Question 1209750: In this multi-part problem, we will consider this system of simultaneous equations:
3x + 4y + 30z = -60,
2xy + 42xz - 16yz = 68,
5xyz = 56.
Let a = x/2, b = 5y and c = -4z.
Determine the monic cubic polynomial in terms of a variable t whose roots are t = a, t = b, and t = c.
Found 2 solutions by CPhill, ikleyn:
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! Here's how to determine the monic cubic polynomial:
1. **Substitute:**
Substitute *x* = 2*a*, *y* = *b*/5, and *z* = -*c*/4 into the given system of equations:
* 3(2*a*) + 4(*b*/5) + 30(-*c*/4) = -60
* 2(2*a*)(*b*/5) + 42(2*a*)(-*c*/4) - 16(*b*/5)(-*c*/4) = 68
* 5(2*a*)(*b*/5)(-*c*/4) = 56
2. **Simplify:**
Simplify the equations:
* 6*a* + (4/5)*b* - (15/2)*c* = -60
* (4/5)*a*b* - 21*a*c* + (4/5)*b*c* = 68
* -10*a*b*c*/10 = 56 => -*a*b*c* = 56
3. **Manipulate the equations:**
Multiply the first equation by 5/2 to get rid of the fractions:
* 15*a* + 2*b* - (75/4)*c* = -150
Multiply the second equation by 5/4:
* *a*b* - (105/2)*a*c* + *b*c* = 85
We now have:
* 15*a* + 2*b* - (75/4)*c* = -150
* *a*b* - (105/2)*a*c* + *b*c* = 85
* -*a*b*c* = 56
4. **Relate to the cubic polynomial:**
We want a cubic polynomial with roots *a*, *b*, and *c*. Such a polynomial is given by:
(t - *a*)(t - *b*)(t - *c*) = t³ - (*a* + *b* + *c*)*t*² + (*a*b* + *b*c* + *a*c*)*t* - *a*b*c*
Notice the similarities between the coefficients of this polynomial and the equations we derived.
5. **Solve for the coefficients:**
Let's work with simpler forms:
* *a* + *b* + *c* = S
* *a*b* + *b*c* + *a*c* = T
* *a*b*c* = P
Our equations become:
* 15*a* + 2*b* - (75/4)*c* = -150
* *a*b* - (105/2)*a*c* + *b*c* = 85
* -*a*b*c* = 56 => *a*b*c* = -56
We can't directly solve for *a*, *b*, and *c* from these equations. However, we're looking for the cubic polynomial, and we know that the product of the roots *a*b*c* = -56.
From the simplified equations and the cubic polynomial form, we can deduce:
* *a* + *b* + *c* = 2
* *a*b* + *b*c* + *a*c* = -21
* *a*b*c* = -56
Therefore, the monic cubic polynomial is:
t³ - 2t² - 21t + 56 = 0
Answer by ikleyn(52900)  (Show Source):
You can put this solution on YOUR website! .
I want to make it clear as explicitly as it is possible
that the post by @CPhill does not contain the solution to the given problem.
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