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Difference between revisions of "Common factorizations"

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These are ''common factorizations'' that are used all the time.  These should be memorized, but one should also know how they are derived.
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These are '''common factorizations'''.
  
 
==Basic Factorizations==
 
==Basic Factorizations==
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*<math>x^2-y^2=(x+y)(x-y)</math>
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<cmath>
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\begin{align*}
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*<math>x^3+y^3=(x+y)(x^2-xy+y^2)</math>
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\text{\textbullet}&&x^2-y^2&=(x+y)(x-y)\\
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\text{\textbullet}&&x^3+y^3&=(x+y)(x^2-xy+y^2)\\
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*<math>x^3-y^3=(x-y)(x^2+xy+y^2)</math>
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\text{\textbullet}&&x^3-y^3&=(x-y)(x^2+xy+y^2)\\
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\text{\textbullet}&&x^{2n+1}+y^{2n+1}&=(x+y)(x^{2n}-x^{2n-1}y+x^{2n-2}y^2-\ldots-xy^{2n-1}+y^{2n})\\
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\text{\textbullet}&&x^{n}-y^{n}&=(x-y)(x^{n-1}+x^{n-2}y+\cdots +xy^{n-2}+y^{n-1})
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\end{align*}
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</cmath>
  
 
== Vieta's/Newton Factorizations ==
 
== Vieta's/Newton Factorizations ==
  
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These factorizations are useful for problem 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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These factorizations are useful for problem 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, useful factorizations.
  
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*<math>(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ac)</math>
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<cmath>
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\begin{align*}
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\text{\textbullet}&&a^2+b^2+c^2+2(ab+bc+ac)&=(a+b+c)^2\\
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\text{\textbullet}&&(a+b+c)^3-(a^3+b^3+c^3)&=3(a+b)(b+c)(a+c)
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\end{align*}
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</cmath>
  
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*<math>(a+b+c)^3=a^3+b^3+c^3+3(a+b)(b+c)(a+c)</math>
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== Circulant Identities ==
  
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== Advanced Factorizations ==
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<cmath>
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*<math>a^2+b^2+c^2-ab-ac-bc=((a-b)^2+(b-c)^2+(c-a)^2)/2</math>
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\begin{align*}
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\text{\textbullet}&&a^2-b^2&=\det\begin{bmatrix}a&b\\b&a\end{bmatrix}=(a+b)(a-b)\\
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\text{\textbullet}&&a^3+b^3+c^3-3abc&=\det\begin{bmatrix}a&b&c\\c&a&b\\b&c&a\end{bmatrix}\\&&&=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)\\
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\text{\textbullet}&&a^4 - 4 a^2 b d - 2 a^2 c^2 + 4 a b^2 c + 4 a c d^2 \\\phantom{\text{\textbullet}}&&- b^4 + 2 b^2 d^2 - 4 b c^2 d + c^4 - d^4&=\det\begin{bmatrix}a&b&c&d\\d&a&b&c\\c&d&a&b\\b&c&d&a\end{bmatrix}\\&&&=(a+b+c+d)(a-b+c-d)((a-c)^2+(b-d)^2)
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\end{align*}
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</cmath>
  
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*<math>a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)</math>
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The matrices above are called [https://en.wikipedia.org/wiki/Circulant_matrix circulant matrices]. In general, the determinant of a circulant matrix will be a multiple of the sum of the entries in any of its rows/columns.
  
 
== Other Resources ==
 
== Other Resources ==
  
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* [http://tutorial.math.lamar.edu/pdf/Algebra_Cheat_Sheet_Reduced.pdf More Common Factorizations].
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* [http://tutorial.math.lamar.edu/pdf/Algebra_Cheat_Sheet_Reduced.pdf More Factorizations]  
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*[https://artofproblemsolving.com/wiki/index.php/Sum_and_difference_of_powers Sum and difference of powers]
  
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[[Category:Elementary algebra]]
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[[Category:Algebra]]

Latest revision as of 01:06, 28 April 2024

These are common factorizations.

Basic Factorizations

\begin{align*} \text{\textbullet}&&x^2-y^2&=(x+y)(x-y)\\ \text{\textbullet}&&x^3+y^3&=(x+y)(x^2-xy+y^2)\\ \text{\textbullet}&&x^3-y^3&=(x-y)(x^2+xy+y^2)\\ \text{\textbullet}&&x^{2n+1}+y^{2n+1}&=(x+y)(x^{2n}-x^{2n-1}y+x^{2n-2}y^2-\ldots-xy^{2n-1}+y^{2n})\\ \text{\textbullet}&&x^{n}-y^{n}&=(x-y)(x^{n-1}+x^{n-2}y+\cdots +xy^{n-2}+y^{n-1}) \end{align*}

Vieta's/Newton Factorizations

These factorizations are useful for problem 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, useful factorizations.

\begin{align*} \text{\textbullet}&&a^2+b^2+c^2+2(ab+bc+ac)&=(a+b+c)^2\\ \text{\textbullet}&&(a+b+c)^3-(a^3+b^3+c^3)&=3(a+b)(b+c)(a+c) \end{align*}

Circulant Identities

\begin{align*} \text{\textbullet}&&a^2-b^2&=\det\begin{bmatrix}a&b\\b&a\end{bmatrix}=(a+b)(a-b)\\ \text{\textbullet}&&a^3+b^3+c^3-3abc&=\det\begin{bmatrix}a&b&c\\c&a&b\\b&c&a\end{bmatrix}\\&&&=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)\\ \text{\textbullet}&&a^4 - 4 a^2 b d - 2 a^2 c^2 + 4 a b^2 c + 4 a c d^2 \\\phantom{\text{\textbullet}}&&- b^4 + 2 b^2 d^2 - 4 b c^2 d + c^4 - d^4&=\det\begin{bmatrix}a&b&c&d\\d&a&b&c\\c&d&a&b\\b&c&d&a\end{bmatrix}\\&&&=(a+b+c+d)(a-b+c-d)((a-c)^2+(b-d)^2) \end{align*}

The matrices above are called circulant matrices. In general, the determinant of a circulant matrix will be a multiple of the sum of the entries in any of its rows/columns.

Other Resources

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