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Difference between revisions of "Factoring"
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Note to readers and editers: Please fix up this page by adding in material from [http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=factoring&t=75738 Joe's awesome factoring page.]
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'''Factoring''' is an essential part of any mathematical toolbox.  To factor, or to break an expression into factors, is to write the expression (often an [[integer]] or [[polynomial]]) as a product of different terms. This often allows one to find information about an expression that was not otherwise obvious.
  
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==Differences and Sums of Powers==
  
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===Why Factor===
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Using the formula for the sum of a [[geometric sequence]], it's easy to derive the general formula for difference of powers:
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Factoring equations is an essintial part of problem solving. Applying number theory to products yields many results.
 
  
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There are many ways to factor.
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<cmath>a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b + \cdots + ab^{n-2} + b^{n-1})</cmath>
  
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===Difference of Squares===
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If <math>n=2</math>, this creates the difference of squares factorization, <cmath>a^2-b^2=(a+b)(a-b)</cmath>
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<math>a^2-b^2=(a+b)(a-b)</math>
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===Difference of Cubes===
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This leads to the difference of cubes factorization, <cmath>a^3-b^3=(a-b)(a^2+ab+b^2)</cmath>
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<math>a^3-b^3=(a-b)(a^2+ab+b^2)</math>
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===Sum of Cubes===
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In addition, if <math>n</math> is [[odd integer | odd]]:
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<math>a^3+b^3=(a+b)(a^2-ab+b^2)</math>
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===Simon's Trick===
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<cmath>a^n+b^n=(a+b)(a^{n-1} - a^{n-2}b + a^{n-3}b^2 - a^{n-4}b^3 + \cdots + b^{n-1})</cmath>
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See [[Simon's Favorite Factoring Trick]]
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(This is not a recognized formula, please do not quote it on the USAMO or similar national proof contests)
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This also leads to the formula for the sum of cubes,
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===Summing Series===
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Also, it is helpful to know how to sum [[Arithmetic Series]] and [[Geometric Series]].
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<cmath>a^3+b^3=(a+b)(a^2-ab+b^2)</cmath>
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Another way to discover these factorizations is the following: the expression <math>a^n - b^n</math> is equal to zero if <math>a = b</math>.  If one factorizes a product which is equal to zero, one of the factors must be equal to zero, so <math>a^n - b^n</math> must have a factor of <math>a - b</math>.  Similarly, we note that the expression <math>a^n + b^n</math> when <math>n</math> is odd is equal to zero if <math>a = -b</math>, so it must have a factor of <math>a - (-b) = a + b</math>.  Note that when <math>n</math> is [[even integer | even]], <math>(-b)^n + b^n = 2b^n</math>, rather than 0, so this gives us no useful information.
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== Vieta's/Newton Factorizations ==
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These factorizations are useful for problems that could otherwise be solved by [[Newton sums]] or problems that give a polynomial and ask a question about the roots.  Combined with [[Vieta's formulas]], these are excellent factorizations that show up everywhere.
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*<math>(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)</math>
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*<math>(a+b+c)^3=a^3+b^3+c^3+3(a+b)(b+c)(c+a)</math>
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*<math>(a+b+c)^5=a^5+b^5+c^5+5(a+b)(b+c)(c+a)(a^2+b^2+c^2+ab+bc+ca) </math>
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*<math>(a+b+c)^7=a^7+b^7+c^7+7(a+b)(b+c)(c+a)((a^2+b^2+c^2+ab+bc+ca)^2+abc(a+b+c))</math>
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== Other Useful Factorizations ==
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*<math>a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)</math>
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* [[Binomial theorem]]
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* [[Simon's Favorite Factoring Trick]]
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* [[Sophie Germain Identity]]
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* [[Factor Theorem]]
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== Practice Problems ==
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* Prove that <math>n^2 + 3n + 5</math> is never divisible by 121 for any positive integer <math>{n}</math>.
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* Prove that <math>2222^{5555} + 5555^{2222}</math> is divisible by 7. - USSR Problem Book
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* Factor <math>(x-y)^3 + (y-z)^3 + (z-x)^3</math>.
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* Factor <math>x^4 + 1</math> into two polynomials with real coefficients.
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* Given that <math>a+b+c=0</math>, prove that <math>abc=\dfrac{a^3+b^3+c^3}{3}</math>.
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== Other Resources ==
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* [http://tutorial.math.lamar.edu/pdf/Algebra_Cheat_Sheet_Reduced.pdf More Common Factorizations].
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[[Category:Algebra]]
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[[Category:Definition]]

Latest revision as of 21:00, 13 January 2024

Factoring is an essential part of any mathematical toolbox. To factor, or to break an expression into factors, is to write the expression (often an integer or polynomial) as a product of different terms. This often allows one to find information about an expression that was not otherwise obvious.

Differences and Sums of Powers

Using the formula for the sum of a geometric sequence, it's easy to derive the general formula for difference of powers:

\[a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b + \cdots + ab^{n-2} + b^{n-1})\]

If $n=2$, this creates the difference of squares factorization, \[a^2-b^2=(a+b)(a-b)\]

This leads to the difference of cubes factorization, \[a^3-b^3=(a-b)(a^2+ab+b^2)\]

In addition, if $n$ is odd:

\[a^n+b^n=(a+b)(a^{n-1} - a^{n-2}b + a^{n-3}b^2 - a^{n-4}b^3 + \cdots + b^{n-1})\]

This also leads to the formula for the sum of cubes,

\[a^3+b^3=(a+b)(a^2-ab+b^2)\]

Another way to discover these factorizations is the following: the expression $a^n - b^n$ is equal to zero if $a = b$. If one factorizes a product which is equal to zero, one of the factors must be equal to zero, so $a^n - b^n$ must have a factor of $a - b$. Similarly, we note that the expression $a^n + b^n$ when $n$ is odd is equal to zero if $a = -b$, so it must have a factor of $a - (-b) = a + b$. Note that when $n$ is even, $(-b)^n + b^n = 2b^n$, rather than 0, so this gives us no useful information.

Vieta's/Newton Factorizations

These factorizations are useful for problems that could otherwise be solved by Newton sums or problems that give a polynomial and ask a question about the roots. Combined with Vieta's formulas, these are excellent factorizations that show up everywhere.

  • $(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)$
  • $(a+b+c)^3=a^3+b^3+c^3+3(a+b)(b+c)(c+a)$
  • $(a+b+c)^5=a^5+b^5+c^5+5(a+b)(b+c)(c+a)(a^2+b^2+c^2+ab+bc+ca)$
  • $(a+b+c)^7=a^7+b^7+c^7+7(a+b)(b+c)(c+a)((a^2+b^2+c^2+ab+bc+ca)^2+abc(a+b+c))$

Other Useful Factorizations

Practice Problems

  • Prove that $n^2 + 3n + 5$ is never divisible by 121 for any positive integer ${n}$.
  • Prove that $2222^{5555} + 5555^{2222}$ is divisible by 7. - USSR Problem Book
  • Factor $(x-y)^3 + (y-z)^3 + (z-x)^3$.
  • Factor $x^4 + 1$ into two polynomials with real coefficients.
  • Given that $a+b+c=0$, prove that $abc=\dfrac{a^3+b^3+c^3}{3}$.

Other Resources

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