# Difference between revisions of "Factoring"

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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. | |

+ | ==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: | |

− | |||

− | + | <cmath>a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b + \cdots + ab^{n-2} + b^{n-1})</cmath> | |

− | == | + | If <math>n=2</math>, this creates the difference of squares factorization, <cmath>a^2-b^2=(a+b)(a-b)</cmath> |

− | <math>a^2-b^2=(a+b)(a-b)</math> | + | |

− | === | + | This leads to the difference of cubes factorization, <cmath>a^3-b^3=(a-b)(a^2+ab+b^2)</cmath> |

− | + | ||

− | ( | + | In addition, if <math>n</math> is [[odd integer | odd]]: |

− | === | + | |

− | + | <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> | |

+ | |||

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

+ | |||

+ | <cmath>a^3+b^3=(a+b)(a^2-ab+b^2)</cmath> | ||

+ | |||

+ | 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. | ||

+ | |||

+ | == 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. | ||

+ | |||

+ | *<math>(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)</math> | ||

+ | |||

+ | *<math>(a+b+c)^3=a^3+b^3+c^3+3(a+b)(b+c)(c+a)</math> | ||

+ | |||

+ | *<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> | ||

+ | |||

+ | *<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> | ||

+ | |||

+ | == Other Useful Factorizations == | ||

+ | *<math>a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)</math> | ||

+ | * [[Binomial theorem]] | ||

+ | * [[Simon's Favorite Factoring Trick]] | ||

+ | * [[Sophie Germain Identity]] | ||

+ | |||

+ | == Practice Problems == | ||

+ | * Prove that <math>n^2 + 3n + 5</math> is never divisible by 121 for any positive integer <math>{n}</math>. | ||

+ | * Prove that <math>2222^{5555} + 5555^{2222}</math> is divisible by 7. - USSR Problem Book | ||

+ | * Factor <math>(x-y)^3 + (y-z)^3 + (z-x)^3</math>. | ||

+ | * Factor <math>x^4 + 1</math> into two polynomials with real coefficients. | ||

+ | * Given that <math>a+b+c=0</math>, prove that <math>abc=\dfrac{a^3+b^3+c^3}{3}</math>. | ||

+ | |||

+ | == Other Resources == | ||

+ | * [http://tutorial.math.lamar.edu/pdf/Algebra_Cheat_Sheet_Reduced.pdf More Common Factorizations]. | ||

+ | |||

+ | [[Category:Algebra]] | ||

+ | [[Category:Definition]] |

## Latest revision as of 13:34, 14 July 2021

**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.

## Contents

## 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:

If , this creates the difference of squares factorization,

This leads to the difference of cubes factorization,

In addition, if is odd:

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

Another way to discover these factorizations is the following: the expression is equal to zero if . If one factorizes a product which is equal to zero, one of the factors must be equal to zero, so must have a factor of . Similarly, we note that the expression when is odd is equal to zero if , so it must have a factor of . Note that when is even, , 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.

## Other Useful Factorizations

## Practice Problems

- Prove that is never divisible by 121 for any positive integer .
- Prove that is divisible by 7. - USSR Problem Book
- Factor .
- Factor into two polynomials with real coefficients.
- Given that , prove that .