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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
• 1997 AIME Problems/Problem 6
  Using [[Simon's Favorite Factoring Trick|SFFT]], {{AIME box|year=1997|num-b=5|num-a=7}}
  3 KB (497 words) - 00:39, 22 December 2018
• 1997 AIME Problems/Problem 3
  ...1000x+y=9xy \Longrightarrow 9xy-1000x-y=0</math>. Using [[Simon's Favorite Factoring Trick|SFFT]], this factorizes to <math>(9x-1)\left(y-\dfrac{1000}{9}\right) {{AIME box|year=1997|num-b=2|num-a=4}}
  2 KB (375 words) - 19:34, 4 August 2021
• 1998 AIME Problems/Problem 3
  Factoring the first one: (alternatively, it is also possible to [[completing the squa {{AIME box|year=1998|num-b=2|num-a=4}}
  1 KB (198 words) - 20:13, 23 February 2018
• 2000 AIME I Problems/Problem 9
  ...<math>\log x, \log y, \log z</math> respectively. Using [[Simon's Favorite Factoring Trick|SFFT]], the above equations become (*) Equating the first and second equations, solving, and factoring, we get <math>a(1-b) = c(1-b) \implies{a = c}</math>. Plugging this result
  4 KB (623 words) - 15:56, 8 May 2021
• 2000 AIME I Problems/Problem 7
  ...</math> and <math>\hspace{0.05cm}\frac{1}{z} = xy</math>. Substituting and factoring, we get <math>x(y+1) = 5</math>, <math>\hspace{0.15cm}y(z+1) = 29</math>, a {{AIME box|year=2000|n=I|num-b=6|num-a=8}}
  5 KB (781 words) - 15:02, 20 April 2024
• 2000 AIME I Problems/Problem 5
  Factoring <math>b_1 \cdot b_2 = 54</math> and saying WLOG that <math>b_1 < b_2 < 25< {{AIME box|year=2000|n=I|num-b=4|num-a=6}}
  7 KB (1,011 words) - 20:09, 4 January 2024
• 2003 AIME I Problems/Problem 8
  We then set <math>400+x^2=y^2</math>, where <math>y</math> is an integer. Factoring using difference of squares, we have ...<math>d</math>, and this must be the only possible answer since this is an AIME problem. We got very lucky in this sense :)
  5 KB (921 words) - 23:21, 22 January 2023
• 2001 AIME II Problems/Problem 6
  ...tion, <math>(b - a/2)^2 + (2b - a - a/2)^2 = a^2/2</math>. Simplifying and factoring, we get <math>2a^2-7ab+5b^2 = (2a-5b)(a-b) = 0.</math> Since <math>a = b</m {{AIME box|year=2001|n=II|num-b=5|num-a=7}}
  4 KB (772 words) - 19:31, 6 December 2023
• 2000 AIME II Problems/Problem 13
  Factoring a bit, we get {{AIME box|year=2000|n=II|num-b=12|num-a=14}}
  6 KB (1,060 words) - 17:36, 26 April 2024
• Mock AIME 1 2006-2007 Problems/Problem 9
  ...ual to <math>2^{10}\binom{10}{0}y^{10}(-1)^0</math> and by simplifying and factoring , we get <math>y^{10}(1024-1)=y^{10}(1023)</math>. Thus, <math>1023</math> *[[Mock AIME 1 2006-2007 Problems/Problem 8 | Previous Problem]]
  5 KB (744 words) - 19:46, 20 October 2020
• Mock AIME 2 2006-2007 Problems/Problem 3
  ...so <math>(n+6)^2 = m^2 + 2043</math>. Subtracting <math>m^2</math> and [[factoring]] the left-hand side, we get <math>(n + m + 6)(n - m + 6) = 2043</math>. < {{Mock AIME box|year=2006-2007|n=2|num-b=2|num-a=4}}
  1 KB (198 words) - 10:50, 4 April 2012
• Mock AIME 3 Pre 2005 Problems/Problem 15
  Factoring the radicand, we have {{Mock AIME box|year=Pre 2005|n=3|num-b=14|after=Last Question}}
  2 KB (312 words) - 10:38, 4 April 2012
• Sophie Germain Identity
  ...out the right side and verifying that it equals the left. To derive the [[factoring]], we begin by [[completing the square]] and then factor as a [[difference ...^4+324)(40^4+324)(52^4+324)}</math>. ([[1987 AIME Problems/Problem 14|1987 AIME, #14]])
  2 KB (222 words) - 15:04, 30 December 2023
• Algebraic manipulation
  ...te perfectly valid solutions. Thus, when possible, it's often better to [[factoring|factor]] than to divide both sides. ...anipulation to get all terms to one side to form a [[quadratic]]. After [[factoring]], we get <math>x = 5, -3</math>.
  4 KB (562 words) - 18:49, 8 November 2020
• Mock AIME 1 Pre 2005 Problems/Problem 5
  ...]{20-x}</math>. Then <math>a+b = 2</math> and <math>a^3 + b^3 = 20</math>. Factoring, <cmath>a^3 + b^3 = (a+b)((a+b)^2-3ab) = 2(4-3ab)= 8-6ab=20 \Longrightarrow {{Mock AIME box|year=Pre 2005|n=1|num-b=4|num-a=6|source=14769}}
  979 bytes (159 words) - 17:46, 21 March 2008
• 2008 AIME I Problems/Problem 13
  Plugging this into the first equation, and factoring, and cancelling <math>(x+1)</math>, and simplifying, we get <math>19x^2 -43 {{AIME box|year=2008|n=I|num-b=12|num-a=14}}
  8 KB (1,218 words) - 00:07, 11 April 2024
• 2008 AIME II Problems/Problem 1
  Factoring this expression yields {{AIME box|year=2008|n=II|before=First Question|num-a=2}}
  4 KB (575 words) - 16:41, 14 April 2024
• 2010 AIME I Problems/Problem 15
  ...s, corresponding to the sides of the triangle and thus degenerate cevians. Factoring those out, we get <math>0 = 4x(x - 15)(3x^2 - 40x + 132) = x(x - 15)(x - 6) {{AIME box|year=2010|num-b=14|after=Last Problem|n=I}}
  14 KB (2,210 words) - 13:14, 11 January 2024
• 2021 AIME II Problems/Problem 3
  ==Solution 5 (Factoring)== {{AIME box|year=2021|n=II|num-b=2|num-a=4}}
  7 KB (1,167 words) - 03:52, 11 March 2023
• 2021 AIME II Problems/Problem 6
  ...|A|+|B|-|A \cap B| = |A \cup B|</math>. Substituting into the equation and factoring, we get that <math>(|A| - |A \cap B|)(|B| - |A \cap B|) = 0</math>, so ther Rearranging and applying Simon's Favorite Factoring Trick give
  8 KB (1,364 words) - 01:02, 29 January 2024
• 2011 AIME I Problems/Problem 15
  Factoring the perfect square, we get: <math>ab-(b+a)^2=-2011</math> or <math>(b+a)^2- {{AIME box|year=2011|num-b=14|after=Last Problem|n=I}}
  8 KB (1,302 words) - 04:07, 24 July 2023
• 2011 AIME I Problems/Problem 7
  ...an integer is quite obviously the number of factors of <math>2010</math>. Factoring <math>2010</math>, we obtain <math>2010 = 2 \times 3 \times 5 \times 67</ma {{AIME box|year=2011|n=I|num-b=6|num-a=8}}
  10 KB (1,673 words) - 19:27, 8 April 2024
• 2011 AIME II Problems/Problem 9
  ...volves products <math>x_i x_j</math> with <math>i \equiv j \pmod 3</math>. Factoring out say <math>x_1</math> and <math>x_4</math> we see that the constraint is {{AIME box|year=2011|n=II|num-b=8|num-a=10}}
  3 KB (466 words) - 15:06, 16 January 2023
• 2012 AIME I Problems/Problem 10
  ...condition implies <math>x^2\equiv 256 \pmod{1000}</math>. Rearranging and factoring, <cmath>(x-16)(x+16)\equiv 0\pmod {1000}.</cmath> {{AIME box|year=2012|n=I|num-b=9|num-a=11}}
  8 KB (1,338 words) - 02:03, 25 November 2023
• 2013 AIME I Problems/Problem 14
  ...{\cos{\theta}(5-4\sin{\theta})}{10-4\sin^2{\theta}-3\sin{\theta}}.</cmath> Factoring the denominator, we have <cmath>10-4\sin^2{\theta}-3\sin{\theta}=(5-4\sin\t {{AIME box|year=2013|n=I|num-b=13|num-a=15}}
  10 KB (1,641 words) - 20:03, 3 January 2024
• 2014 AIME I Problems/Problem 9
  ...quadratic factor clearly does not have any integer solutions and since the AIME has only positive integer answers, we know that this must be the answer the ...imate the roots of the polynomial. (This strategy normally doesn't work on AIME #9, but playing around with a function is often good strategy for getting a
  10 KB (1,653 words) - 00:30, 27 January 2024
• 2015 AIME I Problems/Problem 3
  Rearrange this expression and factor the left side (this factoring can be done using <math>(a^3-b^3) = (a-b)(a^2+a b+b^2)</math> or synthetic Recognizing that AIME answers are <math>0</math> through <math>999</math>, the numbers whose cube
  6 KB (1,031 words) - 23:19, 23 January 2024
• 2015 AIME I Problems/Problem 11
  ...ath>. Simplifying, we have <math>y^2+y^2\frac{x^2-y^2}{(x+y)^2}=64</math>. Factoring out the <math>y^2</math>, we have <math>y^2\left(1+\frac{x^2-y^2}{(x+y)^2}\ [[File:2015 AIME I 11.png|270px|right]]
  5 KB (906 words) - 17:43, 27 September 2023
• 2015 AIME II Problems/Problem 14
  ...^2)}{(x - y)^2} = 13</math>. Now, we clean this up to obtain the following factoring of <math>0 = 2 \cdot (2x - 3y) \cdot (3x - 2y)</math>. This implies that <m {{AIME box|year=2015|n=II|num-b=13|num-a=15}}
  10 KB (1,751 words) - 22:21, 26 November 2023
• 2016 AIME II Problems/Problem 1
  ...have <math>a(k^2+k+1)=444</math>. Adding <math>k</math> to both sides and factoring, ...5)=ak^2-25-(ak-9)</math>. Simplifying, <math>k^2-2k+1=\frac{12}{a}</math>. Factoring,
  5 KB (788 words) - 02:50, 1 March 2024
• 2017 AIME I Problems/Problem 6
  We actually do not need to spend time factoring <math>x^2 - 120x + 3024</math>. Since the problem asks for <math>|x_1 - x_2 ...math> which using trivial algebra gives you <math>x^2-120x+3024</math> and factoring gives you <math>(x-84)(x-36)</math> and so your answer is <math>\boxed{048}
  4 KB (726 words) - 16:18, 5 January 2024
• 2017 AIME II Problems/Problem 6
  ==Solution 7 (alternative factoring)== {{AIME box|year=2017|n=II|num-b=5|num-a=7}}
  7 KB (1,096 words) - 21:03, 12 March 2021
• 2018 AIME I Problems/Problem 1
  The factoring condition is equivalent to the discriminant <math>a^2-4b</math> being equal {{AIME box|year=2018|n=I|before=First Problem|num-a=2}}
  10 KB (1,670 words) - 16:38, 15 January 2024
• 2018 AIME I Problems/Problem 5
  ...>x^2+xy-2y^2=0</cmath> upon simplification and division by <math>3</math>. Factoring <math>x^2+xy-2y^2=0</math> gives <cmath>(x+2y)(x-y)=0</cmath> Then, <cmath> {{AIME box|year=2018|n=I|num-b=4|num-a=6}}
  3 KB (543 words) - 12:52, 30 July 2023
• 2018 AIME I Problems/Problem 6
  ...f the roots of such a polynomial are distinct? One can proceed as follows. Factoring gives us that <math>(z^{840}+1)(z^{600}-1)=0,</math> so this implies that < {{AIME box|year=2018|n=I|num-b=5|num-a=7}}
  7 KB (1,211 words) - 00:23, 20 January 2024
• 2019 AIME II Problems/Problem 11
  ...frac{7 \cdot 189 \cdot 5}{8 \sqrt{5} \cdot 8\sqrt{5}}\right)}</cmath> Then factoring yields <cmath>\sqrt{\frac{21^2(8^2-15)}{(8\sqrt{5})^2}} =\frac{147}{8\sqrt{ [[File:AIME-II-2019-11.png|500px|right]]
  12 KB (1,985 words) - 19:52, 28 January 2024
• 2019 AIME I Problems/Problem 8
  Now factoring <math>m^5+n^5</math> as solution 4 yields <math>m^5+n^5=(m+n)(m^4-m^3n+m^2n {{AIME box|year=2019|n=I|num-b=7|num-a=9}}
  10 KB (1,878 words) - 13:19, 1 February 2024
• 2019 AIME I Problems/Problem 12
  ...c{(z^2-19z)-19(z^2-19z)-(z^2-19z)}{z^2-19z-z}</math> being pure imaginary. Factoring and simplifying, we find that this is simply equivalent to <math>(z-19)(z+1 {{AIME box|year=2019|n=I|num-b=11|num-a=13}}
  8 KB (1,534 words) - 22:17, 28 December 2023
• 2023 AIME I Problems/Problem 10
  Factoring our approximation gives <math>U \approx \frac {\frac{(n)(n+1)(2n+1 - 3a)}{6 {{AIME box|year=2023|n=I|num-b=9|num-a=11}}
  10 KB (1,578 words) - 09:48, 24 April 2024
• 2021 AIME I Problems/Problem 2
  ...ractions by multiplying both sides by <math>11^2=121,</math> then solve by factoring: {{AIME box|year=2021|n=I|num-b=1|num-a=3}}
  8 KB (1,294 words) - 00:59, 23 August 2022
• 2021 AIME I Problems/Problem 5
  ...h>d+k</math> and <math>d-k</math> are of the same parity. Now, we solve by factoring. We can find that <math> d=5 </math> and <math> d=7 </math> are the only po {{AIME box|year=2021|n=I|num-b=4|num-a=6}}
  7 KB (1,150 words) - 03:15, 1 February 2024
• 2021 AIME I Problems/Problem 10
  We cross-multiply, rearrange, and apply Simon's Favorite Factoring Trick to get <cmath>\left(k'+1\right)(t-18)=m-18.</cmath> {{AIME box|year=2021|n=I|num-b=9|num-a=11}}
  6 KB (1,015 words) - 16:30, 26 January 2024
• 2022 AIME I Problems/Problem 15
  ...math>, <math>y</math>, and <math>z</math> are positive real numbers. After factoring out accordingly from each terms one of <math>\sqrt{x}</math>, <math>\sqrt{y [[File:2022 AIME I 15.png|400px|right]]
  15 KB (2,208 words) - 01:25, 1 February 2024
• 2021 AIME II Problems/Problem 7
  ...t}=14.</cmath> Multiplying both sides by <math>3t,</math> rearranging, and factoring give <math>(t+18)(t-60)=0.</math> Substituting back and completing the squa ...ectively. Multiplying both sides of <math>(3)</math> by <math>d</math> and factoring some terms gives us <math>abcd + d^2(ab + ac + bc) = 14d</math>. Substituti
  10 KB (1,680 words) - 00:20, 28 April 2024
• 2023 AIME II Problems/Problem 4
  multiplying both sides by <math>(16-x^2)</math> and factoring, we get: we note that <math>x = 6</math> is a root. Factoring, we get the other roots, -10 and 11.
  10 KB (1,593 words) - 07:23, 9 May 2024
• 2024 AIME I Problems/Problem 1
  Factoring gives us {{AIME box|year=2024|n=I|before=First Problem|num-a=2}}
  2 KB (389 words) - 21:14, 2 February 2024
• Problems Collection
  ==AIME styled== .../math>. Adding them gives <math>2p+2q=pq+1</math> so by [[Simon's Favorite Factoring Trick]], <math>(p-2)(q-2)=3 \implies (p,q)=(3,5)</math> in some order. Henc
  43 KB (7,006 words) - 14:24, 19 February 2024