Introduction
 Factoring a polynomial means expressing it as a product of its factors.
 For example, $p\left(x\right)={x}^{4}2{x}^{3}$ can be expressed as $p\left(x\right)={x}^{3}\left(x2\right)$.
 There are several methods of factoring polynomials. Some of them are described in the subsequent sections.
Factoring Polynomials by GCF
 Factoring polynomials using this method can be accomplished in two steps.

 First, finding the greatest common factor (GCF) of all the terms.
 Then, rewrite the polynomial by taking out the GCF.
 Let us understand this with the help of an example below.
Question: Factorize the polynomial $p\left(x\right)=3{x}^{2}6x$.
Solution: $3{x}^{2}=3\xb7x\xb7x$ and $6x=2\xb73\xb7x$
So, the GCF of $3{x}^{2}6x$ is $3x$.
Now, $3{x}^{2}6x=3x\left(x2\right)$
 In simple terms, this method is equivalent to the distributive property in reverse.
Factoring Polynomials by Grouping
 In this method, the given polynomial is grouped in pairs to find the factors.
 Let us understand this method through an example.
Question: Factorize the polynomial $p\left(x\right)={x}^{4}+3{x}^{3}+3x+9$.
Solution: $p\left(x\right)={x}^{4}+3{x}^{3}+3x+9$
$=\left({x}^{4}+3{x}^{3}\right)+\left(3x+9\right)$
$={x}^{3}\left(x+3\right)+3\left(x+3\right)$
$=\left({x}^{3}+3\right)\left(x+3\right)$
 Note: Sometimes, the terms need to be rearranged before grouping.
Factoring Polynomials by using Identities
 This method can be applied to a polynomial expression that resembles one of the algebraic identities or can be modified to match with algebraic identities.
 Some algebraic identities helpful in factoring polynomials are listed below.
 ${x}^{2}{y}^{2}=\left(x+y\right)\left(xy\right)$
 ${x}^{3}{y}^{3}=\left(xy\right)\left({x}^{2}+xy+{y}^{2}\right)$
 ${x}^{3}+{y}^{3}=\left(x+y\right)\left({x}^{2}xy+{y}^{2}\right)$
 ${x}^{2}+2xy+{y}^{2}={\left(x+y\right)}^{2}$
 ${x}^{2}2xy+{y}^{2}={\left(xy\right)}^{2}$
 Let us factorize ${x}^{2}25$.
${x}^{2}25$$={x}^{2}{\left(5\right)}^{2}$$=\left(x+5\right)\left(x5\right)$
Factoring Trinomials (Quadratic Polynomials)
 The process of factoring trinomials of the general form $p\left(x\right)={x}^{2}+bx+c$ is explained through an example below.
Question: Factorize the polynomial, $p\left(x\right)={x}^{2}+10x24$
Solution: Since the first term is ${x}^{2}$, we know that factoring will take the form.
${x}^{2}+10x24=\left(x+\overline{)}\right)\left(x+\overline{)}\right)$
Now, we need to find two appropriate numbers for the blank spots.
When multiplied, these two numbers must yield 24 as a result and produce the coefficient of the middle term when added. The list of all the possible options are:
$\left(x1\right)\left(x+24\right)={x}^{2}+23x24$ (Incorrect)
$\left(x+1\right)\left(x24\right)={x}^{2}23x24$ (Incorrect)
$\left(x+2\right)\left(x12\right)={x}^{2}10x24$ (Incorrect)
$\left(x2\right)\left(x+12\right)={x}^{2}+10x24$ (Correct)
$\left(x3\right)\left(x+8\right)={x}^{2}+5x24$ (Incorrect)
$\left(x+3\right)\left(x8\right)={x}^{2}5x24$ (Incorrect)
$\left(x4\right)\left(x+6\right)={x}^{2}+2x24$ (Incorrect)
$\left(x+4\right)\left(x6\right)={x}^{2}2x24$ (Incorrect)
Thus the correct solution is ${x}^{2}+10x24=\left(x2\right)\left(x+12\right)$.
 Likewise, the process of factoring trinomials of the general form $p\left(x\right)=a{x}^{2}+bx+c$ is explained through an example below.
Question: Factorize the polynomial, $p\left(x\right)=4{x}^{2}+4x3$.
Solution: The coefficient of ${x}^{2}$ term has more than one positive factor. Thus, the factoring can take one of the following forms.
$4{x}^{2}+4x3=\left(4x+\overline{)}\right)\left(x+\overline{)}\right)$
$4{x}^{2}+4x3=\left(2x+\overline{)}\right)\left(2x+\overline{)}\right)$
Now, we need to find two appropriate numbers for the blank spots.
When multiplied, these two numbers must yield 3 as a result and produce the coefficient of the middle term when added. The list of all the possible options are:
$\left(4x+1\right)\left(x3\right)=4{x}^{2}11x3$ (Incorrect)
$\left(4x1\right)\left(x+3\right)=4{x}^{2}+11x3$ (Incorrect)
$\left(2x+1\right)\left(2x3\right)=4{x}^{2}4x3$ (Incorrect)
$\left(2x1\right)\left(2x+3\right)=4{x}^{2}+4x3$ (Correct)
Thus the correct solution is $4{x}^{2}+4x3=\left(2x1\right)\left(2x+3\right)$.
Cheat Sheet
 The first approach for factoring polynomials should be to factor using the distributive property by taking the GCF out.
 Polynomials in which terms can be rearranged & grouped, factoring by grouping approach can be used to find factors.
 If the given polynomial can be modified to resemble any standard algebraic identities, factoring by using the identities method can be used.
 Factoring quadratic polynomials means factoring them into two linear polynomials.
Blunder Areas
 ${x}^{2}+{y}^{2}\ne {\left(x+y\right)}^{2}$
 By factoring a polynomial, its value remains unchanged; only its form changes.
 Abhishek Tiwari
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