers or polynomial expression that may be multiplied to obtain the given number or simple polynomials are called factors of the given number or a polynomial expression.

For example $3$ and $7$ are the factors of $21$, as $21= 3*7$

The simplest type of factoring is when there exists a common factor for every term in the polynomial taken.

By definition, if every term given in the polynomial comprises of several factors and if each of these factors has at least one factor that is same, then that factor is known as common factor. That common term, then can be factored out using the reverse law of distribution property.

According to the distributive law $a (b+c) = ab+ac$ and $(b+c) a = ba+ca$

So, its reverse law is $ab+ac = a (b+c) $ and $ba+ca = (b+c) a$

For example:

$5x^{3}+25x$

Here first find the factor each term

$5x^{3}= 1*5*x*x*x$

$25x = 1*5*5*x$

So, common factor = $1*5*x=5x$

Hence by

using the reverse of distribution law,

$5x^{3}+25x = 1*5*x*x*x + 1*5*5*x $

= $5x(x^{2}+5)$

An expression is said to be completely factorized, if none of the factors can be further factorized. In general, we use the word ‘factorization’ in the sense of completely factorizing a given expression.

The following identities can be used to work out certain polynomials

$a^{2}+2ab+b^{2}= (a+b)^{2}$

$a^{2}-2ab+b^{2}= (a-b)^{2}$

$a^{2}-b^{2}=(a-b)(a+b)$

Factorize each of the following expression

$p^{2}+\frac{q^{2}}{4}+1+pq+q+2p= p^{2}+(\frac{q}{2})^{2}+1^{2}+2*p*\frac{q}{2}+2*\frac{q}{2}*1+2*1*p$

Using the identity$(a+b+c)^{2}= a^{2}+b^{2}+c^{2}+2ab+2bc+2ac$

So,

$p^{2}+\frac{q^{2}}{4}+1+pq+q+2p=(p+\frac{q}{2}+1)^{2}$

=$(p+\frac{q}{2}+1)* (p+\frac{q}{2}+1)$

2. $2x^{2}+y^{2}+8z^{2}-2\sqrt{2}xy-4\sqrt{2}yz+8xz$

Solution:

$2x^{2}+y^{2}+8z^{2}-2\sqrt{2}xy-4\sqrt{2}yz+8xz=(\sqrt{2}x)^{2}+(-y)^{2}+(2\sqrt{2}z)^{2}-2(\sqrt{2}x)(-y)+2(-y)(2\sqrt{2}z)+2(2\sqrt{2}z(\sqrt{2}x)$

=$(\sqrt{2}x-y+2\sqrt{2}z)^{2}$

=$(\sqrt{2}x-y+2\sqrt{2}z)* (\sqrt{2}x-y+2\sqrt{2}z)$

Quadratic Trinomial

They are polynomial which are both quadratic and has three terms in it. Its general form is $ax^ {2} +bx+c$

Let check how this can be factored when the coefficient of $x^ {2} $ is $1$, that is $a=1$.

For example $x^{2}+7x+12$

To factor this we need two numbers, such that its sum is $7$ and its product is $12$.

To find this first lets see different ways of factoring positive $12$

$12 = 1*12$

= $-1*-12$

= $6*2$

= $-6*-2$

= $3*4$

= $-3*-4$

In this the only numbers such that its sum is $7$ is $3+4= 7$

So

$x^{2}+7x+12=(x+3)(x+4)$

For example $3$ and $7$ are the factors of $21$, as $21= 3*7$

The simplest type of factoring is when there exists a common factor for every term in the polynomial taken.

By definition, if every term given in the polynomial comprises of several factors and if each of these factors has at least one factor that is same, then that factor is known as common factor. That common term, then can be factored out using the reverse law of distribution property.

According to the distributive law $a (b+c) = ab+ac$ and $(b+c) a = ba+ca$

So, its reverse law is $ab+ac = a (b+c) $ and $ba+ca = (b+c) a$

For example:

$5x^{3}+25x$

Here first find the factor each term

$5x^{3}= 1*5*x*x*x$

$25x = 1*5*5*x$

So, common factor = $1*5*x=5x$

Hence by

using the reverse of distribution law,

$5x^{3}+25x = 1*5*x*x*x + 1*5*5*x $

= $5x(x^{2}+5)$

An expression is said to be completely factorized, if none of the factors can be further factorized. In general, we use the word ‘factorization’ in the sense of completely factorizing a given expression.

The following identities can be used to work out certain polynomials

$a^{2}+2ab+b^{2}= (a+b)^{2}$

$a^{2}-2ab+b^{2}= (a-b)^{2}$

$a^{2}-b^{2}=(a-b)(a+b)$

Factorize each of the following expression

- $p^{2}+\frac{q^{2}}{4}+1+pq+q+2p$

$p^{2}+\frac{q^{2}}{4}+1+pq+q+2p= p^{2}+(\frac{q}{2})^{2}+1^{2}+2*p*\frac{q}{2}+2*\frac{q}{2}*1+2*1*p$

Using the identity$(a+b+c)^{2}= a^{2}+b^{2}+c^{2}+2ab+2bc+2ac$

So,

$p^{2}+\frac{q^{2}}{4}+1+pq+q+2p=(p+\frac{q}{2}+1)^{2}$

=$(p+\frac{q}{2}+1)* (p+\frac{q}{2}+1)$

2. $2x^{2}+y^{2}+8z^{2}-2\sqrt{2}xy-4\sqrt{2}yz+8xz$

Solution:

$2x^{2}+y^{2}+8z^{2}-2\sqrt{2}xy-4\sqrt{2}yz+8xz=(\sqrt{2}x)^{2}+(-y)^{2}+(2\sqrt{2}z)^{2}-2(\sqrt{2}x)(-y)+2(-y)(2\sqrt{2}z)+2(2\sqrt{2}z(\sqrt{2}x)$

=$(\sqrt{2}x-y+2\sqrt{2}z)^{2}$

=$(\sqrt{2}x-y+2\sqrt{2}z)* (\sqrt{2}x-y+2\sqrt{2}z)$

Quadratic Trinomial

They are polynomial which are both quadratic and has three terms in it. Its general form is $ax^ {2} +bx+c$

Let check how this can be factored when the coefficient of $x^ {2} $ is $1$, that is $a=1$.

For example $x^{2}+7x+12$

To factor this we need two numbers, such that its sum is $7$ and its product is $12$.

To find this first lets see different ways of factoring positive $12$

$12 = 1*12$

= $-1*-12$

= $6*2$

= $-6*-2$

= $3*4$

= $-3*-4$

In this the only numbers such that its sum is $7$ is $3+4= 7$

So

$x^{2}+7x+12=(x+3)(x+4)$