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In this multi-part problem, we will consider this system of simultaneous equations:
3x+5y-6z =2, 
5xy-10yz-6xz = -41, 
xyz = 6. 

Let a=3x, b=5y and c=-6z .

Determine the monic cubic polynomial in terms of a variable t whose roots are ,t=a, t=b, and t=c.

 Jan 18, 2018
 #1
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x  = a/3      y   =  b/5       z   =  - c/6

 

a  + b  +  c   =  2

5 (a/3)(b/5)  -  10(b/5)(-c/6)  - 6(a/3)(-c/6)  = -41

abc / -90 =  6

 

a +  b  +  c  =  2       ⇒   a + b  =  2 - c       (1)   

ab  + bc  +  ac  =  -123    ⇒    ab  + c ( a + b)  =  -123    (2) 

abc   =  -540   ⇒   ab  =  -540/c        (3)

 

Sub  (1)  and (3)  into (2)

 

-540/ c  + c (2 - c)  =  -123

-540/c   + 2c - c^2  = -123

-540 + 2c^2 - c^3  = -123c

c^3 - 2c^2  - 123c + 540  = 0     re-write as

c^3 -  2c^2  - 63c  -  60c + 540  = 0

c(c^2 - 2c - 63)   -  60c + 540  = 0

c(c -9)(c + 7) -  60 (c - 9) =  0

(c - 9) [ c(c + 7) - 60 ]  = 0

(c - 9) [ c^2 + 7c - 60] =  0

(c - 9) ( c + 12)(c - 5)   =  0

 

So...possible values of c =   5, 9  and - 12

z  is an integer if  c = -12

And  ab  =  45

And  a + b  =  c  - (- 12)  ⇒  a +  b  =  14

And x is an ineger if a  = 9

And  y is an integer if b  =  5

 

So the cubic monomial we need is

 

(t + 12) (t -9) (t - 5)   =  

 

t^3 - 2 t^2 - 123 t + 540

 

 

cool cool cool

 Jan 18, 2018

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