A nice Math Olympiad Problem || Algebra Simplification

[@ronaldnoll3247]

I once made a graphical solution using Excel and came up with the following solutions:
X(1) = -0.435 and x(2) = 0.768

[@boguslawszostak1784]

I do not like this.
I preffer this no trick solution.
Let's try to factor the polynomial with integer coefficients by using the method of expansion and comparing coefficients.
Since the constant term of the product is equal to -1, one of the constant terms of the factors is 1, and the other is -1. Without loss of generality, we can write it as follows:
6x^4 + x^3 - 2x - 1 = (1 + B1*x + C1*x^2) * (-1 + B2*x + C2*x^2)
Expanding and comparing coefficients, we get the following equations:
1. C1 * C2 = 6
2. B1 + B2 = 1
3. B1 * C2 + B2 * C1 = 1
4. C2 - C1 = -B1^2 + 2 * B1
To solve this system, we can express B2 from the second equation:
B2 = 1 - B1
Substitute B2 = 1 - B1 into equations 3 and 4:
Equation 3: B1 * C2 + (1 - B1) * C1 = 1
Equation 4: C2 - C1 = -B1^2 + 2 * B1
Now let's express B1 from Equation 3:
B1 = (1 + 2 * C1) / (C1 + C2)
Since C1 * C2 = 6, both coefficients must have the same sign. If they are negative, we can factor out (-1) from each factor, which gives (-1) * (-1) = 1, so we only need to consider factorizations where C1 and C2 are positive.
So we consider possible positive integer values for C1 and C2 that satisfy C1 * C2 = 6:
1. C1 = 1 and C2 = 6
2. C1 = 2 and C2 = 3
3. C1 = 3 and C2 = 2
4. C1 = 6 and C2 = 1
Let's calculate B1 for each pair:
1. For C1 = 1 and C2 = 6:
B1 = (1 + 2 * 1) / (1 + 6) = 3 / 7 (not an integer)
2. For C1 = 2 and C2 = 3:
B1 = (1 + 2 * 2) / (2 + 3) = 5 / 5 = 1 (integer) {Job has been done}
3. For C1 = 3 and C2 = 2:
B1 = (1 + 2 * 3) / (3 + 2) = 7 / 5 (not an integer)
4. For C1 = 6 and C2 = 1:
B1 = (1 + 2 * 6) / (6 + 1) = 13 / 7 (not an integer)
Conclusion:
The only values that give an integer for B1 are C1 = 2, C2 = 3, and B1 = 1.
Final Factorization:
6x^4 + x^3 - 2x - 1 = (1 + x + 2*x^2) * (-1 - x + 3*x^2)
From the determined factors, it is easy to calculate two real solutions and two complex solutions.
Real solutions:
x3 = (1 + sqrt(13)) / 6
x4 = (1 - sqrt(13)) / 6
Complex solutions:
x1 = (-1 + i*sqrt(7)) / 4
x2 = (-1 - i*sqrt(7)) / 4

[@superacademy247]

Thanks 👍👍💯😊 for your awesome 😎 approach. Very resourceful

[@ГогенЦоллерн]

u^4 + 2U^3 = u + 6
1 + 2x = x^3 + 6x^4 or
6x^4 + x^3 = 1 + 2x
Where the difference? Where is simplification?

[@ГогенЦоллерн]

Flipping is great DOLBOYOBISCHE