Editing Factorization (section)

===General methods===
The following methods apply to any expression that is a sum, or that may be transformed into a sum. Therefore, they are most often applied to [[polynomial]]s, though they also may be applied when the terms of the sum are not [[monomial]]s, that is, the terms of the sum are a product of variables and constants.

====Common factor====
It may occur that all terms of a sum are products and that some factors are common to all terms. In this case, the [[Distributive property|distributive law]] allows factoring out this common factor. If there are several such common factors, it is preferable to divide out the greatest such common factor. Also, if there are integer coefficients, one may factor out the [[greatest common divisor]] of these coefficients.

For example,<ref>{{harvnb|Fite|1921|p=19}}</ref>
<math display="block">6 x^3 y^2 + 8 x^4 y^3 - 10 x^5 y^3 = 2 x^3 y^2(3 + 4xy - 5 x^2 y),</math>
since 2 is the greatest common divisor of 6, 8, and 10, and <math>x^3y^2</math> divides all terms.

====Grouping====
Grouping terms may allow using other methods for getting a factorization.

For example, to factor 
<math display="block">4x^2 + 20x + 3xy + 15y, </math>
one may remark that the first two terms have a common factor {{mvar|x}}, and the last two terms have the common factor {{mvar|y}}. Thus
<math display="block">4x^2+20x+3xy+15y = (4x^2+20x) + (3xy+15y) = 4x(x+5) + 3y(x+5). </math>
Then a simple inspection shows the common factor {{math|''x'' + 5}}, leading to the factorization 
<math display="block">4x^2+20x+3xy+15y = (4x+3y) (x+5).</math>

In general, this works for sums of 4 terms that have been obtained as the product of two [[binomial (polynomial)|binomials]]. Although not frequently, this may work also for more complicated examples.

====Adding and subtracting terms====
Sometimes, some term grouping reveals part of a [[#Recognizable patterns|recognizable pattern]]. It is then useful to add and subtract terms to complete the pattern.

A typical use of this is the [[completing the square]] method for getting the [[quadratic formula]].

Another example is the factorization of <math>x^4 + 1.</math> If one introduces the non-real [[square root of –1]], commonly denoted {{mvar|i}}, then one has a [[difference of squares]]
<math display="block">x^4+1=(x^2+i)(x^2-i).</math>
However, one may also want a factorization with [[real number]] coefficients. By adding and subtracting <math>2x^2,</math> and grouping three terms together, one may recognize the square of a [[binomial (polynomial)|binomial]]:
<math display="block">x^4+1 = (x^4+2x^2+1) - 2x^2 = (x^2+1)^2 - \left(x\sqrt2\right)^2 = \left(x^2+x\sqrt2+1\right) \left(x^2-x\sqrt2+1\right).</math>
Subtracting and adding <math>2x^2</math> also yields the factorization:
<math display="block">x^4+1 = (x^4-2x^2+1)+2x^2 = (x^2-1)^2 + \left(x\sqrt2\right)^2 = \left(x^2+x\sqrt{-2}-1\right) \left(x^2-x\sqrt{-2}-1\right).</math>
These factorizations work not only over the complex numbers, but also over any [[field (mathematics)|field]], where either –1, 2 or –2 is a square. In a [[finite field]], the product of two non-squares is a square; this implies that the [[polynomial]] <math>x^4 + 1,</math> which is [[irreducible polynomial|irreducible]] over the integers, is reducible [[modular arithmetic|modulo]] every [[prime number]]. For example,
<math display="block">x^4 + 1 \equiv (x+1)^4 \pmod 2;</math>
<math display="block">x^4 + 1 \equiv (x^2+x-1)(x^2-x-1) \pmod 3,</math>since <math>1^2 \equiv -2 \pmod 3;</math>
<math display="block">x^4 + 1 \equiv (x^2+2)(x^2-2) \pmod 5,</math>since <math>2^2 \equiv -1 \pmod 5;</math>
<math display="block">x^4 + 1 \equiv (x^2+3x+1)(x^2-3x+1) \pmod 7,</math>since <math>3^2 \equiv 2 \pmod 7.</math>