A Cool Polynomial Equation | Math Olympiads

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Опубликовано:

16 окт 2024

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Комментарии : 41

@goldfing5898 Год назад

x^3 - (x - 1)^2 = 23 x^3 - (x^2 - 2x + 1) = 23 x^3 - x^2 + 2x - 1 = 23 x^3 - x^2 + 2x - 24 = 0 So we try the divisors of 24, which are plus or minus 1,2,3,4,6,8,12,24. It is easier to write x^3 + 2x = x^2 + 24 in order to find the positive integer solutions. We find x1 = 3, because 3^3 + 2*3 = 3^2 + 24 27 + 6 = 9 + 24 33 = 33 Then we divide the polynomials: (x^3 - 1x^2 +2x - 24):(x - 3) = x^2 + 2x + 8 - (x^3 - 3x^2) ----------------------------------- 2x^2 + 2x - 24 - (2x^2 - 6x) -------------------------- 8x - 24 - (8x - 24) --------------- 0 The remaining two solutions are complex: x^2 + 2x + 8 = 0 Subtract 7: x^2 + 2x + 1 = -7 (x + 1)^2 = -7 x + 1 = +- sqrt(-7) x_{2,3} = -1 +- sqrt(7)*i

@spiderjerusalem4009 Год назад

use horner's method if you want factorize efficiently

@pedrovargas2181 Год назад

GENIUS 💪

@rickdesper Год назад

It's only a cubic, and x=3 is obviously a root. Shouldn't be too hard. Why is 3 a root? Looking at the LHS, it's dominated by the x^3. (x-1)^2 is relatively small. So, what perfect cube is closest to 23? Obviously 27. So, try x=3. It works!

@quark67000 Год назад

I used the same method.Then x³-x²+2x-24=0 becomes (x-3)(x²+2x+8)=0 which is so easy (and quick) to resolve. I don't know the theory used in the 2 method used in this video (a link to an explanation can be useful, perhaps SyberMath can provide one?). It's usually a good thing to try to guess obvious solutions so we can reduce the degree of the polynomial.

@appybane8481 Год назад

that perfect cube must be bigger than 23 because (x-1)^2>=0

@n_eros Год назад

I'm glad I was able to solve this with the smart (second) method myself. By the way, I don't count mistakes in arithmetic as mistakes. If there is logic issue or some case missed, then it is mistake.

@harlanpygman1167 Год назад

Same answer but a different technique using the expected format (x+a)(x+b)(x-3)=0 knowing that ab=8 since -3ab = -24. Then solving for a & b by expanding the expected form above. Expansion yields the coefficient for the x^2 term: -3+a+b. Since we know by expanding the original equation x^3-x^2+2x-24=0, the coefficient is -1. Hence -3+a+b=-1 so a+b=2. Solving for a and substituting into ab=8 yields a quadratic in b. The quadratic formula gives us b=1+/-sqrt(7)i . Using the coefficient of the x term yields the same result for verification. Solving for a is simple and yields the complex conjugate. When these values for a and b are inserted in the expanded equation above we easily see our other two roots.

@박성준-z9l 2 месяца назад

My Solution: x^3 - (x-1)^2 = 23 --> x^3 - 3^3 - ((x-1)^2 - 4) = 0 --> (x-3)(x^2+3x+9) - (x-1-2)(x-1+2) = 0 --> (x-3)(x^2+3x+9-x-1) = (x-3)(x^2+2x+8) = 0 --> (Apply quadratic formula): x = {3, -1-sqrt(7)i, -1+sqrt(7)i}

@DonRedmond-jk6hj 6 месяцев назад

You could use Descartes Rule of Sign to see that there must be a positive root. This narrows what to check somewhat.

@HoSza1 Год назад

3rd method: check for solutions mod small numbers: mod 2 doesn't help but mod 3 shows any solutions need to be divisible by 3. So 3 turns out to be a solution when testing, etc. etc. etc.

@DonRedmond-jk6hj 2 месяца назад

Using Descartes' Rule of Signs you can limited the number of integer roots the rrt has you check, You find 3 and see rhere are no larger positive roots. and no negative roots.

@mathswan1607 Год назад

x^3-x^2+2x-24=0 (x-3)(x^2+2x+8)=0 x=3 or x^2+2x+8=0(rejected as it has no real roots) Hence answer: x=3

@Ssilki_V_Profile Год назад

Third power equetion, with 1 root, that you can just guess and check - simple enough

@kobalt4083 Год назад

It doesn't have one root. There are three roots, you should have watched the full video(two imaginary solutions and one real number solution)... Yes, of course you can guess and check the integer solutions, but not all of them.

@Ssilki_V_Profile Год назад

@@kobalt4083 of course, it have 3 roots, but because 1 root you can gues - you can just extract it as a common factor, and solve quadratci equetion

@Ghaith7702 Год назад

i like the bloopers

@DonRedmond-jk6hj Год назад

You might also want to mention Descartes' Rule of Signs occasionally.

@jaii5955 Год назад

So easy question

@yeganaazadaliyeva5660 Год назад

i love this videos

@SyberMath Год назад

Glad to hear that!

@scottleung9587 Год назад

Got 'em all!

@kimba381 Год назад

3 is an obvious solution. divide by x-3, solve the quadratic.

@s1ng23m4n Год назад

[27] - [4] = 23 x = (3) (3) - 1 = 2 x^3 = [27] (x(=3) - 1)^2 = [4] first solution found by guess)) Rest we have a quadratic and it easy to solve

@_lonelywolf Год назад

@pas6295 2 месяца назад

Xis √24.

@MathematikTricks Год назад

I could solve this problem in less than a minute 🤩😍

@molostone8946 Год назад

I solved it by simply guessing and way faster

@havrulenkoevgeniy158 Год назад

❤❤❤

@NickC1966 Год назад

all that work just to get 3.

@barakathaider6333 Год назад

👍

@mrityunjaykumar4202 Год назад

total number of factors of 24 is 8 not 16

@mbmillermo Год назад

@@user-jj4dw6kv5u -- Thanks! I also was thrown by this. I'm usually thinking of integer factors that are neither 1 nor the number itself, so I was getting 6. But for this case -- polynomial roots -- you're absolutely right: We need negatives and we need 1 and 24.

@shay9365 Год назад

Can't we just get 3 by putting it in the equation, and by synthetic division, we can get the equation x2+2x+8, which gonna have imaginary roots, so 3 is the only answer? (Noob alert)