Question 1209750
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