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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
• 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
• 1996 AIME Problems/Problem 3
  Using [[Simon's Favorite Factoring Trick]], we rewrite as <math>[(x+7)(y-3)]^n = (x+7)^n(y-3)^n</math>. Both [ {{AIME box|year=1996|num-b=2|num-a=4}}
  3 KB (515 words) - 04:29, 27 November 2023

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