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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 (1,076 words) - 14:13, 12 June 2024
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
User:Littlefox amc
Simon's Favorite Factoring Trick 1987 AIME Problem 5

263 bytes (34 words) - 17:13, 24 October 2022
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
User:Ddk001
[[2016 AIME I Problems/Problem 10]] Solution 3 [[2017 AIME I Problems/Problem 14]] Solution 2

42 KB (6,740 words) - 17:00, 2 June 2024

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