How to get out of this vicious circle?

By Vieta's formula, you can see that these are the coefficients of a polynomial in form of kx^3 + mx^2 + nx + r = 0.

Then, if we set a, b, and c as the three zeros of this polynomial, then r/k = -abc = -60, n/k = ab + bc + ac = 47, m/k = -(a + b + c) = -12. For the simplicity's sake, we can let k = 1.

Then, (x-a)(x-b)(x-c) = x^3 - 12x^2 + 47x - 60 = 0. Now, by rational roots theorem, we know there are 24 possible roots.

We can now use synthetic division to find a solution for this equation. You'll find that 3 is an easy solution.

Then, you get (x-3)(x^2 - 9x + 20)=0 and from here on, you can solve the quadratics by factoring.

note: in case you don't know what Vieta's formula is, it states that for a polynomial ax^3 + bx^2 + cx + d = a(x-r1)(x-r2)(x-r3), b/a = -(r1 + r2 + r3), c/a = r1*r2 + r1*r3 + r2*r3, and d/a = -r1*r2*r3

Define each variable in terms of the others.

a = 12 - b - c

a = 60/bc

ab + ca = 47 - bc

a(b + c) = 47 - bc

a = (47 - bc)/(b + c)

b = 12 - a - c

b = 60/ac

ab + bc = 47 - ac

b(a + c) = 47 - ac

b = 47 - ac/(a + c)

c = 12 - a - b

c = 60/ab

c(b + c) = 47 - ab

c = (47 - ab)/(a + b)

(12 - b - c)(bc) = 60

12bc - b^2c - bc^2 = 60

Now write b in terms of c.

b = 47 - ac/(a + c)

b = (47 - (12 - b - c)) / (12 - b - c + c)

b = (35 + b + c) / (12 - b)

b = (35 + b + c) / 12 - b

12b - b^2 = 35 + b + c

-b^2 + 11b + 35 = c

Now you have c solved, so put it into one equation and solve for another variable (in this case b)

(12 - b - (-b^2 + 11b + 35))(b)(-b^2 + 11b + 35) = 60

(12 - b + b^2 - 11b - 35)(b)(-b^2 + 11b + 35) = 60

(b^2 - 12b - 23)(b)(-b^2 + 11b + 35) = 60

(b^3 - 12b^2 - 23b)(-b^2 + 11b + 35) = 60

(-b^5 + 11b^4 + 35b^3 + 12b^4 - 121b^3 - 420b^2 + 23b^3 - 251b^2 - 805b) = 60

(-b^5 + 23b^4 - 86b^3 - 671b^2 - 805b) = 60

-b^5 + 23b^4 - 86b^3 - 671b^2 - 805b - 60 = 0

Solve that, and then you'll get b (You can just graph it). Then you substitute it into 2 of the equations, and you'll be able to simplify it so it's just the a & c terms. Then you can solve it like a regular system of 2 equations.

I hope this information was very helpful.

A=3 b=4 c=5

Just find the numbers that each go into for 12,47&60

I had a few more then just 3 4 & 5 like 1 but, that wouldnt even make since to try.

I put my numbers into your equations and boom. Idk. I just get number, it hard for me to explain what went on in my head.

there is an obvious solution of (3,4,5), but i'm thinking there mat be multiple solution sets beyond the 6 permutations ....

345

354

435

453

534

543