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• Completing the square
  Factoring the LHS gives ==Applications of Adding and Factoring==
  2 KB (422 words) - 16:20, 5 March 2023
• Simon's Favorite Factoring Trick
  ...ity/user/1233 Complex Zeta]) is often used in a Diophantine equation where factoring is needed. The most common form it appears is when there is a constant on o ...ivide the coefficient off of the equation.). According to Simon's Favorite Factoring Trick, this equation can be transformed into: <cmath>(x+k)(y+j)=a+jk</cmath
  7 KB (1,107 words) - 07:35, 26 March 2024
• Talk:Completing the square
  The AIME Problem looks like an example of Simon's Favorite Factoring trick, which is related to, but not really the same thing as completing the Hmm, then maybe the problem should you moved to that section? Or maybe the [[Factoring]] article even. -[[User:Eryaman|Eryaman]]
  1 KB (183 words) - 20:13, 17 June 2006
• Euclidean algorithm
  ...he [[nonnegative]] [[integer]]s <math>\mathbb{Z}{\geq 0}</math>, without [[factoring]] them. * [[1985 AIME Problems/Problem 13]]
  6 KB (924 words) - 21:50, 8 May 2022
• Prime factorization
  ...n these cases, a good strategy is to choose the number accordingly to make factoring easier. * [[2000 AIME I Problems/Problem 1]]
  3 KB (496 words) - 22:14, 5 January 2024
• Geometric sequence
  .../math>. Then <cmath>S = a_1 + a_1r + a_1r^2 + \cdots + a_1r^{n-1}.</cmath> Factoring out <math>a_1</math>, mulltiplying both sides by <math>(r-1)</math>, and us * [[2005_AIME_II_Problems/Problem_3 | 2005 AIME II Problem 3]]
  4 KB (644 words) - 12:55, 7 March 2022
• Newton's Sums
  ...polynomial]] raised to a power. They can also be used to derive several [[factoring]] [[identity|identities]]. ...rtofproblemsolving.com/wiki/index.php/2003_AIME_II_Problems/Problem_9 2003 AIME II Problem 9]
  4 KB (690 words) - 13:11, 20 February 2024
• 2006 AIME I Problems/Problem 5
  Since this is the AIME and you do not have a calculator solving <math> abc = \sqrt{52 \cdot 234 \c Factoring, we see that <math>52=13\cdot4</math>, <math>234=13\cdot18</math>, and <mat
  3 KB (439 words) - 18:24, 10 March 2015
• 2005 AIME II Problems/Problem 1
  We must find a [[factoring|factorization]] of the left-hand side of this equation into three consecuti {{AIME box|year=2005|n=II|before=First Question|num-a=2}}
  1 KB (239 words) - 11:54, 31 July 2023
• 2005 AIME II Problems/Problem 5
  ...\log b}=5 </math> Multiplying through by <math>\log a \log b </math> and factoring yields <math>(\log b - 3\log a)(\log b - 2\log a)=0 </math>. Therefore, <ma {{AIME box|year=2005|n=II|num-b=4|num-a=6}}
  3 KB (547 words) - 19:15, 4 April 2024
• 2004 AIME II Problems/Problem 7
  ...h> using supplementary and double angle identities. Multiplying though and factoring yields {{AIME box|year=2004|n=II|num-b=6|num-a=8}}
  9 KB (1,501 words) - 05:34, 30 October 2023
• 1984 AIME Problems/Problem 5
  ...\ln b}{\ln 8} + \frac{2 \ln a}{\ln 4} = 7</math>. Adding the equations and factoring, we get <math>(\frac{1}{\ln 8}+\frac{2}{\ln 4})(\ln a+ \ln b)=12</math>. Re {{AIME box|year=1984|num-b=4|num-a=6}}
  5 KB (782 words) - 14:49, 1 August 2023
• 1985 AIME Problems/Problem 13
  Factoring, we get <cmath>n(n-200)</cmath> {{AIME box|year=1985|num-b=12|num-a=14}}
  4 KB (671 words) - 20:04, 6 March 2024
• 1985 AIME Problems/Problem 6
  [[Image:AIME 1985 Problem 6.png]] Factoring this gives <math>\left(\frac ba-2\right)\left(10\cdot \frac ba - 8\right)=0
  5 KB (789 words) - 03:09, 23 January 2023
• 1987 AIME Problems/Problem 5
  ...math> term to the left side, it is factorable with [[SFFT|Simon's Favorite Factoring Trick]]: {{AIME box|year=1987|num-b=4|num-a=6}}
  1 KB (160 words) - 04:44, 21 January 2023
• 1988 AIME Problems/Problem 12
  ...es us <math>a'b'c'=3a'b'+3b'c'+3c'a'\to a'b'c'-3a'b'-3b'c'-3c'a'=0.</math> Factoring yields <math>(a'-3)(b'-3)(c'-3)=9(a'+b'+c')-27,</math> and the left hand si This problem is quite similar to the 1992 AIME problem 14, which also featured concurrent cevians and is trivialized by th
  4 KB (727 words) - 23:37, 7 March 2024
• 1988 AIME Problems/Problem 7
  ...Cross-multiplying and simplifying, we get <math>11h^2-70h-561 = 0</math>. Factoring, we get <math>(11h+51)(h-11) = 0</math>, so we take the positive positive s {{AIME box|year=1988|num-b=6|num-a=8}}
  1 KB (178 words) - 23:25, 20 November 2023
• 1990 AIME Problems/Problem 15
  Using factoring formulas, the terms can be grouped. First take the first three terms and su {{AIME box|year=1990|num-b=14|after=Last question}}
  4 KB (644 words) - 16:24, 28 May 2023
• 1990 AIME Problems/Problem 12
  [[Image:1990 AIME-12.png]] ...on the 12-gon. Now, <math>d=\sqrt{288-288 \cos \theta}</math>. Instead of factoring out <math>288</math> as in solution 2, factor out <math>\sqrt{576}</math> i
  6 KB (906 words) - 13:23, 5 September 2021
• 1996 AIME Problems/Problem 11
  ...e <math>z^2+z-1</math> and it reminds us of the sum of two cubes. Cleverly factoring, we obtain that.. {{AIME box|year=1996|num-b=10|num-a=12}}
  6 KB (1,022 words) - 20:23, 17 April 2021

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