Editing Factorization (section)

===Recognizable patterns===
Many [[identity (mathematics)|identities]] provide an equality between a sum and a product. The above methods may be used for letting the sum side of some identity appear in an expression, which may therefore be replaced by a product.

Below are identities whose left-hand sides are commonly used as patterns (this means that the variables {{mvar|E}} and {{mvar|F}} that appear in these identities may represent any subexpression of the expression that has to be factorized).<ref>{{harvnb|Selby|1970|p=101}}</ref> 
[[File:Difference_of_squares_and_cubes_visual_proof.svg|thumb|Visual proof of the differences between two squares and two cubes]]
*;[[Difference of two squares]]
::<math> E^2 - F^2 = (E+F)(E-F)</math>
:For example,
:::<math>\begin{align}
a^2 + &2ab + b^2 - x^2 +2xy - y^2 \\
&= (a^2 + 2ab + b^2) - (x^2 -2xy + y^2) \\
&= (a+b)^2 - (x -y)^2 \\
&= (a+b + x -y)(a+b -x + y).
\end{align} </math>
*;Sum/difference of two cubes
{{anchor|Sum/difference of two cubes}}
<!-- [[File:Differenceofcubes.jpg|thumb|A visual representation of the factorization of cubes using volumes. For a sum of cubes, simply substitute z=-y.]] -->
::<math> E^3 + F^3 = (E + F)(E^2 - EF + F^2)</math>
::<math> E^3 - F^3 = (E - F)(E^2 + EF + F^2)</math>
:*;[[Cauchy]] identity
:::<math> a^3 + b^3 + 3ab(a+b) = (a+b)^3 </math>
:::<math> a^3 - b^3 - 3ab(a-b) = (a-b)^3 </math>
*;Difference of two fourth powers
::<math>\begin{align}
E^4 - F^4 &= (E^2 + F^2)(E^2 - F^2) \\
&= (E^2 + F^2)(E + F)(E - F)
\end{align}</math>
*;Sum/difference of two {{mvar|n}}th powers
:In the following identities, the factors may often be further factorized:
:*;Difference, even exponent
::<math>E^{2n}-F^{2n}= (E^n+F^n)(E^n-F^n)</math>
:*;Difference, even or odd exponent
::<math> E^n - F^n  = (E-F)(E^{n-1} + E^{n-2}F + E^{n-3}F^2 + \cdots + EF^{n-2}  + F^{n-1} )</math>
::This is an example showing that the factors may be much larger than the sum that is factorized.
:*;Sum, odd exponent
::<math> E^n + F^n  = (E+F)(E^{n-1} - E^{n-2}F + E^{n-3}F^2 - \cdots - EF^{n-2}  + F^{n-1} )</math>
::(obtained by changing {{mvar|F}} by {{math|–''F''}} in the preceding formula)
:*;Sum, even exponent
::If the exponent is a power of two then the expression cannot, in general, be factorized without introducing [[complex numbers]] (if {{mvar|E}} and {{mvar|F}} contain complex numbers, this may be not the case). If ''n'' has an odd divisor, that is if {{math|1=''n'' = ''pq''}} with {{mvar|p}} odd, one may use the preceding formula (in "Sum, odd exponent") applied to <math>(E^q)^p+(F^q)^p.</math>

*;Trinomials and cubic formulas
:::<math>
\begin{align}
 &x^2 + y^2 + z^2 + 2(xy +yz+xz)= (x + y+ z)^2 \\
 &x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - xz - yz)\\
 &x^3 + y^3 + z^3 + 3x^2(y + z) +3y^2(x+z) + 3z^2(x+y) + 6xyz = (x + y+z)^3 \\
 &x^3 + y^3 + z^3 + 3(x + y)(y + z)(x + z) = (x + y + z)^3 \\
\end{align}
</math>
:*;[[Jean-Robert Argand | Argand]] identity
:::<math> x^4 + x^2y^2 + y^4 = (x^2 + xy +y^2)(x^2 - xy + y^2)</math>
:::<math> x^4 + x^2 + 1 = (x^2 + x + 1)(x^2 - x + 1)</math>

*;Binomial expansions
[[File:binomial_expansion_visualisation.svg|thumb|300px|Visualisation of binomial expansion up to the 4th power]]
:The [[binomial theorem]] supplies patterns that can easily be recognized from the integers that appear in them 
:In low degree:
::<math> a^2 + 2ab + b^2 = (a + b)^2</math>
::<math> a^2 - 2ab + b^2 = (a - b)^2</math>
::<math> a^3 + 3a^2b + 3ab^2 + b^3 = (a+b)^3 </math>
::<math> a^3 - 3a^2b + 3ab^2 - b^3 = (a-b)^3 </math>
:More generally, the coefficients of the expanded forms of <math>(a+b)^n</math> and <math>(a-b)^n</math> are the [[binomial coefficient]]s, that appear in the {{math|''n''}}th row of [[Pascal's triangle]].

====Roots of unity====
The {{mvar|n}}th [[roots of unity]] are the [[complex number]]s each of which is a [[zero of a function|root]] of the polynomial <math>x^n-1.</math> They are thus the numbers 
<math display="block">e^{2ik\pi/n} = \cos \tfrac{2\pi k}n + i\sin \tfrac{2\pi k} n </math>
for <math>k=0, \ldots, n-1.</math>

It follows that for any two expressions {{mvar|E}} and {{mvar|F}}, one has:
<math display="block">E^n-F^n = (E-F) \prod_{k=1}^{n-1} \left(E-F e^{2ik\pi/n}\right)</math>
<math display="block">E^n + F^n = \prod_{k=0}^{n-1} \left(E-F e^{(2k+1)i\pi/n}\right) \qquad \text{if } n \text{ is even}</math>
<math display="block">E^{n}+F^{n}=(E+F) \prod_{k=1}^{n-1}\left(E+F e^{2ik\pi/n}\right) \qquad \text{if } n \text{ is odd}</math>

If {{mvar|E}} and {{mvar|F}} are real expressions, and one wants real factors, one has to replace every pair of [[complex conjugate]] factors by its product. As the complex conjugate of <math>e^{i\alpha}</math> is <math>e^{-i\alpha},</math> and 
<math display="block">\left(a-be^{i\alpha}\right) \left(a-be^{-i\alpha}\right)=
a^2 - ab\left(e^{i\alpha}+e^{-i\alpha}\right) + b^2e^{i\alpha}e^{-i\alpha} =
a^2 - 2ab\cos\,\alpha + b^2,  </math>
one has the following real factorizations (one passes from one to the other by changing {{mvar|k}} into {{math|''n'' − ''k''}} or {{math|''n'' + 1 − ''k''}}, and applying the usual [[trigonometric formulas]]:
<math display="block">\begin{align}
E^{2n}-F^{2n}&= 
(E-F)(E+F)\prod_{k=1}^{n-1} \left(E^2-2EF \cos\,\tfrac{k\pi}n +F^2\right)\\
&=(E-F)(E+F)\prod_{k=1}^{n-1} \left(E^2+2EF \cos\,\tfrac{k\pi}n +F^2\right)
\end{align}</math>
<math display="block"> \begin{align}
E^{2n} + F^{2n} &= 
\prod_{k=1}^n \left(E^2 + 2EF\cos\,\tfrac{(2k-1)\pi}{2n}+F^2\right)\\
&=\prod_{k=1}^n \left(E^2 - 2EF\cos\,\tfrac{(2k-1)\pi}{2n}+F^2\right)
\end{align}</math>

The [[cosine]]s that appear in these factorizations are [[algebraic number]]s, and may be expressed in terms of [[nth root|radicals]] (this is possible because their [[Galois group]] is cyclic); however, these radical expressions are too complicated to be used, except for low values of {{mvar|n}}. For example,
<math display="block"> a^4 + b^4 = (a^2 - \sqrt 2 ab + b^2)(a^2 + \sqrt 2 ab + b^2).</math>
<math display="block"> a^5 - b^5 = (a - b) \left(a^2 + \frac{1-\sqrt 5}2 ab + b^2\right) \left(a^2 +\frac{1+\sqrt 5}2 ab + b^2\right),</math>
<math display="block"> a^5 + b^5 = (a + b) \left(a^2 - \frac{1-\sqrt 5}2 ab + b^2\right) \left(a^2 -\frac{1+\sqrt 5}2 ab + b^2\right),</math>

Often one wants a factorization with rational coefficients. Such a factorization involves [[cyclotomic polynomial]]s. To express rational factorizations of sums and differences or powers, we need a notation for the [[homogenization of a polynomial]]: if <math>P(x)=a_0x^n+a_ix^{n-1} +\cdots +a_n,</math> its ''homogenization'' is the [[bivariate polynomial]] <math>\overline P(x,y)=a_0x^n+a_ix^{n-1}y +\cdots +a_ny^n.</math> Then, one has
<math display="block">E^n-F^n=\prod_{k\mid n}\overline Q_n(E,F),</math>
<math display="block">E^n+F^n=\prod_{k\mid 2n,k\not\mid n}\overline Q_n(E,F),</math>
where the products are taken over all divisors of {{mvar|n}}, or all divisors of {{math|2''n''}} that do not divide {{mvar|n}}, and <math>Q_n(x)</math> is the {{mvar|n}}th cyclotomic polynomial.

For example, 
<math display="block">a^6-b^6= \overline Q_1(a,b)\overline Q_2(a,b)\overline Q_3(a,b)\overline Q_6(a,b)=(a-b)(a+b)(a^2-ab+b^2)(a^2+ab+b^2),</math>
<math display="block">a^6+b^6=\overline Q_4(a,b)\overline Q_{12}(a,b) = (a^2+b^2)(a^4-a^2b^2+b^4),</math>
since the divisors of 6 are 1, 2, 3, 6, and the divisors of 12 that do not divide 6 are 4 and 12.